1 Di erential Equations - University of Texas at Dallasx+1 to transform the equation into a separable equation and then solve it dy dx = y+ 3 x+ 1 + y 3x x+ 1 2 34. Use the substitution

1

Differential Equations

By: Patrick Bourque

Designed for students of MATH 2420 at The University of Texas at Dallas.

2

Contents

1 First Order Equations. 5

1.1 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 First Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.3 Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.4 Homogenous Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.5 Shift to Homogenous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.6 The zα Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

1.7 Equations of the form: y’=G(ax+by+c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1.8 Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

1.9 Integrating Factors for non-exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1.10 Orthogonal Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2 Second Order Equations. 57

2.1 Wronskian, Fundamental Sets and Able’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

2.2 Reduction of Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

2.3 Equations of the form y”=f(x,y’) and y”=f(y,y’) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

2.4 Homogenous Linear Equations with Constant Coefficients. . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

2.5 The Method of Undetermined Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

2.6 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

2.7 Cauchy Euler Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

2.8 Everyone Loves a Slinky: Springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

2.9 Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

3 Series Solution 105

3.1 Series Solutions Around Ordinary Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

3.2 Method of Frobenius: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

3.3 The Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

3.4 Bessel’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4 Laplace Transform 123

4.1 Calculating Laplace and Inverse Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

4.2 Solving Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

4.3 Unit Step Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

4.4 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

3

4 CONTENTS

4.5 Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

5 First Order Systems of Differential Equations 149

5.1 Homogenous Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

5.2 Non Homogenous Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

5.3 Locally Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

5.4 Linear Systems and the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

Chapter 1

First Order Equations.

1.1 Separable Equations

A Differential Equation is Separable if it can be written as:

f(x)dx = g(y)dy

The Solution is found by integrating both sides.

An Example: Solve:

exydx = (e2x + 1)dy y(0) = 1

Solution: ∫ex

e2x + 1dx =

∫dy

y

Using the substitution u = ex, du = exdx on the integral on the left∫du

u2 + 1=

∫dy

y

arctan(u) + C = ln |y| arctan(ex) + C = ln |y|

Applying our initial conditions

arctan(1) + C = ln(1)π

4+ C = 0 C =

−π4

arctan(ex)− π

4= ln |y|

Solving for y gives the solution to the differential equation:

5

6 CHAPTER 1. FIRST ORDER EQUATIONS.

y = earctan(ex)−π4

♠

An Application:

A parachutist falling toward Earth is subject to two forces: the parachutist weight (w = 32m) and the drag of the

parachute. The drag of the parachute the drag is proportional to the velocity of the parachute and in this case is equal

to 8|v|. The parachutist weight is 128lb and initial velocity is zero. Find formulas for the parachutists velocity v(t) and

distance x(t).

Since x(t) increases as the parachutist falls, the downward direction is the positive direction. The force from the

parachutist’s weight acts in the positive direction while the drag from the parachute acts in the negative direction. Since

the parachutist falls down (the positive direction) velocity is always positive so |v| = v. The resultant force will be the

force of the weight of the parachutist minus the force of the drag of the parachute:

F = 128− 8v

And the mass of the parachutist is:

128 = 32m m = 4

By Newton’s second law:

F = ma = 4dv

dtsince a =

dv

dt

This gives a differential equation:

4dv

dt= 128− 8v

This equation is Separable:

dv

32− 2v= dt

Integrating

−1

2ln(32− 2v) = t+ C or ln(32− 2v) = −2t+ C

Applying the initial condition v(0) = 0

C = ln(32)

Solving for v(t) gives:

v(t) = 16− 16e−2t

The parachutist terminal velocity is given by:

1.1. SEPARABLE EQUATIONS 7

limt→∞

v(t) = limt→∞

16− 16e−2t = 16ft/sec

We can now find an equation: x(t) for how far the parachutist has fallen:

x(t) =

∫v(t)dt =

∫(16− 16e−2t)dt = 16t+ 8e−2t +K

Since x(0) = 0 we see that K = −8. Thus,

x(t) = 16t+ 8e−2t − 8

1.

Show

y = cosh(x) =ex + e−x

2

is a solution to

K =1

y2

where K is the curvature

K =|y′′|(

1 + (y′)2) 3

2

2.

Find the values of r so that y = erx is a solution to

y′′′ − 6y′′ + 11y′ − 6y = 0

Then find a differential equation that has solutions

y1 = e2x y2 = e3x y3 = e4x

3.

Find the values of r so that y = erx is a solution to

y′′′ − 9y′′ + 26y′ − 24y = 0

4.

Find the values of k so that y = sin(kx) is a solution to

y′′ + 100y = 0

5.

Find the values of k and r so that y = erx sin(kx) is a solution to


y′′ + 4y′ + 13y = 0

6.

Find the values of n so that y = xn is a solution to

x2y′′ + 7xy′ + 8y = 0

7.

The Clairaut Equation is

y = xy′ + f(y′)

Show y = kx+ f(k) is a solution for some constant k.

Use the above result to find a solution to

y = xy′ + (y′)3

8.

Show

y =

{ex − 1 x ≥ 0

1− e−x x < 0

is a solution to

y′ = |y|+ 1

Remember, you must use the definition of the derivative to calculate y′(0).

9.

Solve the differential equation

(x2 + 1)dy = (4x+ xy2)dx y(0) = 2

10.


(2xy + 2x)dx = e−x2

dy y(√

ln(5)) = 0

11.


dy

dx=√

16x2y − 4x2y2 y(2) = 1

12.


dy

dx= ln

(x+

√x2 − 1

)yy(1) = e


13.


xdy = (y2 + 4y + 5)√x3 − 1dx y(0) = −2

14.


(x3 − x2 + x− 1)dy = (3x2y − 2xy + y)dx y(0) = e

15.


√y + xydy = arcsin(x)dx y(0) = 1

16.

A


cos(2x)dy = (1 + sin(2x))(cos(x)− sin(x))(1 + y2)dx y

(π

4

)= 0

17.

Solve the initial value problem

√1− x2 = 2xy

dy

dxy(1) = 2

18.


4x ln(x) + 4xy ln(x) = ydy

dxy(e) = 0

19.


15e−y sin3(x) = cos6(x)dy

dxy(0) = ln(2)

20.


1 +√y = (1 + sin(x))

dy

dxy(0) = 0

21.

Solve

(ex − e−x)(y2 + 1) = (2yex + 2ye−x)dy

dx

22.

Solve


(y2 + 1)dx = (x34 + x

54 )dy

23.

Solve:

dx

sin(x)=

dy

cos2(x) + cos2(x)√y

24.

Solve:

x2(y +√y)dx = (x4 + 2x2 + 1)dy

25.

Solve:

dy

dx=xex(y2 + 4y + 5)

x2 + 2x+ 1y(0) = −1

26.

Solve

xe3x

y= 2(3x+ 1)2

dy

dx

27.


√x− 2√y − 1

dy =

√y + 1√x+ 2

dx

28.


(xy + 2x+ y + 2)dx = (x2y2 + 2xy2)dy

29.


4

xy + xy2 + x2y + x2y2dy

dx= 1

30.

Use the Second Fundamental Theorem of Calculus to verify

y = Ce−∫ xag(u)du

is a solution to

y′ + g(x)y = 0

31.


Use the substitution u = yex to transform the equation into a separable equation and then solve it

ydx+ (1 + y2e2x)dy = 0

32.

Use the substitution y = zex to transform the equation into a separable equation and then solve it

dy

dx= y +

√e2x − y2

33.

Use the substitution z = y+3x+1 to transform the equation into a separable equation and then solve it

dy

dx=y + 3

x+ 1+

(y − 3x

x+ 1

)2

34.

Use the substitution z = y+1x2+1 to transform the equation into a separable equation and then solve it

dy

dx= 2x

y + 1

x2 + 1+x2 + 1

y − x2

35.

Use the substitution z = y2 + x− 1 to transform the equation into a separable equation and then solve it

2ydy

dx= y2 + x− 1

Some times it is useful to convert a differential equation to polar coordinates before solving it. The conversions to

polar coordinates is:

x = r cos θ y = r sin θ

Calculating the total differential of both x and y we get:

dx = cos θdr − r sin θdθ dy = sin θdr + r cos θdθ

Making

dy

dx=

sin θdr + r cos θdθ

cos θdr − r sin θdθ

Use this conversion to polar coordinates to solve the next two problems:

36.

Solve by converting to polar coordinates

x+ y = xdy

dx

37.



x(x+ y)dy = (y − x2)dx

38.


(2xy + 3y2)dx = (2xy + x2)dy

39.


dy

dx=y3 + x2y − x− yx3 + xy2 − x+ y

40.

Salt water containing .25 pounds of salt per gallon is being pumped into a tank initially containing 100 gallons of

water and 10 pounds of salt at a rate of 4 gallons per minute. The mixture in the tank is kept well stirred and fluid flows

out of the tank at a rate of 4 gallons per minute. Find a formula that represents the amount of salt in the tank at any

time.

41.

The logistic differential equation that models the size of a population of species in an environment of fixed size is given

by the following differential equation:

dP

dt= kP (M − P )

where M is the carrying capacity of the environment: that is, M is the maximum population of the species that can

fit in the environment and k > 0 is a constant depending on the reproduction rate of the species. For example if P

represent the population of a bacteria in a petri dish then M would be the maximum population of the bacteria in the

dish. We also see from the differential equation if a population P is less than M then dPdt > 0 and the population will

increase and approach the carrying capacity and if P is greater than M then dPdt < 0 and the population will decrease

and approach the carrying capacity. Note: This is an example of an Autonomous Differential Equation. With an

Autonomous Differential Equation the right hand side of the equation is a function of the dependent variable alone; that

is dPdt = F (P ) (there are no t) on the right hand side.

Show the population is increasing fastest when the population is half the carrying capacity then solve this differential

equation for the population P (t) by separating and applying partial fractions and then and show:

limt→∞

P (t) = M

Then solve this differential equation with the substitution z = 1P

42.

It has been calculated that the world population cannot exceed 20 billion people. In 1970 the population was 3.7 billion

and in 2014 the population grew to 6.8 billion. Write a logistic differential equation representing the world population

and solve it for the world population as a function of time. When will the population exceed 10 billion?

43.

Another type of population model is the Gopertz growth model. It is similar to the logistic equation in that the model

assumes the population will increase at a rate proportional to the size of the population.That means the population will


increase a a rate of kP (t). The like the logistic model the Gopertz growth model also takes into account the maximum

population a species can have in an environment of fixed size and resources. Instead of using (M − P (t)) as a factor like

the logistic model does the Gopertz growth model uses ln

(MP (t)

)as a factor, with M being the maximum population

(carrying capacity). The Gopertz growth model is

dP

dt= kP ln

(M

P

)k > 0

We also see from the differential equation if a population P is less than M then dPdt > 0 and the population will

increase and approach the carrying capacity and if P is greater than M then dPdt < 0 and the population will decrease

and approach the carrying capacity.

Show the population is increasing fastest when the population is Me and then solve this differential equation for the

population P (t) as a function of time and show:

limt→∞

P (t) = M

44.

200 fish of a particular species are introduced to a lake which can sustain no more than 3000 fish. After 2 years the fish

population had increased to 800 fish. If the population follows the Gopertz growth model how long after the introduction

of the fish to the lake will the population reach 2000 fish. Repeat the calculation using the logistic model and compare

the results.

45.

Differential equations can also be used to model the genetic change or evolution of a species. A commonly used hybrid

selection model is

dy

dt= ky(1− y)(a− by)

With y represents the portion of a population that has a certain characteristic and a, b, k constants and t is time

measured in generations.

At the beginning of a study of a population of a particular species it is found that half population had the advantageous

trait T and three generations later 60 percent of the population had trait T. Use the hybrid selection model with a = 2

and b = 1 to determine the number of generations it will take until more than 80 percent of the population has trait T.

46.

Newton’s law of cooling states that an object with temperature T in a medium of constant temperature M will

experience a change in temperature proportional to the difference in the temperature of the object and the medium

(M − T ). This gives the differential equation:

dT

dt= k(M − T )

A cup of 170◦ coffee is place in a 75◦ room. After 10 minutes the coffee is measured to have a temperature of 150◦.

How long will it take for the coffee to cool to 120◦?

47.

In the study of learning it has been shown that a person’s ability to learn a task is governed by the differential equation


dy

dt=

2py32 (1− y)

32

√n

Where y represents the level that a student has learned a skill as a function of time and n and p are constants

depending on the person learning the skill and the diffuculty of learning the skill. Solve this differential equation with

the initial condition y(0) = 0 and p = 1, n = 4.

48.

Torricelli’s Law states that water draining from a tank of volume V (t) through a hole of area a in the bottom will

have an exiting velocity of

v(t) =√

2gy(t)

where y(t) measures the height of the water level above the hole in the tank. The change in volume in the tank is

given by

dV

dt= −av(t) = −a

√2gy(t)

If A(y) denotes the area of the cross section of the tank at height y then for any slice of water at a height of y and

thickness dy will have volume

V (y) =

∫ y

0

A(y)dy

Using the second fundamental theorem of calculus to differentiate this integral gives

dV

dt= A(y)

dy

dt

equating this result to the previous formula for dVdt gives Torricelli’s Law:

A(y)dy

dt= −a

√2gy(t)

Use the above results to find how long it takes a spherical tank with radius of 60 inches to be drained through a 1

inch hole in the bottom.

49.

Show that if y1 and y2 are solutions to

y′ + P (x)y = Q1(x)

and

y′ + P (x)y = Q2(x)

respectively then y = y1 + y2 is a solution to

y′ + P (x)y = Q1(x) +Q2(x)

50.

1.2. FIRST ORDER LINEAR EQUATIONS 15

Show that if y1 and y2 are solutions to

y′ + P (x)y2 = Q1(x)

and

y′ + P (x)y2 = Q2(x)

respectively then y = y1 + y2 is not a solution to

y′ + P (x)y2 = Q1(x) +Q2(x)

51.

There are about 3300 families of human languages spoken in the world. Assuming that all languages have evolved

from a single language and that one family of languages evolves into 1.5 families of language every 6000 years how long

ago was the original language spoken?

52.

For every point P(x, y) on a curve in the first quadrant, the rectangle containing the points O(0, 0) and P(x, y) as

vertices is divided by the curve into two regions: upper region A and lower region B. If the curve contains the point Q(1,

3), and region A always has twice the area of region B, find the equation of the curve.

53.

Find a function f(x) with the following properties: The average value of f on [1, x] is equal to twice the functions

value at x and f(2) = 1.

54.

Find a function F (x, y) with the following properties: The normal line to the curve always contains the point (0, 0).

1.2 First Order Linear Equations

A Differential Equation is First Order Linear if it has the form:

dy

dx+ P (x)y = Q(x)

To solve this equation we recognize the left hand side: dydx + P (x)y looks close to the derivative of the product of

some function times y. Idea: multiply both sides of the equation by some function I(x) to make the left hand side the

derivative of the product of I(x) times y. Multiplying both sides by I(x) gives:

I(x)dy

dx+ I(x)P (x)y = I(x)Q(x)

If the left hand side is the derivative of the product I(x) · y:

d

dx

(I(x) · y

)= I(x) · dy

dx+ I ′(x) · y

Then:

I(x)dy

dx+ I(x)P (x) · y = I(x) · dy

dx+ I ′(x) · y

So


I(x)P (x) · y = I ′(x) · y

I ′(x)

I(x)= P (x)

Integrating gives:

ln |I(x)| =∫P (x)dx

Solving for the Integrating Factor I(x) gives:

I(x) = e∫P (x)dx

After multiplying both sides of the original differential equation by I(x) the left hand side is the derivative of the

product I(x) · y so the equation:

I(x)dy

dx+ I(x)P (x)y = I(x)Q(x)

Becomes:

d

dx

(I(x) · y

)= I(x)Q(x)

Integrating gives: (I(x) · y

)=

∫I(x)Q(x)dx

And the solution is given by:

y =1

I(x)

(∫I(x)Q(x)dx+ C

)An Example: Solve:

cos(x)dy

dx+ sin(x)y = sec2(x)

Writing the differential equation in standard form:

dy

dx+ tan(x)y = sec3(x)

Creating the integrating factor

I = e∫tan(x)dx = eln(sec(x)) = sec(x)

Our Solution is:

y =1

sec(x)

(∫sec(x) sec3(x)dx+ C

)

y = cos(x)

(∫sec4(x)dx+ C

)= cos(x)

(∫sec2(x)(1 + tan2(x))dx+ C

)


y = cos(x)

(∫sec2(x)dx+

∫tan2(x) sec2(x)dx+ C

)Using the substitution u = tan(x), du = sec2(x)dx on the second integral

y = cos(x)

(∫sec2(x)dx+

∫u2du+ C

)

y = cos(x)

(tan(x) +

u3

3+ C

)

y = cos(x)

(tan(x) +

1

3tan3(x) + C

)♠

Example

In this next example we will transform a nonlinear differential equation into a linear equation by converting it to polar

coordinates.

(x2 + y2 + x)dy

dx= y

Let x = r cos θ, y = r sin θ

Therefore dx = cos θdr − r sin θdθ and dy = sin θdr + r cos θdθ

Under this substitution our differential equation becomes:

(r2 + r cos θ)(sin θdr + r cos θdθ) = r sin θ(cos θdr − r sin θdθ)

Multiplying things out gives

r2 sin θdr + r3 cos θdθ + r cos θ sin θdr + r2 cos2 θdθ = r sin θ cos θdr − r2 sin2 θdθ

Which reduces to

r cos θdθ + sin θdr + dθ = 0

dr

dθ+ cot θr = − csc θ

This is first order linear with integrating factor

I = e∫

cot θdθ = sin θ

And the solution is...


r =1

sin θ

(−∫

sin θ csc θdθ + C

)which reduces to

r =C − θsin θ

Or

r sin θ = C − θ

Converting back to rectangular coordinate system gives the solution:

y = C − arctan

(y

x

)55.

Solve:

xy′ − 3y = x4 y(1) = 1

56.

Solve:

y′ + exy = ex y(0) = 2e

57.

Solve:

y′ + tan(x)y = tan(x) y

(π

4

)= 1

58.

Solve:

y′ + 4 sec(x)y = sec(x)(sec(x) + tan(x))

59.

Solve:

y′ + cos(x)y = sin(2x)

60.

Solve:

y′ + exy = ex

61.

Solve:


√1− x2 dy

dx+ y = 1 y(0) = 4

62.

Solve:

xdy

dx+ 3y = xex

4

y(0) = 1

63.

Solve:

dy

dx− cos(x)

1 + sin(x)y = 1 y(0) = 1

64.

Solve:

dy

dx+

6x2 − 4x+ 8

x3 − x2 + 4x− 4y =

ex3+12x

(x− 1)2(x2 + 4)

65.

Solve:

dy

dx+

cos(x)− sin(x)

cos(x) + sin(x)y = sec3(x) y(0) = 4

66.

Solve:

(1− x2)dy

dx− xy = 1 y

(1

2

)=

√3

2

67.

Solve:

dy

dx+

4x+ 1

xy = ex y(1) = 0

68.

Solve:

(1 + x2)dy

dx+ (4x2 − 4x+ 2)y = 9 ln(x) y(1) = 0

69.

Solve:

dy

dx+ sin(x)y = sin(2x)

70.

Solve:

dy

dx+

y

1 + e−x=

1

e2x + 2xex + x2y(0) = 0


71.

Solve:

dy

dx− 2xy = (2 + x−2) y(1) = 0

72.

Solve:

(1 + x2)dy

dx− 2xy = (1) y(0) = 1

73.

Solve:

cos2(x)y′ + y = 1 y(0) = −3

74.

Solve:

sin(x)dy

dx+ cos(x)y = ln(x)

75.

Solve:

(ex + e−x)dy

dx+

((ex + e−x)2

(ex − e−x)

)y = 1

76.

Solve:

(1 + x4)dy

dx− 4x3y = (x5 + x) arctan(x2) y(1) = π

77.

Solve:

(1 + x2)dy

dx+ xy = (x2 + 1)

52 y(0) = 1

78.

Solve:

(1 + x2)dy

dx− 4xy = x2

79.

Solve:

dy

dx+

6x

x4 + 5x2 + 4y = x

80.

Solve:


dy

dx+

x2 − 4x− 1

x3 − 2x2 + x− 2y = (x− 2) arctan(x)

81.

Solve:

y − xdydx

= y2eydy

dx

82.

Solve:

dy

dx+

2 + tan2(x)

x+ tan(x)y = cos(x)

83.

Solve:

dy

dx+ (3x2 + 2x)y = 3x5 + 5x4 + 2x3

84.

dy

dx+

1 + cos3(x)

sin(x) cos(x)(1 + cos(x))y = cos2(x)

85.

Solve:

dx

dy=

cos2(x)

y + tan(x)

86.

Find all values of k so that the solution y approaches 0 as x approaches ∞

y′ +k

xy = x2

87.

Express the solution to

dy

dx= 1 + 2xy

in terms of the error function:

erf(x) =2√π

∫ x

0

e−t2

dt

88.

The solution to the differential equation

dy

dx+ P (x)y = Q(x)

is

y(x) = Ce−2x + t+ 1


Find functions P (x) and Q(x).

89.

Salt water containing .25 pounds of salt per gallon is being pumped into a tank initially containing 100 gallons of

water and 10 pounds of salt at a rate of 4 gallons per minute. The mixture in the tank is kept well stirred and fluid flows

out of the tank at a rate of 2 gallons per minute. Find a formula that represents the amount of salt in the tank at any

time.

90.

At t = 0 one unit of a drug is administered to a patient who is hooked up to an IV drip supplying him with more of

the drug so that one unit of the drug is always present in his system. If the patient’s liver removes 15 percent of the drug

each hour how much of the drug must be administered by the IV each hour to keep 1 unit of the drug present?

91.

The following equation is not separable or linear. Use the substitution u = e2y to transform it into a linear equation

and solve it.

2xe2ydy

dx= 3x4 + e2y

92.

The following equation is not separable or linear. Use the substitution u = ey to transform it into a linear equation

and solve it.

dy

dx+

2

x=

e−y

1 + x3

93.

The following equation is not separable or linear. Use the substitution u = tan(y) to transform it into a linear equation

and solve it.

sec2(y)dy

dx− 3

xtan(y) = x4

94.

The following equation is not separable or linear. Use the substitution u = 14+y to transform it into a linear equation

and solve it.

1

(4 + y)2dy

dx− 2

x(4 + y)= ln(x)

95.

The following equation is not separable or linear. Use the substitution y = eu to transform it into a linear equation

and solve it.

xdy

dx− 4x2y + 2y ln(y) = 0

96.

The following equation is not separable or linear. Use the substitution z = ln(y) to transform it into a linear equation

and solve it.

xdy

dx+ 2y ln(y) = 4x2y

97.


The following equation is not separable or linear. Use the substitution z = ln(y) to transform it into a linear equation.

dy

dx+ f(x)y = g(x)y ln(y)

98.

The following equation is not separable or linear. Use the substitution u = y2 to transform it into a linear equation

and solve it.

2xydy

dx+ 2y2 = 3x− 6

99.

The differential equation governing the velocity v of a falling object subject to air resistance is

mdv

dt= mg − kv k > 0 v(0) = v0

Solve this differential equation and determine the limiting velocity of the object.

100.

If A(t) represents the amount of money in an account then the change in the amount in the account is given by:

dA

dt= Deposits−Withdraws + Interest

With a constant interest rate r the interest on the account is rA (remember A is the amount of money you will be

getting interest on). This makes the differentia equation:

dA

dt= Deposits−Withdraws + rA

In first order linear form:

dA

dt− rA = Deposits−Withdraws

Use this differential equation to solve the following:

A person opens an account yielding 3 percent interest is opened with an initial investment of 1000 dollars. On the

first year they deposit 100 dollars, the year month 110 dollars, the third year 120 dollars.... So each year they deposit 10

dollars more than the year before. How much will they have in the account in ten years?

101.

An equation of the form:

y′ + P (x)y = 0

is called a first order linear homogenous (Q(x)=0) differential equation. It can be solve by separation of variables

while

y′ + P (x)y = Q(x)

cannot. Show that if yh(x) is the solution to the homogenous equation

y′ + P (x)y = 0


then

y(x) = yh(x)

∫Q(x)

yh(x)dx

is the solution to the nonhomogenous equation

y′ + P (x)y = Q(x)

Use this technique to solve:

xy′ + y = e4x

The Riccati Differential Equation is an equation of the form:

dy

dx= P (x)y2 +Q(x)y +R(x)

If u(x) is a solution to the equation then the substitution y = u + 1v will transform the equation into a first order

linear equation.

102.

Solve the Riccati Equation:

dy

dx= −8xy2 + 4x(4x+ 1)y − (8x3 + 4x2 − 1) u(x) = x is one solution

103.

Solve the Riccati Equation:

dy

dx= x3(y − x)2 +

y

xu(x) = x is one solution

104.

Show the nonlinear differential equation

(y′)2 + y · y′ = x2 + xy

can be factored into

(y′ + y + x)(y′ − x) = 0

Set each factor to zero and solve each of the differential equations. Then show each solution you obtain is also a

solution to the original differential equation.

105.

Find a function f such that its average value on [0, x] is equal to the function squared.

1.3. BERNOULLI EQUATION 25

1.3 Bernoulli Equation

The Bernoulli Differential Equation is a nonlinear equation of the form:

dy

dx+ P (x)y = Q(x)yn

After the substitution z = y1−n the Bernoulli Equation will be First Order Linear. Differentiating gives:

dz

dx= (1− n)y−n

dy

dx

Multiplying both sides of the Bernoulli equation by (1− n)y−n gives:

(1− n)y−ndy

dx+ (1− n)P (x)y1−n = (1− n)Q(x)

Under the substitution this equation becomes:

dz

dx+ (1− n)P (x)z = (1− n)Q(x)

Which is a First Order Linear differential equation.

An Example: Solve:

dy

dx+ 2xy = e3x

2+2xy4

Let z = y1−4 = y−3 Sodz

dx= −3y−4

dy

dx

Multiplying both sides of the differential equation by −3y−4 gives:

−3y−4dy

dx+ (−6x)y−3 = −3e3x

2+2x

Under our substitution our differential equation becomes:

dz

dx+ (−6x)z = −3e3x

2+2x

This is First Order Linear. Our Integrating Factor is:

I = e∫(−6x)dx = e−3x

2

The solution is:

z =1

e−3x2

(∫e−3x

2

(−3e3x2+2x)dx+ C

)

z = e3x2

(− 3

∫e2xdx+ C

)= e3x

2

(−3

2e2x + C

)Back substituting gives:

1

y3=−3

2e3x

2+2x + Ce3x2

=Ce3x

2 − 3e3x2+2x

2


y3 =2

Ce3x2 − 3e3x2+2x

So

y = 3

√2

Ce3x2 − 3e3x2+2x

♠

106.

Solve:

ydy

dx+

xy2

x2 + 1= x

107.

Solve:

dy

dx+ cot(x)y = cos3(x)

√y

108.

Solve:

xdy

dx− y

2= x arcsin(x)y5

109.

Solve:

dy

dx= y4 cos(x) + y tan(x)

110.

Solve:

dy

dx+

9x+ 2

9x2 + 4xy = ln(x)y−1

111.

Solve:

dy

dx+

x

2 + 2x2y =

1

xy

112.

Solve:

dy

dx− sec(x) tan(x)

1 + sec(x)y =

sin3(x)

cos4(x)y2

113.

Solve:

1.3. BERNOULLI EQUATION 27

4 cos2(x)dy

dx+ y = 4 sin(x) cos(x)y−3

114.

Solve:

dy

dx+

1

sin(2x)y = (1− cos(2x))y3

115.

Solve:

dy

dx+

cos(x)

4 + 4 sin(x)y = cos3(x)y−3

116.

Solve:

xdy

dx− y = ex

2

y5

117.

Solve:

dy

dx+

5x

x2 + 1y = (x2 + 1)y2

118.

Solve:

xdy

dx+ y =

−y2

x

119.

Solve:

dy

dx+ tan(x)y = sec4(x)y3

120.

Solve:

2dy

dx+

cos(x)

1 + sin(x)y = cos3(x)y−1

121.

Solve:

dy

dx+

cos(x)− sin(x)

3 cos(x) + 3 sin(x)y = (cos(x)− sin(x))y−2

122.

Solve:

dy

dx=

1

xy + x2y3


123.

Use the substitution u = ey to transform the equation into a Bernoulli equation and then solve it

dy

dx+

1

x= 6xe3y

Another differential equation that can be transformed into a first order linear differential equation is Lagrange’s Equation.

This is an equation of the form:

y = xF (y′) +G(y′)

To simplify notation let y′ = p and our equation becomes

y = xF (p) +G(p)

Differentiating with respect to x gives

y′ = xF ′(p)dp

dx+ F (p) +G′(p)

dp

dx

Or

p = xF ′(p)dp

dx+ F (p) +G′(p)

dp

dx

Solving for dxdp

p− F (p) = (xF ′(p) +G′(p))dp

dx

dp

dx=

p− F (p)

xF ′(p) +G′(p)

dx

dp=

(F ′(p)

p− F (p)

)x+

(G′(p)

p− F (p)

)In standard form

dx

dp+

(F ′(p)

F (p)− p

)x =

(G′(p)

p− F (p)

)This is now a first order linear equation

124.

Solve the Lagrange Equation

y = 2xy′ + 4(y′)3

You may leave your solution in parametric form with p the parameter.

125.

Solve the Lagrange Equation

y = 2xy′ − 9(y′)2

You may leave your solution in parametric form with p the parameter.

1.4. HOMOGENOUS EQUATION 29

1.4 Homogenous Equation

A differential equation is Homogenous if it has the form:

dy

dx= f

(y

x

)The substitution z = y

x or y = xz will transform the Homogenous equation into a Separable equation.

dy

dx= z + x

dz

dx

Under this substitution the Homogenous equation becomes:

z + xdz

dx= f(z)

This reduces to the Separable equation

dz

f(z)− z=dx

x

An Example: Solve

dy

dx=y3 + 2x2y

xy2 + x3

Solution:

Multiplying the numerator and denominator on the right hand side by 1x3 gives:

dy

dx=

y3

x3 + 2 yxy2

x2 + 1

Let z =y

xso y = xz

dy

dx= z + x

dz

dx


z + xdz

dx=z3 + 2z

z2 + 1

xdz

dx=

z

z2 + 1

z + 1

zdz =

dx

x∫ (z +

1

z

)=

∫dx

x

z2

2+ ln(z) = ln(x) + C

1

2

y2

x2+ ln

(y

x

)= ln(x) + C

Since we cannot solve for y this implicit solution is our final answer.


♠Notice the sum of the exponents in both terms in the numerator and both terms in the denominator is 3. Whenever

the sum of the exponents in each term in the problem is the same constant you should consider using the homogenous

substitution to solve the differential equation. The first practice problem should help reinforce this idea.

126.

Solve:

dy

dx=y4 + x2y2 + x4

x3y

127.

Solve:

dy

dx=

2y5 + x2y3 + 3x4y

2xy4 + 3x5

128.

Solve:

dy

dx=

13y6 + 18x2y4 + 3x5y

12xy5 + 16x3y3 + 2x6

129.

Solve:

dy

dx=y

x+ cos2

(y

x

)130.

Solve:

dy

dx=y

x+

√y

x+ 1

131.

Solve:

dy

dx=

6y3 − 5xy2 + 2x2y − x3

5xy2 − 4x2y + x3

132.

Solve:

dy

dx=y4 + xy3 + 2x2y2 + x4

xy3

133.

Solve:

dy

dx=

(y

x

) 32

+y

x+

√y

x

134.

Solve:

1.4. HOMOGENOUS EQUATION 31

dy

dx=

y2 ln

(yx

)+ x2

xy ln

(yx

)135.

Solve:

x sin

(y

x

)dy

dx= y sin

(y

x

)+ x

136.

Solve:

dy

dx=

((y

x

)3

+

(y

x

) 32

+ 1

)(x

y

) 12

137.

Solve:

dy

dx=y

x+ tan4

(y

x

)138.

Solve:

dy

dx=

(√x+√y

√x

)2

139.

Solve: (x− y arctan

(y

x

))dx+ x arctan

(y

x

)dy = 0

140.

Solve:

dy

dx=

y2 tan2

(yx

)+ x2

xy tan2

(yx

)141.

Solve:

x sin

(y

x

)dy

dx= y sin

(y

x

)− xe

yx

142.

Solve:

dy

dx=

2y2 + 2xy + 2x2

xy


143.

Solve:

dy

dx=y3 + 2xy2 + x2y + x3

x(x+ y)2

144.

Solve:

dy

dx=y(ln(y)− ln(x) + 1)

x

145.

Solve:

xdy

dx= y +

√x2 + y2 y(1) = 0

146.

Solve:

dy

dx= csc2

(y

x

)+y

x

147.

Solve:

dy

dx=

2xye(xy )

2

y2 + y2e(xy )

2

+ 2x2e(xy )

2

148.

Solve:

dy

dx=

y

x+√xy

149.

Solve:

(√x+ y +

√x− y)dx+ (

√x− y −

√x+ y)dx = 0 y(1) = 1

150.

Solve:

(2(y − 1)(x+ y − 1) + (y − 1)2)dx+ (4(y − 1)(x+ y − 1) + (x+ y − 1)2)dy = 0

151.

Solve:

exy (y − x)

dy

dx+ y(1 + e

xy ) = 0

152.

Use the substitution z = yx to solve the following differential equation

1.5. SHIFT TO HOMOGENOUS 33

xdy = (x2 + y2 + y)dx

153.

Use the substitution z = yx2 to solve the following differential equation

dy

dx=

2y

x+ cos

(y

x2

)

154.

Use the substitution z = y√x

to solve the following differential equation

dy

dx=

y

2x+y2

x+ 1

155.

Use the substitution z = yx2 to solve the following differential equation

dy

dx=

2y

x+ cos

(y

x2

)

156.

Use the substitution u = x2 + y2 v = xy to solve the following differential equation

(x2 + y2)(xdy + ydx)− xy(xdx+ ydy) = 0

157.

Use the substitution u = x3 v = y2 to solve the following differential equation

3x5 − y(y2 − x3)dy

dx= 0

158.

Use the substitution u = xy to solve the following differential equation

xdy + ydx = x2y2dx

1.5 Shift to Homogenous

A differential equation can shifted to produce a Homogenous differential equation if it is of the form:

dy

dx=a1x+ b1y + c1a2x+ b2y + c2


If the constant terms c1 and c2 are both zero then the above equation is already Homogenous. This provides the

motivation of shifting x and y by a constant so that the resulting equation has constant terms c1 and c2 both equal to

zero. To shift this into a Homogenous differential equation use the substitutions:

x = x+ h y = y + k

With h and k chosen so that the two equations:

a1x+ b1y + c1 = 0 a2x+ b2y + c2 = 0

reduce to:

a1x+ b1y = 0 a2x+ b2y = 0

The idea is motivated by the fact that if the constants c1 and c2 are both zero then:

dy


is Homogenous.

So choose h and k to be the solutions to:

a1h+ b1k + c1 = 0 a2h+ b2k + c2 = 0

An Example: Solve:

dy

dx=x− 2y − 2

2x+ y + 6x, y > −2

Solving system of equations:

x− 2y − 2 = 0 2x+ y + 6 = 0

x = 2 + 2y 2(2 + 2y) + y = −6 5y = −10 y = −2 x = −2

We will use the substitution x = x− 2 and y = y − 2 making dydx = dy

dx

Our differential equation becomes:

dy

dx=x− 2y

2x+ y

Multiplying the numerator and denominator on the right hand side by 1x gives:

dy

dx=

1− 2 yx2 + y

x

Letz =y

xy = zx

dy

dx= z + x

dz

dx


z + xdz

dx=

1− 2z

2 + z

1.5. SHIFT TO HOMOGENOUS 35

xdz

dx=−z2 − 4z + 1

z + 2

z + 2

−z2 − 4z + 1dz =

dx

x

Integrating gives:

−1

2ln | − z2 − 4z + 1| = ln |x|+ C

Multiplying by −2 and negating the contents inside the absolute value bars on the left gives

ln |z2 + 4z − 1| = ln |x|−2 + C

z2 + 4z − 1 =C

x2

Completing the square on the left gives

(z + 2)2 =C

x2+ 5

(z + 2)2 =C + 5x2

x2

Extracting a square root gives

|z + 2| =√C + 5x2

|x|Since both x and y are greater than -2, x > 0 and z + 2 > 0. So we get

z + 2 =

√C + 5x2

x

z =

√C + 5x2

x− 2x

x

y

x=

√C + 5x2

x− 2x

x

y =√C + 5x2 − 2x

y + 2 =√C + 5(x+ 2)2 − 2(x+ 2)

and our final solution is

y =√C + 5(x+ 2)2 − 2x− 6

The same differential equation can be solved using the substitution u = x− 2y− 2 and v = 2x+ y+ 6 but the solution

is a bit longer (try it and see which solution you prefer).


159.

Solve:

dy

dx=−8x+ 3y + 2

−9x+ 5y − 1

160.

Solve:

dy

dx=x− y − 3

x+ y − 1

161.

Solve:

dy

dx=

x+ 3y + 4

4y − 3x+ 1

162.

Solve:

dy

dx=x− y + 1

x+ y

163.

Solve:

dy

dx=

x+ y + 3

2y − x+ 3

164.

Solve:

dy

dx=

2x+ y

−y − x+ 1

165.

Solve:

dy

dx=x+ y + 1

x− y + 3

166.

Solve:

dy

dx=

3− 2x− yx+ y − 1

167.

Solve:

dy

dx=

5x+ 4y − 3

6x− y + 8

168.

Solve:

dy

dx=

1

2

(x+ y − 1)2

(x+ 2)2

1.6. THE Zα SUBSTITUTION 37

1.6 The zα Substitution

In the previous section we learned how to shift an equation into a homogenous equation but our shifting method only

worked for equations of the form:

dy


where the exponents of the x and y term were 1. Now we will study a substitution that works when the exponents

are not 1.

An Example: Solve:

(x2y2 − 2)dy + (xy3)dx = 0

This differential equation is almost homogenous: the sum of the exponents in x2y2 and xy3 are both 4. In fact, if the

−2 was not involved in the equation it would be homogenous. This is a good candidate for a zα substitution.

y = zα dy = αzα−1dz

Under this substitution the differential equation becomes:

(x2z2α − 2)αzα−1dz + xz3αdx = 0

α(x2z3α−1 − 2zα−1)dz + xz3αdx = 0

We will now choose α so that the exponents on each term sum to the same value. So we choose α so that:

3α+ 1 = α− 1 α = −1

If α = −1 our equation becomes:

−(x2z−4 − 2z−2)dz + xz−3dx = 0

−(x2z−4 − 2z−2)dz

dx+ xz−3 = 0

dz

dx=

xz−3

x2z−4 − 2z−2

dz

dx=

xz

x2 − 2z2

Dividing both numerator and denominator by x2 gives

dz

dx=

zx

1− 2 z2

x2

This is now a homogenous equation so we make the following substitution:

u =z

xz = ux

dz

dx= u+ x

du

dx


Under this substitution our equation becomes:

u+ xdu

dx=

u

1− 2u2

This is now a separable differential equation

xdu

dx=

2u3

1− 2u2

1− 2u2

2u3du =

dx

x

Integrating gives

−1

4u2− ln|u| = ln |x|+ C

Since u = zx and z = y−1 u = 1

xy the solution becomes

−x2y2

4− ln | 1

xy| = ln |x|+ C

Or

−x2y2

4+ ln |xy| = ln |x|+ C

169.

Solve:

(2x2y − 1)dy + 2xy2dx = 0

170.

Solve:

(x2y + 1)dy + xy2dx = 0

171.

Solve:

(xy3 + 1)dy + y4dx = 0

172.

Solve:

(x2y4 + 4)dy + xy5dx = 0

173.

Solve:

(xy + 2)dy − y2dx = 0

1.7. EQUATIONS OF THE FORM: Y’=G(AX+BY+C) 39

1.7 Equations of the form: y’=G(ax+by+c)

If a differential equation is of the form:

dy

dx= G(ax+ by + c)

Then use the following substitution to transform the equation into a Separable Differential Equation:

z = ax+ by + cdy

dx=

1

b

(dz

dx− a)

Under this substitution our differential equation becomes

1

b

(dz

dx− a)

= G(z)

which reduces to the Separable Differential Equation:

dz

a+ bG(z)= dx

An Example: Solve:

dy

dx=

(x+ y

)(1 + ln(x+ y)

)− 1

Let z = x+ ydy

dx=dz

dx− 1


dz

dx− 1 = z(1 + ln(z))− 1

∫dz

z(1 + ln(z))=

∫dx

ln(1 + ln(z)) = x+ C

1 + ln(x+ y) = Cex

ln(x+ y) = Cex − 1

x+ y = eCex−1

And our final solution is:

y = eCex−1 − x

174.

Solve:


dy

dx= (x+ y − 4)2

175.

Solve:

3dy

dx= (2x+ 3y − 1) + 4(2x+ 3y − 1)−3 − 2

176.

Solve:

2dy

dx=

1

(x+ 2y + 1)e(x+2y+1)2− 1

177.

Solve:

dy

dx= tan2(x+ y)

178.

Solve:

dy

dx= sin2(y − x)

179.

Solve:

dy

dx=

(x+ y)4 + (x+ y)2 + 1

(x+ y)2

180.

Solve:

dy

dx= csc2(4x+ y + 1)− 4

181.

Solve:

dy

dx= sin(x+ y)

182.

Solve:

dy

dx=

1√x+ y

183.

Solve:

dy

dx=√e2x+2y − 1− 1

184.

1.7. EQUATIONS OF THE FORM: Y’=G(AX+BY+C) 41

Solve:

2dy

dx= sec(2y − 4x+ 1) + tan(2y − 4x+ 1) + 4

185.

Solve:

dy

dx=x− y +

√1 + (x− y)2√

1 + (x− y)2

186.

Solve:

dy

dx=

4(x− y) ln(x− y)− 1

4(x− y) ln(x− y)

187.

Solve:

dy

dx= 1 +

√e2y−2x − 1

188.

Solve:

dy

dx=

2e−x−y

ex+y − e−x−y189.

Solve:

dy

dx= (4 + (4x+ y)2)

32 − 4

190.

Solve:

e−y(dy

dx+ 1

)= xex

191.

Solve:

dy

dx=

(cos3(x+ y)− 1

)(cos3(x+ y) + 1

)192.

Solve:

dy

dx= sin(2x+ 2y)− sin2(x+ y)

193.

Use the substitution z = y + x to solve

dy

dx+

2y

x+ 3 = x2(x+ y)3


194.

Use the substitution z = y + x to solve

dy

dx+

2x+ y

x=

4x

x+ y

195.

Use the substitution z = y − x to solve

dy

dx+ x(y − x) + x3(y − x)2 = 1

1.8 Exact Equations

In Multivariable we analyzed the functions of the form z = f(x, y) by studying their level sets. The level sets of this three

dimensional function can be graphed in two dimensional space by replacing z with different constants and analyzing the

resulting two dimensional graphs. The total differential of:

f(x, y) = C is fxdx+ fydy = 0

We also know that mixed partials are equal meaning:

fxy = fyx for all f(x, y) = C

We will now study the Exact Differential Equation. An Exact Differential Equation is an equation that was created by

calculating the total differential of some function of two or more variable set equal to a constant. In calculus we learned

to go from

f(x, y) = C to fxdx+ fydy = 0

In differential equations we will learn to go from

fxdx+ fydy = 0 to f(x, y) = C

The equation

Mdx+Ndy = 0 is Exact if My = Nx

If an equation is exact then M is the partial derivative of some function f(x, y) = C with respect to x and N is the

partial derivative of the same function with respect to y. So to find f up to a constant we need to integrate M with

respect to x (obtaining a constant which will be a function g(y)) or we need to integrate N with respect to y (obtaining

a constant which will be a function g(x)). That is:

f(x, y) =

∫Mdx+ g(y)

We then solve for g(y) using the fact that fy = N . That is:

1.8. EXACT EQUATIONS 43

fy =∂

∂y

∫Mdx+ g′(y) = N

And the solve for g(y) and insert it into our formula for f(x, y) and write the final answer as a level set f(x, y) = C

An Example: Solve:

(8xy3 + 2xy + 3x2)dx+ (12x2y2 + x2 + 4y3)dy = 0

Solution:

M = 8xy3 + 2xy + 3x2 N = 12x2y2 + x2 + 4y3

Test for exactness:

My = 24xy2 + 2x Nx = 24xy2 + 2x

Since My = Nx our differential equation is exact, making M = 8xy3 + 2xy+ 3x2 the partial derivative of the function

F we are solving for with respect to x. That is:

Fx = 8xy3 + 2xy + 3x2

So

F =

∫(8xy3 + 2xy + 3x2)dx = 4x2y3 + x2y + x3 + g(y)

Since this equation is exact Fy = N . This creates the equation:

Fy = 12x2y2 + x2 + g′(y) = 12x2y2 + x2 + 4y3

So

g′(y) = 4y3 g(y) = y4

Our final solution is:

4x2y3 + x2y + x3 + y4 = C

196.

Solve: (6xy + 6y4 − 24x2

)dx+

(3x2 + 24xy3 +

6

y

)dy = 0

197.

Solve: (−y

x2 + y2

)dx+

(x

x2 + y2

)dy = 0 y(1) = 1

198.

Solve:


(y cos(y) + y sin(x) + xy cos(x)

)dx+

(x sin(x) + x cos(y)− xy sin(y)

)dy = 0

199.

Solve: (sin(y) + y sin(x) + 2e2x

)dx+

(x cos(y)− cos(x) + sin(y)

)dy = 0

200.

Solve:

(2xexy + x2yexy)dx+ (x3exy)dy = 0 y(1) = ln(4)

201.

Solve: (3x2 sin(y) + y cos(y)

)dx+

(x3 cos(y) + x cos(y)− xy sin(y)

)dy = 0

202.

Solve: (xy2 cos(xy) + 1

)dx+

(x2y cos(xy) + x sin(xy)

)dy = 0

203.

Solve: (xy(exy + 1) + y

)dx+

(x2exy + x

)dy = 0

204.

Solve: (xex − ex + y

)dx+

(1 + ln(y) + x

)dy = 0

205.

Solve: (y(y2 − x2)

(x2 + y2)2

)dx+

(x(x2 − y2)

(x2 + y2)2

)dy = 0

206.

Find the function M(x, y) that makes the following differential equation exact.

M(x, y)dx+

(xexy + 2xy +

1

x

)dy = 0

207.

The following differential equation arises from the total differential of a function F with variables: x, y and z set equal

to a constant. Find this function F (x, y, z)

(2yz2 + 6xz)dx+ (2xz2 + 30y2z)dy + (4xyz + 3x2 + 10y3)dz = 0

1.9. INTEGRATING FACTORS FOR NON-EXACT EQUATIONS 45

208.

The following differential equation arises from the total differential of a function F with variables: x, y and z set equal

to a constant. Find this function F (x, y, z)

(yz + 1)dx+ (xz + 2y)dy + (xy + 3z2)dz = 0

1.9 Integrating Factors for non-exact Equations

If a differential equation is not exact: My 6= Nx sometimes we can multiply both sides of the equation by an integrating

factor I to make it exact. We need a formula for this integrating factor. Starting with:

Mdx+Ndy = 0 and multiplying both sides by I gives IMdx+ INdy = 0

For this to be exact ∂∂y (IM) = ∂

∂x (IN). Calculating these partials gives:

IyM +MyI = IxN +NxI

This is a partial differential equation that we cannot solve. So we will solve a special case: the case where I is a

function of one variable. Case 1: I is a function of x making Iy = 0. Now our partial differential equation is a bit easier

to solve:

MyI = IxN +NxI

Or

I(My −Nx) = IxN

Which becomes:

IxI

=My −Nx

N

Integrating both sides with respect to x gives:

ln |I| =∫ (

My −NxN

)dx

Making our integrating factor:

I(x) = e

∫ (My−NxN

)dx

Case 2: I is a function of y making Ix = 0. In this case the integrating factor is:

I(y) = e

∫ (Nx−MyM

)dy

An Example: Solve:

(2y7 + y4)dx+ (6xy6 − 3)dy = 0


M = 2y7 + y4 N = 6xy6 − 3

Test for exactness:

My = 14y6 + 4y3 Nx = 6y6

Since My 6= Nx our equation is not exact. We will look for an integrating factor. Since

Nx −My

M=

6y6 − 14y6 − 4y3

2y7 + y4=−4y3(2y3 + 1)

y4(2y3 + 1)=−4

y

is a function of only y our integrating factor is:

I = e∫ Nx−My

M dy = e∫ −4

y = y−4

Multiplying both sides of the differential equation by y−4 gives:

(2y3 + 1)dx+ (6xy2 − 3y−4)dy = 0

M = 2y3 + 1 N = 6xy2 − 3y−4

Now

My = 6y2 = Nx

Since My = Nx our differential equation is exact, making M = 2y3 + 1 the partial derivative of the function F we are

solving for with respect to x. That is:

Fx = 2y3 + 1

So

F =

∫(2y3 + 1)dx = 2xy3 + x+ g(y)

And Fy = N

Fy = 6xy2 + g′(y) = 6xy2 − 3y−4 g′(y) = −3y−4 g(y) = y−3

And our final solution is:

F (x, y) = 2xy3 + x+ y−3 = C

209.

Solve: (3y + 5x2y3 + 4x

)dx+

(x+ 3x3y2

)dy = 0

210.

Solve:


(2x4 + x3yexy − 2

)dx+

(2x3y + x4exy

)dy = 0

211.

Solve:

(2x2 + y)dx+ (x2y − x)dy = 0 y(1) = 1

212.

Solve:

(16x2y2 + 64x4 + 1)dx+ (4xy3 + 16x3y) = 0 y(1) = 1

213.

Solve:

(xey2

)dx+ (yey2

)dy = 0 y(0) = 0

214.

Solve:

(2xy7 + y)dx+ (3x2y6 − 3x)dy = 0 y(1) = 1

215.

Solve:

(1 + y + x2y)dx+ (x3 + x)dy = 0 y(1) = 0

216.

Solve:

(3x2y + y4)dx+ (−2x3 + xy3)dy = 0 y(1) = 0

217.

Solve:

(cos(y) + sin(y) + tan(x))dx+ (tan(x)(cos(y)− sin(y)))dy = 0

218.

Solve: (2e2x + ey

)dx+

(3e2x + 4xey

)dy = 0

219.

Solve: (3 +

6xy

1 + x2

)dx+ 3dy = 0

220.


Solve: (ex

2

(2x2 + 1) + e−y2

(2y3 + 1)

)dx+

(2xyex

2

+ e−y2

(6xy2 + 1)

)dy = 0

221.

Solve:

3xydx+ (x2 + 1)dy = 0

Sometimes we cannot find an integrating factor of just a single variable using the above formulas. In this case we

guess the form of the integrating factor and try to find constants that make it work.

222.

Solve by finding an integrating factor of the form I = xnym:(3y4 + 18y−1

)dx+

(5xy3 + 2x−2

)dy = 0

223.

Solve by finding an integrating factor of the form I = xnym:

y3

x(x+ y)2dx+

xy

(x+ y)2dy = 0

224.


2 cos(x2y2)(ydx+ xdy) = 0

225.


2 cos

(1

x2y4

)(y5dx+ 2x3dy) = 0

226.

If I(x) = x is an integrating factor for

f(x)dy

dx+ x2 + y = 0

Find all functions f(x)

227.

Solve by finding an integrating factor of the form I = ekx cos(y):

dy

dx= tan(y)− ex sec(y)

228.

Solve by finding an integrating factor of the form I = sinn(x) cosm(y):(4 cos(x) + 3 cot(x)

)dx+

(− 2 sin(x) tan(y)− 3 tan(y)

)dy = 0

229.


Solve:

Show that I = 1x2+y2 is an integrating factor for(

y + xf(x2 + y2)

)dx+

(yf(x2 + y2)− x

)dy = 0

Use this result to solve (y + x(x2 + y2)2

)dx+

(y(x2 + y2)2 − x

)dy = 0

230.

Show that

I =1

Ax2 +Bxy + Cy2

is an integrating factor for

xdy − ydx = 0

231.

Show that

I1 = x−3 I2 = y−3 I3 =1

(x2 + y2)2

are all integrating factors of

(y2 − xy)dx+ (x2 − xy)dy = 0

232.

Show that

G(x, y) = ln(x+ y)− 1

x+ y

is a solution to the exact equation(1

x+ y+

1

(x+ y)2

)dx+

(1

x+ y+

1

(x+ y)2

)dy = 0

Now multiply both sides of this exact equation by (x+ y)2 producing

(x+ y + 1)dx+ (x+ y + 1)dy = 0

which has a solution

F (x, y) =1

2x2 +

1

2y2 + xy + x+ y

What is the relationship between F and G

Now show in general that if F (x, y) = C is a solution to

Fxdx+ Fydy = 0


and G(x, y) = C is a solution to

I(x, y)Fxdx+ I(x, y)Fydy = 0

then

FxGy = GxFy

233.

In the study of first order linear differential equations:

dy

dx+ P (x)y = Q(x)

we learned that multiplying by the integrating factor:

I(x) = e∫P (x)dx

will transform the equation into a separable differential equation. Show that by multiplying both sides of(P (x)y −Q(x)

)dx+ dy = 0

by the same integrating factor will transform the above equation into an exact equation.

Bernoulli’s Spread of Smallpox:

Let x(t) represent the population of all living people.

Let y(t) represent the population of all living people who have not contracted smallpox.

Let a > 0 be the rate at which population y(t) contracts smallpox.

Therefore

dy

dt= −ay

Let 0 < b < 1 be the precentage of population y(t) that get and die from smallpox.

Therefore

dx

dt= −aby

Let d(t) be the average death rate of populations x(t) and y(t) from causes other than smallpox.

Without smallpox we have

dx

dt= −d(t)x(t)

dy

dt= −d(t)y(t)

With smallpox we have

dx

dt= −d(t)x(t)− aby(t)

dy

dt= −d(t)y(t)− ay(t)

To eliminate the average death rate d(t) we multiply dxdt by y(t) and dy

dt by x(t) and subtract

ydx

dt− xdy

dt= axy − aby2

1.10. ORTHOGONAL TRAJECTORIES 51

Recognizing the left hand is the numerator of ddtxy we now divide both sides by y2

y dxdt − xdydt

y2= a

x

y− ab

This equation is now

d

dt

x

y= a

x

y− ab

The substitution z = xy yields the seperable equation

dz

dt= az − ab

dz

z − b= adt

ln(z − b) = at+ C

z − b = Ceat

y = x(Ceat + b)

1.10 Orthogonal Trajectories

A common geometric problem in many applications involves finding a family of curves: Orthogonal Trajectories, that

intersect a given family of curves orthogonally at each point. If you are given a family of curves in the form F (x, y) = K

then the slope of this family of curves is given by the derivitive:

dy

dx= −Fx

Fy

Since we are looking for an orthogonal family of curves, and in R2 orthogonal lines have negative reciprocal slopes,

we will be for a family of curves that satisfy the following differential equation:

y′ =FyFx

An Example:

Find the orthogonal trajectories for the circle:

x2 + y2 = r2

To find the orthogonal trajectories we take F (x, y) = x2 + y2 we will need to solve the differential equation:

y′ =2y

2x=y

x

This equation is separable


dy

y=dx

x

So

ln(y) = ln(x) + C

The family of curves orthogonal to the circle is:

y = kx

Conversely, the family of curves orthogonal to the lines y = kx is given by the circle x2 + y2 = r2.

An Example: Find the orthogonal trajectories for the circle:

x2 + y2 = Cx

Here we will take F (x, y) = x2 + y2 − Cx and we are looking for a family of curves that satisfy the following:

y′ =2y

2x− CThe problem with this differential equation is that it involves the constant C. To eliminate this constant we will solve

for C in the original equation of the circle and insert it into the differential equation.

C =x2 + y2

x


y′ =2xy

x2 − y2

Recognizing that the sum of the exponents in each term in both the numerator and denominator add to the same

value of 2 we see that this is a homogenous equation. Remember, to solve the homogenous equation you must first write

it as:

dy

dx= f

(y

x

)and make the substitution

z =y

x

Multiplying both the numerator and denominator by 1x2 we get the homogenous equation:

dy

dx=

2 yx1− ( yx )2

Making the substitutions


z =y

x

dy

dx= x

dz

dx+ z


xdz

dx+ z =

2z

1− z2

This equation is now separable:

xdz

dx=z3 + z

1− z2∫1− z2

z(z2 + 1)dz =

∫dx

x

The integral on the left requires partial fraction decomposition. After applying partial fraction we get:∫ (1

z− 2z

1 + z2

)dz =

∫dx

x

Integrating

ln(z)− ln(1 + z2) = ln(x) + C

ln

(z

1 + z2

)= ln(x) + C

z

1 + z2= Cx

z = Cx+ Cxz2

Expressing this quadratic with the coefficient of z2 being 1 gives

z2 − 1

Cxz + 1 = 0

To solve for z in this equation we must compleat the square by adding 14C2x2 to both sides:

z2 − 1

Cxz +

1

4C2x2+ 1 =

1

4C2x2

Factoring the first 3 terms on the left we get

(z − 1

2Cx

)2

=1

4C2x2− 1

z − 1

2Cx= ±

√1− 4C2x2

4C2x2

z =1

2Cx±√

1− 4C2x2

2Cx

Solving for y remembering that z = yx gives:


y = x

(1±√

1− 4C2x2

2Cx

)The orthogonal family of curves we desire is:

y =1±√

1− 4C2x2

2C

Sometimes we are concerned with finding a family of curves that make an angle of α 6= 90◦ with a given family of

curves. These curves are called Oblique Trajectories.

Given a family of curves: F (x, y) = K with its derivative (slope) given by:

dy

dx= f(x, y)

Treating dy as the change in y and dx as the change in x and creating a triangle gives:

dx

dy

√(dx)2 + (dy)2

θ

From the triangle we see, tan(θ) = dydx = f(x, y) so the tangent line has an angle of inclination of arctan(f(x, y)). So

the tangent line of an oblique trajectory that intersects this curve at an angle of α will have an angle of inclination of:

arctan(f(x, y)) + α

Making the slope of the oblique trajectory

tan

(arctan(f(x, y)) + α

)=

f(x, y) + tan(α)

1− f(x, y) tan(α)

Thus the differential equation of this family of oblique trajectories is given by:

y′ =f(x, y) + tan(α)

1− f(x, y) tan(α)

An Example:

Find the family of oblique trajectories that intersect the family of straight lines y = Cx at an angle of 45◦.

Here we take F (x, y) = y − Cx and compute the slope of the tangent line:

dy

dx= −Fx

Fy=C

1= C

Solving for C we get


dy

dx=y

x= f(x, y)

Using this function for f(x, y) and α = 45◦ the differential equation of this family of oblique trajectories is

y′ =yx + tan(45◦)

1− yx tan(45◦)

=x+ y

x− y

Multiplying both numerator and denominator by 1x produces

dy

dx=

1 + yx

1− yx

This is a homogenous equation so we make the following substitution

z =y

x

dy

dx= z + x

dz

dx

The differential equation becomes

z + xdz

dx=

1 + z

1− zThis equation is now separable

1− z1 + z2

dz =dx

x

Integrating gives

arctan(z)− 1

2ln(1 + z2) = ln(x) + C

Treating C as ln(K) and using some properties of logs we get

2 arctan(z) = ln(K2x2(1 + z2)

Converting back to x and y we get the family of oblique trajectories

2 arctan

(y

x

)= ln(K2(x2 + y2))

234.

Find the orthogonal trajectories for the family of straight lines.

y = mx+ 1

235.

Find the orthogonal trajectories for each given family of curves.

y = Cx3

236.



x2 + y2 = Cx3

237.


y = Ce2x

238.


y = x− 1 + Ce−x

239.


x− y = Cx2

240.

Find the value of n so that the curves xn + yn = C are the orthogonal trajectories of

y =x

1 +Kx

241.

Show that the following family of curves is self orthogonal.

y2 = 4C(x+ C)

242.

Show that the following family of curves is self orthogonal.

y2 = 2Cx+ C2

243.

Find a family of oblique trajectories that intersect the family of circles x2 + y2 = r2 at an angle of 45◦

244.

Find a family of oblique trajectories that intersect the family of circles y2 = Cx2 at an angle of 60◦

245.

Let O be the origin and P be a point on the curve P (x). Let N be the point on the x-axis where the normal line to

P (x) intersects the x-axis. If OP = ON what is the equation of P (x)?

Chapter 2

Second Order Equations.

2.1 Wronskian, Fundamental Sets and Able’s Theorem

In this section we will mostly be dealing with the second order linear differential equation:

y′′ + P (x)y′ +Q(x)y = 0

If we want to find all solutions to this equation it can be shown that we are looking for two solutions y1 and y2 to

the equations with the one restriction that y1 and y2 cannot be scalar multiples of each other. But in order to expand

our knowledge to third order and higher order equations we replace the restriction that the solutions cannot be scalar

multiples of each other with the restriction that the set of solutions must be linearly independent.

The Set of functions{y1, y2, ..., yn}is Linearly Independent if the only solution to

C1y1 + C2y2 + ...+ Cnyn = 0 is C1 = C2 = ... = Cn = 0

Although this is the formal definition of Linearly Independent sets we will not be using it. Instead we will be using

the Wronskian to determine if a set is linearly independent. Since we are considering only second order equations in this

chapter we will limit our study to the linear independence or dependence of two functions y1 and y2

The Wronskian of two functions y1 and y2 is given by the determinate:

W (y1, y2) =

∣∣∣∣∣y1 y2

y′1 y′2

∣∣∣∣∣ = y1y′2 − y2y′1

If two functions y1 and y2 are linearly dependent then, from the formal definition, is possible to express one function

as a scalar multiple of the other. That is:

y2 = Cy1

Making the Wronskian:

57

58 CHAPTER 2. SECOND ORDER EQUATIONS.

W (y1, y2) = W (y1, Cy1) =

∣∣∣∣∣y1 Cy1

y′1 Cy′1

∣∣∣∣∣ = Cy1y′1 − Cy1y′1 = 0

So if two functions are linearly dependent their Wronskian is identically zero.

If y1 and y2 are both solutions to y′′ + P (x)y′ + Q(x)y = 0 and {y1, y2} is linearly independent then {y1, y2} is a

Fundamental Solution Set of the differential equation.

246.

Show that the set is a fundamental solution set of the differential equation

{y1, y2} = {e3x, xe3x} y′′ − 6y′ + 9y = 0

247.


{y1, y2} = {x2, x3} x2y′′ − 4xy′ + 6y = 0

248.


{y1, y2} = {sinh(x), cosh(x)} y′′ − y = 0

Remember

sinh(x) =ex − e−x

2cosh(x) =

ex + e−x

2

249.


{y1, y2, y3} = {ex, e2x, e3x} y′′′ − 6y′′ + 11y′ − 6y = 0

250.


{y1, y2} = {eax sin(bx), eax cos(bx)} y′′ − 2ay′ + (a2 + b2)y = 0

251.

Show that the two sets are both fundamental solution set of the differential equation. Which one would you rather

work with?

S1 = {y1, y2} = {ex, e2x} S2 = {y1, y2} = {4ex, e2x − 6ex} y′′ − 3y′ + 2y = 0

252.

Show that if y1 and y2 both have a relative extrema at x = x0 then they cannot be a fundamental solution set to

y′′ + P (x)y′ +Q(x)y = 0 on an interval containing x = x0

253.

Show that y1 = cos(2x) and y2 = cos2(x)− sin2(x) are Linearly Dependent

2.1. WRONSKIAN, FUNDAMENTAL SETS AND ABLE’S THEOREM 59

254.

Show for a, b Constants that

W (ay1, by2) = abW (y1, y2)

255.

Show for f , g, h differentiable functions that

W (fg, fh) = f2W (g, h)

256.

Calculate and simplify the following

e∫W(fg ,gf

)dx

257.

Show

W (y1 + a, y2 + a) = W (y1, y2) + ad

dx(y2 − y1)

258.

Show

W (y1, y2) = y22d

dx

(y1y2

)259.

Show

y1 ·W(y2y1, y1

)+ y2 ·W

(y1y2, y2

)=

d

dx(y1 · y2)

260.

If W (y1, y2) = e4x and y1 = ex find y2 if y2(0) = 2

261.

Show:

W (f, g + h) = W (f, g) +W (f, h)

262.

If

W (f, g) = e5x

Find

W (f + g, f − g)

263.

If

W (x, y) = W (x2, y)

Find y


264.

If

W (y, y2) = e3x y(0) = 1

Find y(x)

265.

Show

W ′(f, g) = W (f, g′) +W (f ′, g)

266.

If W (f(x), g(x))|x=0 = 10 Find W (f(3x), g(3x))|x=0

267.

If W (sin(x), y) = y2 Find y

268.

If W (y, 1y ) = 1 Find y

269.

If f and g are even functions show that W (f, g) is an odd function

270.

If

W (f, g) = W (g, h)

Show

g(x) = C

(f(x) + h(x)

)271.

If

W (f, g) = f · g

Find

g

f

272.

Show:

W

(f(x) cos(x), f(x) sin(x)

)=

(f(x)

)2

273.

If

W (f(g), g) = 0


Show

f(x) = Cx

Interpret the results in terms of linear independence

274.

Solve for f(x) if

W (x, f) =

√1 + x2

x2

275.

Find

e

∫W

(f, 1f

)dx

276.

Show:

W (x · f, f) ≤ 0

277.

Let y1 and y2 be solutions to

y′′ + P (x)y′ +Q(x)y = 0

Show:

P (x) = −y1y′′2 − y2y′′1

W (y1, y2)and Q(x) =

y′1y′′2 − y′2y′′1

W (y1, y2)

Use this result to show the differential equation with solutions y1 = xn and y2 = xm with n 6= m has the form

ax2y′′ + bxy′ + cy = 0

278.

Show the second order linear homogenous equation

y′′ + P (x)y′ +Q(x)y = 0

with fundamental solution set {y1, y2} and Wronskian W can be written as

1

W

∣∣∣∣∣∣∣y y1 y2

y′ y′1 y′2

y′′ y′′1 y′′2

∣∣∣∣∣∣∣ = 0

279.

Use the results of either of the previous problem to find a differential equation with the following Fundamental Solution

Set


{y1, y2} = {sin(kx), cos(kx)}

280.

Show

W

(1, x, x2, x3, ..., xn

)= 1! · 2! · 3! · ... · n!

281.

The differential equation

y′′ + P (x)y′ +Q(x)y = 0

can be converted into Normal Form:

u′′ + f(x)u = 0

with the substitution

y(x) = u(x) · v(x) v(x) = e−12

∫P (x)dx

Use this to convert Bessel’s equation of order v to normal form

x2y′′ + xy′ +

(x2 − v2

)y = 0

Another way of calculating the Wronskian of the two solutions:y1 and y2 of y′′ + P (x)y′ + Q(x)y = 0 on (a, b) is to

use Abel’s Identity:

W (y1, y2) = Ce−

∫ xx0P (t)dt

x0 ∈ (a, b) P and Q continuous on (a, b)

This can be easily derived by noticing that if {y1, y2} is a fundamental solution set to y′′ + P (x)y′ + Q(x)y = 0 on

(a, b) then:

y′′1 + Py′1 +Qy1 = 0 y′′2 + Py′2 +Qy2 = 0

Multiplying the first equation by y2 and the second by y1 gives:

y2y′′1 + Py2y

′1 +Qy2y1 = 0 y1y

′′2 + Py1y

′2 +Qy1y2 = 0

Subtracting the second equation from the first gives:

y2y′′1 − y1y′′2 + P (y′1y2 − y2y′1) = 0

Remembering

W (y1, y2) = y′1y2 − y2y′1 and calculating W ′(y1, y2) = y2y′′1 − y1y′′2


The above equation becomes:

W ′(y1, y2) + PW (y1, y2) = 0 orW ′(y1, y2)

W (y1, y2)= −P

Integrating and solving for W (y1, y2)

ln |W (y1, y2)| = −∫ x

x0

P (t)dt+ C

W (y1, y2) = Ce−

∫ xx0P (t)dt

This result is known as Abel’s Identity

282.

Use Abel’s Identity to find the Wronskian up to a constant for

(1 + x2)y′′ + 2xy′ + y = 0 on (−∞,∞)

283.


(x2 + 3x+ 2)y′′ + y′ + y = 0 on (0,∞)

284.


cos2(x)y′′ +

(cos3(x) + 1

)y′ + y = 0 on

(−π4,π

4

)285.


xy′′ + (2x2 + 1)y′ + xy = 0 on (0,∞)

286.


y′′ +6x

x4 + 5x2 + 4y′ + y = 0 on

(−∞,∞

)287.


y′′ − 16x

4x− 1y′ +

16

4x− 1y = 0 on

(1

4,∞)

288.

Show y1 = 1x−1 and y2 = 1

x+1 are solutions to the given differential equation

(x2 − 1)y′′ + 4xy′ + 2y = 0


and calculate the Wronskian of y1 and y2 and then confirm your answer with Abel’s Identity.

289.


(P (x)y′)′ +Q(x)y = 0

290.

If the Wronskian of the solutions to

y′′ + P (x)y′ +Q(x)y = 0

is a constant, what does it say about P (x)

2.2 Reduction of Order

Question: given a second order differential equation of the form:

y′′ + P (x)y′ +Q(x)y = 0

and one solution to the differential equation can we find a second solution? If y1 is a solution to y′′+P (x)y+Q(x)y = 0

then we know y′′1 +P (x)y′1 +Q(x)y1 = 0. Let us try to find a second solution of the form y2 = vy1. Differentiating gives:

y′2 = vy′1 + v′y1 y′′2 = vy′′1 + 2v′y′1 + v′′y1

Substituting these into the original differential equation gives:

vy′′1 + 2v′y′1 + v′′y1 + P (x)(vy′1 + v′y1) +Q(x)vy1 = 0

Which reduces to:

v(y′′1 + P (x)y′1 +Q(x)y1) + 2v′y′1 + v′′y1 + P (x)v′y1 = 0

v′′y1 + 2v′y′1 + P (x)v′y1 = 0

v′′y1 + v′(2y′1 + P (x)y1) = 0

v′′y1 = −v′(2y′1 + P (x)y1)v′′

v′= −2y′1

y1− P (x)

Integrating gives:

ln |v′| = −2 ln |y1| −∫P (x)dx

Solving for v′

2.2. REDUCTION OF ORDER 65

v′ = eln |y−21 |−

∫P (x)dx =

e−∫P (x)dx

y21

Integrating again give the formula for v

v =

∫e−

∫P (x)dx

y21dx

An Example: Find a second linearly independent solution:

xy′′ − (x+ 1)y′ + y = 0 y1 = ex

Writing the equation is standard form gives:

y′′ +

(− 1− 1

x

)y′ +

1

xy = 0

v =

∫e

∫−

(−1− 1

x

)dx

e2xdx =

∫ex+ln(x)

e2xdx

v =

∫xe−xdx = −(x+ 1)e−x

y2 = y1v = ex(−(x+ 1)e−x) = −x− 1

Making the homogenous solution

yh = C1y1 + C2y2

yh = C1ex + C2(x+ 1)

♠291.

y1 = e2x is one solution to y′′ − 6y′ + 8y = 0. Use reduction of order to find a second linearly independent solution.

292.

y1 = x−2 is one solution to x2y′′+6xy′+6y = 0. Use reduction of order to find a second linearly independent solution.

293.

y1 = 1x is one solution to xy′′ + (2x + 2)y′ + 2y = 0. Use reduction of order to find a second linearly independent

solution.

294.

y1 = ex is one solution to xy′′ − (x + 1)y′ + y = 0. Use reduction of order to find a second linearly independent

solution.

295.

y1 = ex is one solution to xy′′+ (1−2x)y′+ (x−1)y = 0. Use reduction of order to find a second linearly independent

solution.

296.


y1 = ex is one solution to (2x − 1)y′′ − (4x2 + 1)y′ + (4x2 − 2x + 2)y = 0. Use reduction of order to find a second

linearly independent solution.

297.

y1 = ex is one solution to (sin(x) − cos(x))y′′ − 2 sin(x)y′ + (sin(x) + cos(x))y = 0. Use reduction of order to find a

second linearly independent solution.

298.

y1 = x2 is one solution to (x2 − 2x)y′′ + (2− x2)y′ + (2x− 2)y = 0. Use reduction of order to find a second linearly

independent solution.

299.

y1 = tan(x) is one solution to y′′−tan(x)y′−sec2(x)y = 0. Use reduction of order to find a second linearly independent

solution.

300.

y1 = x sin(x) is one solution to x2y′′−2xy′+(2+x2)y = 0. Use reduction of order to find a second linearly independent

solution.

301.

y1 = x + 1 is one solution to (x2 + 2x− 1)y′′ − (2x + 2)y′ + 2y = 0. Use reduction of order to find a second linearly


302.

y1 = x2 + 1 is one solution to y′′ − 2xx2−1y

′ + 2x2−1y = 0. Use reduction of order to find a second linearly independent

solution.

303.

y1 = sin(x) is one solution to y′′ − 3 cot(x)y′ + 3−2 sin2(x)sin2(x)

y = 0. Use reduction of order to find a second linearly


304.

y1 = 1x−2 is one solution to (x2 − 4)y′′ + 4xy′ + 2y = 0. Use reduction of order to find a second linearly independent

solution.

305.

y1 = x2 is one solution to

y′′ − x3 − 3x+ 1

x3 − 3xy′ +

2x3 − 2x2 + 2

x4 − 3x3y = 0

Use reduction of order to find a second linearly independent solution.

306.

y1 = x is one solution to

y′′ − x

x− 1y′ +

1

x− 1y = 0


307.

y1 = sin(x)√x

is one solution to the Bessel equation of order 12 :

x2y′′ + xy′ +

(x2 − 1

4

)y = 0


2.2. REDUCTION OF ORDER 67

308.

y1 = ln(x) is one solution to

x2(

ln(x)

)2

y′′ − 2x ln(x)y′ +

(2 + ln(x)

)y = 0 x > 0


309.

The Hermite equation is an equation of the form:

y′′ − 2xy′ + λy = 0

Find the homogenous solution for the given values of λ and y1

A) λ = 4 and y1 = 1− 2x2

B) λ = 6 and y1 = 3x− 2x3

310.

The Legendre equation is an equation of the form:

(1− x2)y′′ − 2xy′ + λ(λ+ 1)y = 0 x ∈ (−1, 1)


A) λ = 1 and y1 = x

B) λ = 2 and y1 = 3x2 − 1

C) λ = 3 and y1 = x3 − 3x

311.

The Laguerre equation is an equation of the form:

xy′′ + (1− x)y′ + λy = 0


A) λ = 1 and y1 = x− 1

B) λ = 2 and y1 = x2 − 4x+ 2

312.

First, use Abel’s Identity to find the Whronskian up to a constant for

y′′ − 16x

4x− 1y′ +

16

4x− 1y = 0 on

(1

4,∞)

Second, notice y1 = x is a solution and apply reduction of order to find y2

313.

The reduction of order algorithm can be applied to third order equations although the formula we derived for v above

will not work. y1 = ex is a solution to:

xy′′′ − xy′′ + y′ − y = 0

Use y2 = vy1 to reduce this equation to a second order equation by letting w = y′


2.3 Equations of the form y”=f(x,y’) and y”=f(y,y’)

For a second order equation of the form:

y′′ = f(x, y′)

the substitution v = y′ will transform the equation into first order equation.

An Example:

Solve:

xy′′ − y′ = 3x2

Using the substitution v = y′, making v′ = y′′ our equation becomes:

xv′ − v = 3x2 in standard form v′ +−1

xv = 3x

This equation is now first order linear. Creating an integrating factor

I(x) = e∫ −1

x = x−1

And now the solution in terms of v

v = x

(∫x3xdx+ C

)

v = x(x3 + C)

Converting back to y

y′ = x4 + Cx

Integrating

y =x5

5+Cx2

2+K

If the differential equation is of the form:

y′′ = f(y, y′)

the substitution v = y′ will transform the equation into a first order equation.

An Example from differential geometry:

The curvature of a circle of radius r is defined to be 1r and the curvature of a straight line defined to be zero. For

other equations in R2 the curvature is given by the second order equation:

K =|y′′|

(1 + (y′)2)32

2.3. EQUATIONS OF THE FORM Y”=F(X,Y’) AND Y”=F(Y,Y’) 69

By replacing K with 1r and solving the second order equation we should obtain the equation of a circle of radius r.

For simplicity we will assume y′′ > 0 so we do not have to deal with the pesky absolute values.

1

r=

y′′

(1 + (y′)2)32

Substituting v = y′ making v′ = y′′ we get:

1

r=

v′

(1 + (v)2)32

ordx

r=

dv

(1 + (v)2)32

This is now a separable and we now need to integrating both sides.∫dx

r=

∫dv

(1 + (v)2)32

The integral on the left is easy while the integral on the right will require a trig substitution.

x

r+ C =

∫dv

(1 + (v)2)32

v = tan θ dv = sec2 θdθ

1

v

√1 + v2

θ

Under this substitution the integral becomes:

x

r+ C =

∫sec2 θdθ

(1 + tan2 θ)32

x

r+ C =

∫cos θdθ

x

r+ C = sin θ

Using the triangle to convert back to v:

x

r+ C =

v√1 + v2

Converting back to y gives another separable equation:

x+ C

r=

y′√1 + (y′)2

Solving for y′

(x+ C)(√

1 + y′2) = ry′

(x+ C)2(1 + (y′)2) = r2(y′)2


(x+ C)2 + (x+ C)2(y′)2 = r2(y′)2

(y′)2 =(x+ C)2

r2 − (x+ C)2

y′ =(x+ C)√

r2 − (x+ C)2or

∫dy =

∫(x+ C)√

r2 − (x+ C)2dx

Substituting u = r2 − (x+ C)2, −12 du = (x+ C)dx for the integral on the right.

y +K =−1

2

∫u−12 du

y +K = −√r2 − (x+ C)2

Squaring both sides and rearranging terms gives the equation of a circle of radius r:

(y +K)2 + (x+ C)2 = r2

314.

Solve:

(1 + x2)y′′ = 2xy′

315.

Solve:

y′′ +1

xy′ =

4x

y′y′(1) = 2

316.

Solve:

xy′′ − y′ = 3x2

317.

Solve:

x2y′′ = 2xy′ + (y′)2

318.

Solve:

y′′ + x(y′)2 = 0

319.

Solve:

x2y′′ = (y′)2

2.3. EQUATIONS OF THE FORM Y”=F(X,Y’) AND Y”=F(Y,Y’) 71

If the independent variable x is missing from the differential equation then you will have an equation of the form:

y′′ = f(y, y′)

and the substitution v = y′ will transform the second order equation into a first order equation.

v =dy

dx

d2y

dx2=dv

dx=dv

dy· dydx

= vdv

dy

Using this substitution our equation becomes:

vdv

dy= f(y, v)

An Example:

Solve

y′′ + k2y = 0

Using the substitution above the second order equation becomes:

vdv

dy+ k2y = 0

This equation is now separable:

vdv = −k2ydy

Integrating and solving for v:

v2

2=−k2y2

2+ C

v =√C − k2y2 or

dy

dx= ±k

√A2 − y2

This equation is now separable: ∫dy√

A2 − y2=

∫±kdx

arcsin

(y

A

)= ±kx+B

y = A sin(±kx+B)

320.

Solve:

yy′′ + (y′)2 = 0

321.


Solve:

yy′′ = y2y′ + (y′)2

322.

Solve:

y′′ + 2yy′ = y

323.

Solve:

yy′′ = (y′)3

324.

Solve:

y′′ = 12y(y′)32

325.

If

{y1, (y1)2}

are solutions to

y′′ − 3y′ + ky = 0

Find y1

326.

Solve:

W (y, y′) = 0

Interprit the results in terms of linear independce or depencence

2.4 Homogenous Linear Equations with Constant Coefficients.

Let us consider the second order equation:

ay′′ + by′ + cy = 0

and look for a solution of the form y = erx. Differentiating twice give:

y = erx y′ = rerx y′′ = r2erx

Substituting these into the differential equation gives:

2.4. HOMOGENOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS. 73

ar2erx + brerx + cerx = 0

Dividing by erx gives the characteristic polynomial:

ar2 + br + c = 0

This quadratic equation can have 3 types of roots: real and distinct, complex or real and repeated.

If the two roots, r1 and r2 of the characteristic polynomial are real and distinct: r1 6= r2 then the solution to the

differential equation is

y = C1er1x + C2e

r2x

If the two roots, r1 and r2 of the characteristic polynomial are complex r1 = α+ βı and r2 = α− βı then the solution

to the differential equation is

y = C1e(α+βı)x + C2e

(α−βı)x or y = C1eαxeβıx + C2e

αxe−βıx

After applying Euler’s equation: eıθ = cos(θ) + ı sin(θ) we get the solution to the differential equation to be:

y = C1eαx cos(βx) + C2e

αx sin(βx)

If the two roots, r1 and r2 of the characteristic polynomial are real and repeated: r1 = r2 = −b2a then one solution to

the differential equation is

y1 = er1x

The second linearly independent solution comes by applying the reduction of order algorithm to the problem. If

y1 = er1x = e−b2a x is a solution to y′′ +

b

ay′ +

c

ay = 0

then the reduction of order algorithm gives:

v =

∫e−

∫badx

(e−b2a x)2

dx =

∫e−bxa

e−bxa

dx = x

Making y2 = xy1 = xer1x and the solution to the differential equation is

y = C1er1x + C2xe

r2x

327.

Solve

y′′ − y′ − 2y = 0 y(0) = 2 y′(0) = 1

328.

Solve

y′′ − 12y′ + 36y = 0 y(0) = 1 y′(0) = 1


329.

Solve

y′′ − 2y′ + 10y = 0 y(0) = 2 y′(0) = 1

330.

Solve

y′′ − 8y′ + 41y = 0 y(0) = 1 y′(0) = 1

331.

Solve

y′′ − 16y′ + 64y = 0 y(0) = 1 y′(0) = 9

332.

Solve

y′′′ − 6y′′ + 11y′ − 6y = 0 y(0) = 1 y′(0) = 1 y′′(0) = 1

333.

Solve

y′′′ − 7y′′ + 15y′ − 9y = 0

334.

Solve

y′′′′ − 8y′′′ + 24y′′ − 32y′ + 16y = 0

335.

Solve

y′′′ − 2y′′ + 4y′ − 8y = 0 y(0) = 2 y′(0) = 4 y′′(0) = 0

336.

Find a third order differential equation with the following solution

yh = C1e3x + C2e

3x sin(2x) + C3e3x cos(2x)

337.

Solve the differential equation for different values of k and sketch the solutions

y′′ + ky′ + 6y = 0 k ∈ {−7, 5, 2, 0}

338.

Solve

y′′ − 2ıy′ + 3y = 0

2.4. HOMOGENOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS. 75

339.

Consider the differential equation:

ay′′ + by′ + cy = 0 a, b, c ∈ R+

Show all solutions y tend to zero as x tends to infinity. Show this is not true if b = 0

340.


ky′′ + (k + 1)y′ + (k + 2)y = 0

Find the values of k so that the characteristic ploynomial has real and distint roots, real and repeated roots and

complex roots. What values of k make the solution y tend to zero as x→∞341.


y′′ + ay′ + a2y = 0

Show the characteristic polynomial has complex roots for all a 6= 0

342.

Show that if y1 and y2 are solution to the second order linear equation

ay′′ + by′ + cy = 0 a, b, c ∈ R

Then y = Cy1 +Ky2 is also a solution.

343.

Find a third order homogenous differential equation with constant coefficients with the given solution

y = C1e2x + C2xe

2x + C3x2e2x

344.

If y = xex is a solution to

y′′ + ay′ + by = 0

Find a and b

345.

Solve

dz

dt= e(a+bı)t

by separating into real and imaginary parts. Then use the fact that∫eat cos(bt)dt =

eat

a2 + b2

(a cos(bt) + b sin(bt)

)+ C

To calculate ∫eat sin(bt)dt


346.

For the given differential equation:

y′′ + 2y′ + Cy = 0 C ∈ R

A) For what values of C does the characteristic equation have two real distinct roots, real repeated roots and complex

roots.

B) For the case where you have two real distinct roots find the values of C that make the solution tend to zero as

x→∞.

347.

Solve the differential equation with the discontinuous coefficient function

y′′ + sgn(x)y = 0

sgn(x) =

{−1 x < 0

1 0 < x

2.5 The Method of Undetermined Coefficients

In this section we will study a way to solve the Linear Differential equation:

y′′ + ay′ + by = g(x)

for different equations g(x).

For example if you were given the differential equation

y′′ − 3y′ + 5y = 3ex

It would be reasonable to think the solution would involve the exponential function ex simply because exponential

functions differentiate into more exponential functions.

If you were given the differential equation

y′′ + y′ + y = cos(x) + sin(x)

It would be reasonable to think the solution would involve the functions sin(x) and cos(x) since sin(x) and cos(x)

differentiate into each other.

So in the method of undetermined coefficients we guess the form of the solution and try to make the constants work.

For

y′′ − 3y′ + 5y = 3ex

we would guess the form of the solution to be y = Aex and solve for the value of A that makes it a solution.

For

y′′ + y′ + y = cos(x) + sin(x)

2.5. THE METHOD OF UNDETERMINED COEFFICIENTS 77

we would guess the form of the solution to be y = A sin(x) +B cos(x) and solve for the values of A and B that makes

it a solution.

One problem that can arise using this method is the problem of finding homogenous solution twice. If you were given

y′′ − 4y′ + 4y = e2x

it would be reasonable to think the solution would be of the form y = Ae2x but we know the homogenous solution to

y′′ − 4y′ + 4y = 0

is y = C1e2x + C2xe

2x. So looking for a particular solution of the form y = Ae2x will again find the homogenous

solution, not the particular solution we are interested in. The solution to this problem is to multiply the particular solution

by xs so that it is linearly independent of the terms in the fundamental solution set. The nonnegative integer s should

be chosen to be the smallest nonnegative integer so that no term in the particular solution appears in the fundamental

solution set. In this problem the fundamental solution set is:

{e2x, xe2x}

So we would choose s = 2 and look for a particular solution of the form y = Ax2e2x.

If you choose the wrong form of the particular solution you’re gonna have a bad time.

In general we choose the form of the particular solution based on the following table:

For y′′ + ay′ + by = g(x)

g(x) yp

pn = anxn + ...+ a1x+ a0 xsPn = xs(Anx

n + ...+A1x+A0)

aeαx Axseαx

a cos(βx) + b sin(βx) xs(A cos(βx) +B sin(βx))

pneαx xs(Pne

αx)

pn cos(βx) + pm sin(βx) xs(PN cos(βx) +QN sin(βx)) N = max(n,m)

aeαx cos(βx) + beαx sin(βx) xs(Aeαx cos(βx) +Beαx sin(βx))

pneαx cos(βx) + qme

αx sin(βx) xs(PNeαx cos(βx) +QNe

αx cos(βx)) N = max(n,m)

The integer s is chosen to be the smallest nonnegative integer so that no term in the particular solution appears in the

homogenous solution.

An Example: Solve:

y′′ − 4y′ + 4y = 6e2x

First we need the homogenous solution, so we form the auxiliary equation:


r2 − 4r + 4 = 0 (r − 2)2 = 0 y1 = e2x y2 = xe2x

The homogenous solution is:

yh = C1e2x + C2xe

2x

The Fundamental Solution Set is:

F.S.S. = {e2x, xe2x}

Due to the right hand side of the differential equation we will choose a particular solution of the form yp = xs(Ae2x)

with s chosen to be the smallest positive integer so that no term in yp is in the fundamental solution set. So we choose

s = 2 making yp = Ax2e2x

y′p = e2x(2Ax+ 2Ax2)

Differentiating and simplifying gives:

y′′p = e2x(4Ax2 + 8Ax+ 2A)

Substituting these derivatives into the original differential equation gives:

e2x(4Ax2 + 8Ax+ 2A)− 4e2x(2Ax+ 2Ax2) + 4Ax2e2x = 6e2x

This simplifies nicely to:

2Ae2x = 6e2x A = 3

So our particular solution is:

yp = 3x2e2x

The general solution to the differential equation is the sum of the homogenous solution and the particular solution:

y = C1e2x + C2xe

2x + 3x2e2x

348.

Solve

y′′ + 3y′ + 2y = 18 sin(2x) + 26 cos(2x) y(0) = 2 y′(0) = 4

349.

Solve

y′′ + 4y = (5x2 + 9x+ 4)ex

350. Solve:

y′′ − 5y′ + 6y = ex(4 sin(x)− 2 cos(x)) y(0) = 3 y′(0) = 3


351.

Solve

y′′ − 4y′ + 4y = 6xe2x y(0) = 2 y′(0) = 4

352.

Solve

y′′ − 4y′ + 5y = (2x+ 2)ex

353.

Solve

y′′ + 5y = (2x3 + 6x2 + 6x)ex

354.

Solve

y′′ + 4y = 14 cos(x) + 12x sin(x) y(0) = 2 y′(0) = 4

355.

Solve

y′′ − 8y′ + 7y = −15xe2x − 32e2x y(0) = 2 y′(0) = 4

356.

Solve

y′′ − 3y′ + 2y = ex sin(x) y(0) = 2 y′(0) = 4

357.

Solve:

y′′ − 4y′ + 4y = 8e2x

358.

Solve

y′′ − 2y′ + y = (x+ 3)ex y(0) = 1 y′(0) = 2

359.

Solve

y′′ + 4y = (5x2 + 4x+ 2)ex y(0) = 4 y′(0) = 3

360.

Solve

y′′ − 3y′ + 2y = 6x3 − 19x2 + 8x− 11


361.

Solve

y′′ − 3y′ + 2y = 2x2 − 2x− 4− ex y(0) = 2 y′(0) = 6

362.

Solve

y′′ − 4y′ + 4y = 16 sin(x) cos(x) + 16 cos2(x)− 16 sin2(x)

363.

Solve

y′′ + 4y = cos3(x)

Hint

cos3(x) =3

4cos(x) +

1

4cos(3x)

364.

Find the form of the solution to

y′′ + 6y′ + 10y = x2e−3x sin(x)

365.


y′′′ − y′′ − y′ + y = x3(ex + e−x)

366.


y′′′ − y′′ + 4y′ − 4y = x2(ex + sin(2x))

367.

Find a solution of the form y = Ax2 +Bx+ C to solve the nonlinear differential equation

y′′ + (y′)2 + y = 5x2 + 5x+ 3

368.

The form of the particular solution to the third order linear differential equation:

ay′′′ + by′′ + cy′ + d = g(x)

is

yp = (Ax4 +Bx3 + Cx2)ex +Dxe2x

what does this tell you about the roots of the characteristic polynomial


ar3 + br2 + cr + d = 0

and the function g(x).

Another method that can be used to solve the above differential equations is the Annihilator Method.

Definition: A linear differential operator A is said to annihilate a function f if

A(f) = 0

The operator D operates on a function by calculating its derivative. The operator D2 operates on a function by

calculating its second derivative. For example

D(x3) = 3x2 D2(sin(x)) = − sin(x)

Since Dn+1(xn) = 0 we say Dn+1 annihilates xn

369.

Show that D − r annihilates erx

370.

Show that (D − r)n+1 annihilates xnerx

371.

Show that (D − α)2 + β2 annihilates both eαx cos(βx) and eαx sin(βx)

372.

Show that ((D − α)2 + β2)n+1 annihilates both xneαx cos(βx) and xneαx sin(βx)

Example:

Solve:

y′′ − 3y′ + 2y = e3x

Multiplying both sides by (D − 3) annihilates the right hand side giving the homogenous equation

y′′′ − 3y′′ + 2y′ − 3y′′ + 9y′ − 6y = 0

y′′′ − 6y′′ + 11y′ − 6y = 0

This has a characteristic polynomial

r3 − 6r2 + 11r − 6 = 0

Factoring

(r − 1)(r − 2)(r − 3) = 0 r = 1, 2, 3

The solution is

y = C1ex + C2e

2x + C3e3x


Now part of this solution is the homogenous solution to the original differential equation and the other part is the

particular solution. Since the right hand side of the original differential involves a e3x term the particular solution must

be

yp = C3e3x

Placing this into the original equation and solving for C3 gives

yp =1

2e3x

y = C1ex + C2e

2x +1

2e3x

373.

Solve by annihilation

y′′ + y = 6xex

374.


y′′ + 4y = ex sin(x)

375.


y′′′ + 2y′′ − y′ − 2y = ex − 1

2.6 Variation of Parameters

In the method of variation of parameters we will develop a solution to the second order linear differential equation:

y′′ + P (x)y′ +Q(x)y = g(x)

As you can see the method of variation of parameters is a far more general method of solving differential equation than

the method of undetermined coefficients since variation of parameters allows for variable coefficients of y′ and y whereas

the method of undetermined coefficients works only for linear equations with constant coefficients and the function g(x)

does not need to be listed in the table in the section on undetermined coefficients.

Given the fundamental solution set {y1, y2} for the homogenous equation:

y′′ + P (x)y′ +Q(x)y = 0

we will look for a particular solution to the nonhomogeneous equation of the form

2.6. VARIATION OF PARAMETERS 83

y = v1y1 + v2y2

Differentiating yields:

y′ = v1y′1 + v′1y1 + v2y

′2 + v′2y2

To avoid second derivatives of the unknown functions v1 and v2 we impose the condition:

v′1y1 + v′2y2 = 0

Making

y′ = v1y′1 + v2y

′2

Differentiating again gives:

y′′ = v1y′′1 + v′1y

′1 + v2y

′′2 + v′2y

′2

Substituting y, y′ and y′′ into the original differential equation gives:

v1y′′1 + v′1y

′1 + v2y

′′2 + v′2y

′2 + P (x)(v1y

′1 + v2y

′2) +Q(x)(v1y1 + v2y2) = g(x)

This equation can be rewritten as:

v1(y′′1 + P (x)y′1 +Q(x)y1) + v2(y′′2 + P (x)y′2 +Q(x)y2) + y′1v′1 + y′2v

′2 = g(x)

Since y1 and y2 are solutions of the homogenous equation

y′′ + P (x)y′ +Q(x)y = 0

the above equation reduces to:

y′1v′1 + y′2v

′2 = g(x)

This equation along with the restriction we made on y′: v′1y1 + v′2y2 = 0 gives us a system of two equations with two

unknown variables: v′1 and v′2 that we will solve using Cramer’s Rule. Writing the system as a matrix equation gives:[y1 y2

y′1 y′2

][v′1

v′2

]=

[0

g(x)

]Solving for the two unknown variables: v′1 and v′2 using Cramer’s Rule gives:

v′1 = − y2g(x)

y1y′2 − y2y′1= − y2g(x)

W (y1, y2)and v′2 =

y1g(x)

y1y′2 − y2y′1=

y1g(x)

W (y1, y2)

Integrating gives:

v1 = −∫

y2g(x)

y1y′2 − y2y′1dx = −

∫y2g(x)

W (y1, y2)dx and v2 =

∫y1g(x)

y1y′2 − y2y′1dx =

∫y1g(x)

W (y1, y2)dx


An Example: Solve:

y′′ + y = sec(x)

First note that this equation cannot be solved by the method of undetermined coefficients due to the fact that the

right hand side: sec(x) is not in the table in the undetermined coefficients section.

First we need the homogenous solution, so we form the auxiliary equation:

r2 + 1 = 0 r = ±ı y1 = cos(x) y2 = sin(x)

The homogenous solution is:

yh = C1 cos(x) + C2 sin(x)

Our fundamental solution set is:

F.S.S = {cos(x), sin(x)}

The Wronskian of the fundamental solution set is:

W (cos(x), sin(x)) =

∣∣∣∣∣ cos(x) sin(x)

− sin(x) cos(x)

∣∣∣∣∣ = cos2(x) + sin2(x) = 1

v1 = −∫

sin(x) sec(x)dx = −∫

tan(x)dx = ln(cos(x))

v2 =

∫cos(x) sec(x)dx = x

yp = v1y1 + v2y2 = ln(cos(x)) cos(x) + x sin(x)

So the general solution is:

y = C1 cos(x) + C2 sin(x) + ln(cos(x)) cos(x) + x sin(x)

376.

Solve:

y′′ − 3y′ + 2y = xe2x

377.

Solve:

y′′ + y = tan(x)

378.


Solve:

y′′ − 4y′ + 5y = xe2x

379.

Solve:

y′′ − 2y′ + y =ex

1 + x2

380.

Solve:

y′′ + 4y = sin3(2x)

381.

Solve:

y′′ − 2y′ + y = ex arcsin(x)

382.

Solve:

y′′ + y′ = ln(x)

383.

Solve:

y′′ + 4y = csc(2x)

384.

Solve:

y′′ + 4y = ln(x+ 1)

385.

The fundamental solution set for the differential equation

xy′′ − y′ − 4x3y = x3ex2

is

{y1, y2} = {ex2

, e−x2

}

Use variation of parameters to solve the differential equation

386.

Show that y1 = tan(x) is a solution to the homogenous differential equation:

y′′ − tan(x)y′ − sec2(x)y = 0


Use reduction of order to find the second homogenous solution y2 and then use variation of parameters to find the

particular and general solution of:

y′′ − tan(x)y′ − sec2(x)y = sin(x)

387.

Show that y1 = x3 + x is a solution to the homogenous differential equation:

y′′ − 4x

x2 + 1y′ +

6x2 − 2

(x2 + 1)2y = 0



y′′ − 4x

x2 + 1y′ +

6x2 − 2

(x2 + 1)2y =

2

1 + x2

388.

Show that y1 = ex is a solution to the homogenous differential equation:

xy′′ − (x+ 1)y′ + y = 0



xy′′ − (x+ 1)y′ + y = x2e2x

389.

Show that y1 = x2 + 1 is one solution to

y′′ − 2x

x2 − 1y′ +

2

x2 − 1y = 0.



y′′ − 2x

x2 − 1y′ +

2

x2 − 1y = 2x.

390.

Show that y1 = x is a solution to the homogenous differential equation:

(x+ 1)y′′ + xy′ − y = 0



(x+ 1)y′′ + xy′ − y = (x+ 1)2

391.

Show that y1 = x is a solution to the homogenous differential equation:


(x2 − 1)y′′ − 2xy′ + 2y = 0



(x2 − 1)y′′ − 2xy′ + 2y = x2 − 1

392.

Show that y1 = 5x− 1 is a solution to the homogenous differential equation:

xy′′ + (5x− 1)y′ − 5y = 0



xy′′ + (5x− 1)y′ − 5y = x2e−5x

393.

Show that y1 = x+ 1 is a solution to the homogenous differential equation:

(x2 + 2x)y′′ − 2(x+ 1)y′ + 2y = 0



(x2 + 2x)y′′ − 2(x+ 1)y′ + 2y = (x+ 2)2

394.

Show that y1 = sin(x2) is a solution to the homogenous differential equation:

xy′′ − y′ + 4x3y = 0



xy′′ − y′ + 4x3y = 2x3

395.

Show that y1 = sin(x) is a solution to the homogenous differential equation:

sin2(x)y′′ − 2 sin(x) cos(x)y′ + (1 + cos2(x))y = 0



sin2(x)y′′ − 2 sin(x) cos(x)y′ + (1 + cos2(x))y = sin3(x)

396.


Show that y1 = sin(x) is a solution to the homogenous differential equation:

y′′ − 3 cot(x)y′ +3− 2 sin2(x)

sin2(x)y = 0



y′′ − 3 cot(x)y′ +3− 2 sin2(x)

sin2(x)y = sin3(x)

397.

Solve by first guessing a solution to the homogenous equation then using reduction of order to find the second

homogenous solution y2 and then use variation of parameters to find the particular and general solution of:

y′′ − 2x

1 + x2y′ +

2

1 + x2y = 1 + x2

398.

The Bessel Equation of order one half is:

x2y′′ + xy′ +

(x2 − 1

4

)y = 0

and has solutions y1 = cos(x)√x

and y2 = sin(x)√x

. Use variation of parameters to solve

x2y′′ + xy′ +

(x2 − 1

4

)y = x

52

399.

One solution to the equation:

y′′ + P (x)y′ +Q(x)y = 0

is (1 + x)2, and the Wronskian of the two solutions is 1. Find the general solution to:

y′′ + P (x)y′ +Q(x)y = 1 + x

2.7 Cauchy Euler Equation

:

The second order Cauchy Euler Equation is:

ax2d2y

dx2+ bx

dy

dx+ cy = 0

This equation can be transformed into a second order liner differential equation with constant coefficients with use of

the substitution:

x = et

2.7. CAUCHY EULER EQUATION 89

Differentiating with respect to t gives:

dx

dt= et So

dx

dt= x

In the Cauchy Euler equation we see the term x dydx which we need to substitute for, so I will multiply both sides ofdxdt = x by dy

dx :

dx

dt

dy

dx= x

dy

dxThis simplifies to

dy

dt= x

dy

dx

Differentiating dydt = x dydx with respect to x gives:

d

dx

dy

dt=dy

dx+ x

d2y

dx2

In the Cauchy Euler equation we see the term x2 d2ydx2 which we need to substitute for, so I will multiply the left side

of the above equation by dxdt and the right hand side by x (Remember they are equal):

d

dx

dy

dt

dx

dt=

(dy

dx+ x

d2y

dx2

)x This simplifies to

d2y

dt2= x

dy

dx+ x2

d2y

dx2

Replacing x dydx with dydt and solving for x2 d

2ydx2 gives:

x2d2y

dx2=d2y

dt2− dy

dt

So under this substitution the Cauchy Euler equation becomes:

a

(d2y

dt2− dy

dt

)+ b

dy

dt+ cy = 0

This simplifies to the second order linear equation with constant coefficients:

ad2y

dt2+ (b− a)

dy

dt+ cy = 0

Which we can solve by finding the roots of the characteristic polynomial:

ar2 + (b− a)r + c = 0

Again there are three cases: the roots are real and distinct, the roots are real and repeated or the roots are complex.

Case 1: we have two real and distinct roots r1 and r2. Then the solutions to the differential equation are:

y1 = er1t and y2 = er2t

Since x = et, t = ln(x) Making the solution:

y1 = er1 ln(x) = eln(xr1 ) and y2 = er2 ln(x) = eln(x

r2 )

So

y1 = xr1 and y2 = xr2

Case 2: The roots are real and repeated r1 = r2. Then the solutions to the differential equation are:


y1 = er1t and y2 = ter1t


y1 = er1 ln(x) = eln(xr1 ) and y2 = ln(x)er1 ln(x) = ln(x)eln(x

r1 )

So

y1 = xr1 and y2 = ln(x)xr1

Case 3: The roots are complex r1 = α+ βı and r2 = α− βı. Then the solutions to the differential equation are:

y1 = eαt cos(βt) and y2 = eαt sin(βt)


y1 = xα cos(β ln(x)) and y2 = xα sin(β ln(x))

An Example: Solve:

x2y′′ − 5xy′ + 13y = 0

Forming the characteristic polynomial

r2 + (−5− 1)r + 13 = 0 r2 − 6r + 13 = 0 (r − 3)2 = −4 r = 3± 2ı

So we have complex roots so the solution is:

yh = C1e3t cos(2t) + C2e

3t sin(2t)

Converting back to x gives:

yh = C1x3 cos(2 ln(x)) + C2x

3 sin(2 ln(x))

An Example: Solve:

sin2(x)d2y

dx2+ tan(x)

dy

dx− k2 cos2(x)y = 0

Although this is not the Cauchy-Euler equation we can transform it into one using the following substitution:

u = sin(x)

dy

dx=dy

du· dudx

= cos(x)dy

du

Using the product rule gives

d2y

dx2=

d

dx

(cos(x)

dy

du

)= − sin(x)

dy

du+ cos(x)

d

dx· dydu


d2y

dx2= − sin(x)

dy

du+ cos(x)

d

dx

(dy

du

du

du

)d2y

dx2= − sin(x)

dy

du+ cos(x)

d2y

du2· dudx

d2y

dx2= − sin(x)

dy

du+ cos2(x)

d2y

du2

Substituting these expressions into the differential equation gives

sin2(x)

(− sin(x)

dy

du+ cos2(x)

d2y

du2

)+ tan(x) cos(x)

dy

du− k2 cos2(x)y = 0

Rearranging the terms gives

sin2 cos2(x)d2y

du2+

(sin(x)− sin3(x)

)dy


sin2 cos2(x)d2y

du2+ sin(x)

(1− sin2(x)

)dy


After a trig identity we see a cos2(x) in each term that we can eliminate yielding

sin2 d2y

du2+ sin(x)

dy

du− k2y = 0

Since u = sin(x) we get the Cauchy-Euler equation

u2d2y

du2+ u

dy

du− k2y = 0

Now the Cauchy-Euler substitution

u = et udy

du=dy

dtu2d2y

du2=d2y

dt2− dy

dt

Our differential equation now becomes

d2y

dt2− k2y = 0

which has the characteristic equation

r2 − k2 = 0 r = ±k

So the solution is

y = C1ekt + C2e

−kt

Converting back to the variable u

y = C1uk + C2u

−k

Converting back to the variable x gives the final solution

y = C1 sink(x) + C2 sin−k(x)


400.

Solve:

x2d2y

dx2+ 7x

dy

dx+ 8y = 0

401.

Solve:

x2d2y

dx2+ 9x

dy

dx+ 12y = 0

402.

Solve:

x2d2y

dx2− 3x

dy

dx+ 20y = 0

403.

Solve:

x2d2y

dx2− 11x

dy

dx+ 36y = 0

404.

Solve:

x2d2y

dx2− 4x

dy

dx+ 6y = x3 ln(x)

405.

Solve:

x2d2y

dx2+ 3x

dy

dx− 8y = (ln(x))3 − ln(x)

406.

Solve:

x2d2y

dx2+ x

dy

dx+ y = x

407.

Solve:

x2d2y

dx2− 5x

dy

dx+ 8y = x3 arctan(x)

408.

Solve:

x2d2y

dx2+ x

dy

dx+ 4y =

1

x

409.

Solve:


x2d2y

dx2+ 3x

dy

dx+ y = 8x

410.

Solve:

x2d2y

dx2− 7x

dy

dx+ 16y = x3

411.

Solve:

x2d2y

dx2− 4x

dy

dx+ 6y =

√1− x2

412.

Solve:

x2d2y

dx2+ x

dy

dx+ 4y = 8

413.

Solve:

x3d2y

dx2+ x2

dy

dx+ xy = 0

414.

Find a Cauchy Euler equation with the following solution

y =C1x

2 + C2

x

415.

Find a Cauchy Euler equation with the following solution

y = C1x2 cos(4 ln(x)) + C2x

2 sin(4 ln(x))

416.

Solve:

xy′′ + y′ = 0

417.

Solve:

2(x− 4)2d2y

dx2+ 5(x− 4)

dy

dx− 2y = 0

418.

Solve:

x3d3y

dx3− 4x2

d2y

dx2+ 8x

dy

dx− 8y = 4 ln(x)

419.


Find the values of α that make the solution: y, to the given equation tend to zero as x→∞

x2d2y

dx2+ αx

dy

dx+

1

4y = 0

420.

Find the value of α that makes y → 0 as x→ 0

x2d2y

dx2− 6y = 0

y(1) = 1 y′(1) = α

421.

The nonlinear equation

y′ + y2 + P (x)y +Q(x) = 0

is an example of a Riccati Equation. Show the substitution y = z′

z transform the equation to

z′′ + P (x)z′ +Q(x)z = 0

422.

Use the results of the previous problem to solve

y′ + y2 − 4

xy +

6

x2= 0

423.

Use the results of the previous problem to solve

y′ + y2 − 4

xy +

12

x2= 0

424.

Use the substitution x =√t to transform the differential equation to an equation with constant coefficients and then

solve.

y′′ − 1

xy′ + 4x2y = 0

425.

Show that if y(x) is a solution to the Cauchy Euler Equation for x > 0 then y(−x) is a solution for x < 0

2.8 Everyone Loves a Slinky: Springs

The goal of this section is to develop a differential equation that governs the motion of a mass connected to an ideal

spring. We will first study the theoretical case of a spring with no damping: (internal resistance of the spring, air friction

ect.). We will also study springs with damping and then with a forcing function attached to the mass.

2.8. EVERYONE LOVES A SLINKY: SPRINGS 95

Newton’s Second law states that force equals mass times acceleration: F = ma. So if y(t) represents the position of

a moving mass on a spring then its acceleration is a = d2ydt2 .

Now consider a mass spring system with a mass m attached and stretched so that the mass is still. This unmoving

system is said to be in equilibrium. We measure the distance, y, to be the displacement of the mass from equilibrium.

When the mass is displaced from equilibrium , the spring is stretched or compressed and it exerts a force in the opposite

direction of the displacement. The force exerted by the spring is given by Hooke’s Law:

Fspring = −ky k > 0

where k is a constant dependent on the stiffness of the spring and y is the displacement of the mass from equilibrium.

All mass spring systems experience some form of internal resistance known as damping which is proportional to the

velocity of the mass. Since the mass has a velocity v = dydt the damping force is given by:

Fdamping = −bdydy

b > 0

where b is the damping coefficient.

All other forces on the system are external making the differential equation governing the mass spring system:

md2y

dx2= −bdy

dy− ky + Fexternal(t)

Or

md2y

dx2+ b

dy

dy+ ky = Fexternal(t)

In the absence of damping and an external force: b = 0, Fexternal(t) = 0 the differential equation becomes:

md2y

dx2+ ky = 0

Which has an auxiliary mr2 + k = 0 which has purely imaginary roots r = ±ωı making the solution:

y = C1 cos(ωt) + C2 sin(ωt) ω =

√k

m

This solution can be written in the form:

y = A sin(ωt+ φ)

by first applying a trig identity to sin(ωt+ φ).

y = A sin(ωt+ φ) = A(sin(ωt) cos(φ) + cos(ωt) sin(φ))

These two solution are equal if

A sin(φ) = C1 and A cos(φ) = C2

We see the amplitude of the solution: A is given by:

A =√C2

1 + C22 and tan(φ) =

C1

C2


We see the solution is a sinusoid with angular frequency ω =√

km and Period T = 2π

ω .

All springs experience some form of damping. To explore the nature of this damping let us consider the equation:

my′′ + by′ + ky = 0

The auxiliary equation is:

mr2 + br + k = 0

with roots:

r =−b±

√b2 − 4mk

2m=−b2m± 1

2m

√b2 − 4mk

The nature of the solution depends on the discriminate b2 − 4mk.

If b2 − 4mk < 0 the roots will be complex and we say the spring system has Underdamped Motion.

Letting α be the real part and β the imaginary part of the roots we have:

α =−b2m

and β =1

2m

√4mk − b2

The solution is:

y = eαt(C1 cos(βt) + C2 sin(βt))

We can express this solution in a alternate form as we did earlier:

y = Aeαt sin(βt+ φ) A =√C2

1 + C22 and tan(φ) =

C1

C2

So our solution is the product of a sinusoid: sin(βt+φ) and an exponential damping factor: Aeαt. As t→∞ Aeαt → 0

and our solution also tends to zero. Further as b→ 0 α = α = −b2m → 0 and the solution tends to the sinusoid:

y = A sin(βt+ φ)

Going back to the discriminate: b2 − 4mk. If b2 − 4mk > 0 the roots will real and distinct and we say the spring

system has Overdamped Motion. The roots to the auxiliary equation are:

r1 =−b2m

+1

2m

√b2 − 4mk r2 =

−b2m− 1

2m

√b2 − 4mk

And the solution is:

y = C1er1t + C2e

r2t

It is clear that r2 is negative; r1 is negative as well since b2 > b2 − 4mk making b >√b2 − 4mk. Subtracting b from

this inequality and dividing by 2m gives r1 = −b2m + 1

2m

√b2 − 4mk < 0.

Since both r1 and r2 are negative the solution y will approach zero as t→∞.

Going back to the discriminate: b2 − 4mk. If b2 − 4mk = 0 the roots will real and repeated and we say the spring

system has Critically Damped Motion. The roots to the auxiliary equation are:

r1 = r2 =−b2m


And the solution is:

y = e−b2m (C1 + C2t)

As t→∞ our solution will tend to zero.

An Example:

A 1kg mass hangs from a spring stretching it .392m from equilibrium. The mass is then pulled down .5m and released.

Find the equation of the motion of the mass if:

1) damping constant b = 0




Solution: We already know mass m and damping constant b so all we need is the spring constant k. Using Hook’s

Law:

9.8 = k(.392) k = 25

The differential equation governing the system is:

y′′ + by′ + 25y = 0 y(0) = −.392 y′(0) = 0

1) If there is no damping then b = 0 and our equation becomes:

y′′ + 25y = 0

The auxiliary equation and roots are:

r2 + 25 = 0 r = ±5ı

The solution is:

y = C1 cos(5t) + C2 sin(5t)

After the laborious task of applying the initial conditions we get:

y = −.392 cos(5t)

We see the Amplitude is A = .392 and φ = π2 The solution can be written in the form:

y = −.392 sin

(5t+

π

2

)2) The damping constant is b = 8. Our equation becomes:

y′′ + 8y′ + 25y = 0



r2 + 8r + 25 = 0 r = −4± 3ı

The solution is:

y = C1e−4t cos(3t) + C2e

−4t sin(3t)


y = −.392e−4t cos(3t)− .52267e−4t sin(3t)

We see the Amplitude is A = .653336 and φ = .643498. Since the initial displacement is down (against the force of

the spring) we will use A = −.653336. The solution can be written in the form:

y = −.65336e−4t sin(3t+ .643498)

3) The damping constant is b = 10. Our equation becomes:

y′′ + 10y′ + 25y = 0


r2 + 10r + 25 = 0 r = −5 is repeated root

The solution is:

y = C1e−5t + C2te

−5t


y = −.392e−5t − 1.96te−5t

4) The damping constant is b = 26. Our equation becomes:

y′′ + 26y′ + 25y = 0


r2 + 26r + 25 = 0 r = −1,−25

The solution is:

y = C1e−t + C2e

−26t


y = .40768e−t − .01568e−26t

426.


A 3kg mass is attached to a spring with stiffness k = 48N/m. The mass is displaced 1/2m to the left of equilibrium

point and given a velocity of 2m/s to the right. The damping constant is 0. Find the equation of motion of the mass

along with the amplitude, period, and frequency.

427.

A 1/8kg mass is attached to a spring with stiffness k = 16N/m. The mass is displaced 3/4m to the left of equilibrium

point and given a velocity of 2m/s to the left. The damping constant is 2Ns/m. Find the equation of motion of the mass.

428.

Consider the following differential equation

y′′ + ty′ + y = 0

Although we cannot solve this differential equation we can determine the nature of the solution for large t by thinking

of the equation in terms of a mass spring system.

Find

limt→∞

y(t)

Now let us consider the mass spring system with a forcing function applied to the system. The differential equation

governing this system is:

my′′ + by′ + ky = F0 cos(ωt) F0 > 0 ω > 0

Let us first explore the underdamped case (0 < b2 < 4mk). From previous discussion we know the homogenous

solution is:

yh = Ae−b

2m t sin

(√4mk − b2

2mt+ φ

)With

A =√C2

1 + C22 tan(φ) =

C1

C2

To find the particular solution we apply the method of undetermined coefficients. We choose the form of the particular

solution to be:

yp = A1 cos(ωt) +A2 sin(ωt)

Making:

y′p = −A1ω sin(ωt) +A2ω cos(ωt) y′′p = −A1ω2 cos(ωt)−A2ω

2 sin(ωt)

Substituting this into the differential equation and simplifying gives:((k −mω2)A1 +A2bω

)cos(ωt) +

((k −mω2)A2 +A1bω

)sin(ωt) = F0 cos(ωt)

Equating corresponding coefficients gives:


(k −mω2)A1 +A2bω = F0 (k −mω2)A2 +A1bω = 0

Solving this system of equations gives:

A1 =F0(k −mω2)

(k −mω2)2 + b2ω2A2 =

F0bω

(k −mω2)2 + b2ω2

Making the solution :

yp =F0

(k −mω2)2 + b2ω2

((k −mω2) cos(ωt) + bω sin(ωt)

)Let

tan(θ) =A1

A2

Drawing a triangle for tan(θ) = yp = k−mω2

bω

bω

k −mω2

√(k −mω2)2 + b2ω2

θ

From the triangle we see:

sin(θ) =k −mω2√

(k −mω2)2 + b2ω2cos(θ) =

bω√(k −mω2)2 + b2ω2

Making:

k −mω2 =√

(k −mω2)2 + b2ω2 sin(θ) bω =√

(k −mω2)2 + b2ω2 cos(θ)

Now our solution is:

yp =F0

(k −mω2)2 + b2ω2

(√(k −mω2)2 + b2ω2 sin(θ) cos(ωt) +

√(k −mω2)2 + b2ω2 cos(θ) sin(ωt)

)

yp =F0√

(k −mω2)2 + b2ω2(sin(θ) cos(ωt) + cos(θ) sin(ωt))

After a trig identity we get:

yp =F0√

(k −mω2)2 + b2ω2sin(ωt+ θ)

So the general solution is:

y = yh + yp = Ae−b

2m t sin

(√4mk − b2

2mt+ φ

)+

F0√(k −mω2)2 + b2ω2

sin(ωt+ θ)

The first term in the solution yh tends to zero as t tends to infinity. So we refer to this term as the transient solution.

As t gets large and yh approaches zero the solution approaches the particular solution yp. Hence we call this term the

steady state solution.


The factor:

1√(k −mω2)2 + b2ω2

in the particular solution represents the ratio of the magnitude of the forcing function F0 to the magnitude of the

sinusoidal response to the input force so we call it: the frequency gain and has units length/force.

An Example:

An 8-kg mass is attached to a spring hanging from the ceiling causing the spring to stretch 1.96m upon coming to

rest at equilibrium. At t = 0 the forcing function F (t) = cos(2t) is applied to the system. The damping constant for the

system is 3 N-sec/m. Find the steady state solution and the frequency gain.

Solution:

At t = 0 the system is equilibrium so it has an initial position y(0) = 0 and initial velocity y′(0) = 0

First we need the spring constant k. Since the 8-kg mass stretched the spring 1.96m Hooks Law gives:

8(9.8) = k(1.96) k = 40

The differential equation governing the system is:

8y′′ + 2y′ + 40y = cos(2t)

We could use the method of undetermined coefficients to find the steady state solution but we have already derived

equations for it:

yp =F0√

(k −mω2)2 + b2ω2sin(ωt+ θ) tan(θ) =

k −mω2

bω

Plugging in the values for m, b, k, F0, θ and ω gives:

tan(θ) =1

2θ = .463648

And the steady state solution is:

yp =1√80

sin(2t+ .463648)

The frequency gain is:

1√(k −mω2)2 + b2ω2

=1

80

429.

An 8-kg mass is attached to a spring hanging from the ceiling causing the spring to stretch 7.84m upon coming to

rest at equilibrium. At t = 0 the forcing function F (t) = 2 cos(2t) is applied to the system. The damping constant for

the system is 1 N-sec/m. Find the steady state solution and the frequency gain.

430.

An 2-kg mass is attached to a spring hanging from the ceiling causing the spring to stretch .2m upon coming to rest

at equilibrium. At t = 0 the mass is displaced .005m below equilibrium and released. At t = 0 the forcing function

F (t) = .3 cos(t) is applied to the system. The damping constant for the system is 5 N-sec/m. Find the steady state

solution and the frequency gain.


431.

Although we cannot solve the following differential equation

y′′ + ety′ + y = 0

we can determine the limiting behavior of the solution by thinking about the differential equation in terms of a spring

equation.

Find

limt→∞

y(t)

For a given mass spring system with known mass: m, damping constant: b and spring constant k with a forcing

function F0 cos(ωt) we know the frequency gain: M(ω), will be a be a function of the variable ω and is given by:

M(ω) =1√

(k −mω2)2 + b2ω2

Often times we want to know the value of ω that maximizes the frequency gain. Differentiating gives:

M ′(ω) =−1

2((k −mω2)2 + b2ω2)

−32 (2(k −mω2)(−2mω) + 2b2ω)

Simplifying:

M ′(ω) = ω

(−2m2ω2 − b2 + 2km

(k −mω2)2 + b2ω2)32

)So M has critical numbers:

ω = 0 in this case the forcing function is constant

The critical number we care about is:

ω =

√k

m− b2

2m2

And the maximum frequency gain occurs when:

ω =

√k

m− b2

2m2

The maximum Amplitude of the steady state response is:

F0M(ω) = F01√

(k −mω2)2 + b2ω2

432.

Let the following differential equation govern the motion of a mass on a spring.

1

2y′′ + by′ + 10y = 3 cos(2t)

Find the value of b that maximizes the amplitude of the steady state response.

2.9. CIRCUITS 103

2.9 Circuits

We now turn our study to the circuit consisting of a resister whose letter representation is R and is measured in Ohms, a

capacitor whose letter representation is C and the inductor whose letter representation is L connected to a voltage source

whose letter representation is E.

This circuit will be governed by Kirchhoff’s loop rules. Here they are:

1. The sum of the currents flowing into any junction point are zero.

2 The sum of the voltage around any closed loop is zero.

From physics we can find the voltage drop by the resistor, capacitor and the inductor. Here they are

1. The voltage drop across a resistor is given by

ER = IR where I is the current passing through the resistor

2. The voltage drop across a capacitor is

EC =1

CQ where Q is the charge on the capacitor

3 The voltage drop across an inductor is

EL = LdI

dt

If a voltage source is connected to the circuit an adds voltage at a level of E(t) then Kirchhoff’s voltage law gives:

EL + ER + EC = E(t)

Or

LdI

dt+RI +

1

CQ = E(t)

Since the change in charge is the current we have

dQ

dt= I

Making

dI

dt=d2Q

dt2


Now our differential equation becomes

Ld2Q

dt2+R

dQ

dt+

1

CQ = E(t)

Sometimes we want to determine the current I(t) in the circuit so we differentiate the above equation to get:

Ld2I

dt2+R

dI

dt+

1

CI = E′(t)

433.

A RLC circuit has a voltage source given by E(t) = 40 cos(2t)V a resistor of 2 ohms, a inductor of .25 henrys and a

capacitor of 113 farads. If the initial current is zero and the initial charge on the capacitor is 3.5 coulombs, determine the

charge on the capacitor as a function of t.

434.

A RLC circuit with no voltage source has a resistor of 20 ohms, a inductor of .1 henrys and a capacitor of 125 farads.

If the initial current is zero and the initial charge on the capacitor is 10 coulombs, determine the charge on the capacitor

as a function of t.

435.

A RLC circuit has a voltage source given by E(t) = 40V a resistor of 10 ohms, a inductor of .2 henrys and a capacitor

of 113 farads. If the initial current is zero and the initial charge on the capacitor is 0, determine the current as a function

of t.

Chapter 3

Series Solution

In calculus we learned that all continuously differentiable function can be represented by a Taylor series. The Taylor

series for a function centered at x = c is:

f(x) = f(c) + f ′(c)(x− c) +f ′′(c)(x− c)2

2!+f ′′′(c)(x− c)3

3!+f (4)(c)(x− c)4

4!+ ...

If the series is centered at zero then we call it a Mclauren series.

Some common power series are:

ex = 1 + x+x2

2!+x3

3!+x4

4!+ ... =

∞∑n=0

xn

n!

sin(x) = x− x3

3!+x5

5!− x7

7!+ ... =

∞∑n=0

(−1)nx2n+1

(2n+ 1)!

cos(x) = 1− x2

2!+x4

4!− x6

6!+ ... =

∞∑n=0

(−1)nx2n

(2n)!

1

1− x= 1 + x+ x2 + x3 + ... =

∞∑n=0

xn

arctan(x) = x− x3

3+x5

5− x7

7=

∞∑n=0

(−1)nx2n+1

2n+ 1

3.1 Series Solutions Around Ordinary Points

In this chapter we will not be looking for some equation f(x) that is the solution to a differential equation, instead we

will be looking for its Power Series (normally a Taylor Series). For the case of the Taylor Series we will look for a solution

to the differential equation of the form:

y =

∞∑n=0

anxn

105

106 CHAPTER 3. SERIES SOLUTION

with the sequence an to be determined by substituting y, y′ and y′′ into the differential equation and developing a

recurrence relation for an and solving the recurrence relation for a formula for an.

An Example:

Find at least the first seven terms in the power series that is a solution to the differential equation

y′′ + 3xy′ + 2y = 0

We will look for a solution of the form:

y =

∞∑n=0

anxn making y′ =

∞∑n=1

nanxn−1 y′′ =

∞∑n=2

n(n− 1)anxn−2

Substituting y, y′ and y′′ into the differential equation produces:

∞∑n=2

n(n− 1)anxn−2 + 3

∞∑n=1

n(n− 1)anxn + 2

∞∑n=0

anxn = 0

I will now shift the first series in the above expression so that it too has an xn factor:

∞∑n=0

(n+ 2)(n+ 1)an+2xn + 3

∞∑n=1

nanxn + 2

∞∑n=0

anxn = 0

Now I will add the zero term in the first and third series so that all indexes will be n = 1

2a2 + 2a0 +

∞∑n=1

(n+ 2)(n+ 1)an+2xn + 3

∞∑n=1

nanxn + 2

∞∑n=1

anxn = 0

Now that all three series have an xn factor and all 3 indexes are the same: n = 1, we can combine the three series

into one series:

2a2 + 2a0 +

∞∑n=1

((n+ 2)(n+ 1)an+2 + (3n+ 2)an

)xn = 0

Setting the constant term on the left: 2a2 + 2a0 equal to the constant term on the right: 0 and setting the coefficient

on xn on the left: (n+ 2)(n+ 1)an+2 + (3n+ 2)an equal to the coefficient of xn on the right: 0 gives:

2a2 + 2a0 = 0 (n+ 2)(n+ 1)an+2 + (3n+ 2)an = 0

a2 = −a0 and our recurrence relation is: an+2 =−(3n+ 2)an

(n+ 2)(n+ 1)

Substituting n = 1, 2, ...5 in to the recurrence relation gives:

a3 =−5

6a1 a4 =

2

3a0 a5 =

11

24a1 a6 =

−14

45a0

The solution is

y = a0 + a1x− a0x2 −5

6a1x

3 +2

3a0x

4 +11

24a1x

5 − 14

45a0x

6

Separating this solution into two linearly independent solutions:

3.1. SERIES SOLUTIONS AROUND ORDINARY POINTS 107

y = a0

(1− x2 +

2

3x4 − 14

45x6)

+ a1

(x− 5

6x3 +

11

24x5)

Since a pattern in the terms cannot be found we shall leave the solution as a sixth degree polynomial.

Sometimes a pattern can be found and the solution can be written much more concisely. Consider the following

example:

Example: Solve using power series about x = 0:

(x+ 1)y′′ − 2xy′ − 4y = 0

First lets note that y = e2x is a solution. We will try to find this solution using power series. The series expansion for

e2x that we are trying to obtain is:

e2x =

∞∑n=0

2nxn

n!= 1 + 2x+ 2x2 +

4

3x3 +

2

3x4 +

4

15x5...

We will look for a solution of the form:

y =

∞∑n=0

anxn making y′ =

∞∑n=1

nanxn−1 y′′ =

∞∑n=2

n(n− 1)anxn−2

Substituting y, y′ and y′′ into the differential equation produces:

∞∑n=2

n(n− 1)anxn−1 +

∞∑n=2

n(n− 1)anxn−2 − 2

∞∑n=1

nanxn − 4

∞∑n=0

anxn = 0

Shifting the first and second series so that they each have an xn factor.

∞∑n=1

(n+ 1)(n)an+1xn +

∞∑n=0

(n+ 2)(n+ 1)an+2xn − 2

∞∑n=1

nanxn − 4

∞∑n=0

anxn = 0

Adding the n = 0 term in he second and fourth series so that all series start with an index of n = 1 and condensing

into one series give:

2a2 − 4a0 +

∞∑n=1

((n+ 1)(n)an+1 + (n+ 2)(n+ 1)an+2 − 2nan − 4an

)xn = 0

2a2 − 4a0 = 0 (n+ 1)(n)an+1 + (n+ 2)(n+ 1)an+2 − 2nan − 4an = 0

a2 = 2a0 an+2 =2(n+ 2)an − n(n+ 1)an+1

(n+ 2)(n+ 1)

So our recurrence relation simplifies to

an+2 =2

n+ 1an −

n

n+ 2an+1


a3 = a1 −1

3a2 = a1 −

2

3a0 a4 =

2

3a2 −

1

2a3 =

5

3a0 −

1

2a1

a5 = a1 −4

3a0 a6 =

14

9a0 −

4

5a1

As of now our solution is:

y = a0 + a1x+ 2a0x2 +

(a1 −

2

3a0

)x3 +

(5

3a0 −

1

2a1

)x4 +

(a1 −

4

3a0

)x5 +

(14

9a0 −

12

15a1

)x6

Or

y = a0

(1 + 2x2 − 2

3x3 +

5

3x4 − 4

3x5 +

14

9x6)

+ a1

(x+ x3 − 1

2x4 + x5 − 4

5x6)

This solution does not look like the series expansion for e2x that we are expecting. Noticing the first and third terms

in the first linearly independent solution: 1 + 2x2 match the first and third terms in the series but we seem to be missing

the second term: 2x. I will rewrite the linear term in the second linearly independent solution so that it has a 2x term

in it. I will do this by letting

a1 = 2a0 +K

Making

a3 = a1 −1

3a2 =

4

3a0 +K a4 =

2

3a2 −

1

2a3 =

2

3a0 −

1

2K

a5 = a1 −4

3a0 =

4

15a0 +

4

5K

Now our solution is

y = a0 + (2a0 +K)x+ 2a0x2 +

(4

3a0 +K

)x3 +

(2

3a0 −

1

2K

)x4 +

(4

15a0 +

4

5K

)x5

Separating this into two linearly independent solutions gives

y = a0

(1 + 2x+ 2x2 +

4

3x3 +

2

3x4 +

4

15x5)

+K

(x+ x3 − 1

2x4 +

4

5x5)

Now the first linearly independent solution has the expansion for e2x so our solution simplifies to

y = a0e2x +K

(x+ x3 − 1

2x4 +

4

5x5)

Now that we have one solution we normally would apply the reduction of order algorithm to find the second solution

but the integral it produces is quite unpleasant so we must accept the series expansion as our second linearly independent

solution.

436. Find the first six terms of the solution to the differential equation:

y′′ + xy′ + y = 0

3.1. SERIES SOLUTIONS AROUND ORDINARY POINTS 109


(x+ 1)y′′ − xy′ − y = 0


y′′ + (3x− 1)y′ − y = 0


(2x+ 3)y′′ − (4x+ 8)y′ + 4y = 0


(3x+ 4)y′′ − (3x+ 4)y′ + (2x+ 3)y = 0


(x2 − 2x)y′′ + (2− x2)y′ + (2x− 2)y = 0


y′′ − x

x− 1y′ +

1

x− 1y = 0


4xy′′ + 3y′ + xy = 0


(x2 + 1)y′′ + y = 0


(2x+ 1)y′′ − 2y′ − (2x+ 3)y = 0


(x2 + 2)y′′ + xy′ − y = 0


447.

Find the first 5 terms in y1 in the series expansion about x = 0 for a solution to:

y′′ − 2xy′ − 2y = 0

448.


y′′ + xy′ − (4 + 2x)y = 0

449.

The Hermite equation is an equation of the form:

y′′ − 2xy′ + λy = 0

Show that one of the solution for the given values of λ is a finite polynomial. Find the polynomial then use reduction

of order to find the second linearly independent solution.

A) λ = 4

B) λ = 6

450.

Show that the solutions to

y′′ = y′ + y

is given by

∞∑n=0

Fnxn

n!

where Fn is the nth Fibonacci number

451.

Use the power series for sin(x), cos(x) and ex to derive Euler’s Equation:

eix = cos(x) + i sin(x)

452.

Use your knowledge of power series to evaluate

∞∑n=0

(−1)n

2n+ 1

3.2 Method of Frobenius:

We will now try to find a power series solution to the differential equation:

y′′ + P (x)y′ +Q(x)y = 0

3.2. METHOD OF FROBENIUS: 111

centered at a point where either P (x) or Q(x) is not analytic. To motivate our procedure lets reconsider the Cauchy

Euler equation.

x2d2y

dx2+ ax

dy

dx+ by = 0

After dividing by x2 we get

d2y

dx2+a

x

dy

dx+

b

x2y = 0

So both

P (x) =a

xand Q(x) =

b

x2

are not analytic at x = 0.

The Cauchy Euler equation has a characteristic equation

r(r − 1) + ar + b = 0

which can be created by the following equation

r(r − 1) + xP (x) + x2Q(x) = 0

This equation is quite similar to the characteristic equation we use in the method of Frobenous.

First a definition:

x = x0 is a Singular Point of the above differential equation if either P (x) or Q(x) is not analytic at x = x0. x = x0

is a Regular Singular Point if the following limits exist:

p0 = limx→x0

(x− x0)P (x) q0 = limx→x0

(x− x0)2Q(x)

To find a series solution about a regular singular point we use the Method of Frobenius. In the Method of Frobenius

we look for a series solution of the form:

y =

∞∑n=0

anxn+r making y′ =

∞∑n=0

(n+ r)anxn+r−1 y′′ =

∞∑n=0

(n+ r)(n+ r − 1)anxn+r−2

The values of r that we will use are the roots of the Indicial Equation:

r(r − 1) + p0r + q0 = 0

There will be three cases we will have to consider based on the roots: r1 and r2 to the indical equation. Note: we

always take r1 to be the larger of the two roots. The first case we will consider is the case when the roots differ by a non

integer. That is:

r1 − r2 6∈ Z

An Example:

Find the first 6 terms in the series expansion for the solution to the following differential equation centered at x = 0.

2x(x− 1)y′′ + 3(x− 1)y′ − y = 0


In standard form this equation is:

y′′ +3

2xy′ +

−1

2x(x− 1)y = 0

p0 = limx→0

x3

2x=

3

2q0 = lim

x→0x2

−1

2x(x− 1)= 0

The indical equation is:

r(r − 1) +3

2r = 0 or r

(r +

1

2

)= 0

The roots are r1 = 0 and r2 = −12 .

Inserting:

y =

∞∑n=0

anxn+r making y′ =

∞∑n=0


∞∑n=0

(n+ r)(n+ r − 1)anxn+r−2

into the original differential equation gives:

2

∞∑n=0

(n+r)(n+r−1)anxn+r−2

∞∑n=0

(n+r)(n+r−1)anxn+r−1+3

∞∑n=0

(n+r)anxn+r−3

∞∑n=0

(n+r)anxn+r−1−

∞∑n=0

anxn+r = 0

Shifting the first, third and fifth summation so that all series will have an xn+r−1 factor:

2

∞∑n=1

(n+ r − 1)(n+ r − 2)an−1xn+r−1 − 2

∞∑n=0

(n+ r)(n+ r − 1)anxn+r−1+

3

∞∑n=1

(n+ r − 1)an−1xn+r−1 − 3

∞∑n=0

(n+ r)anxn+r−1 −

∞∑n=1

an−1xn+r−1 = 0

Adding the n = 0 terms in the second and forth series and condensing into a single series gives:

(−2r(r−1)a0−3ra0)xr−1+∞∑n=1

(2(n+r−1)(n+r−2)an−1−2(n+r)(n+r−1)an+3(n+r−1)an−1−3(n+r)an−an−1

)xn+r−1 = 0

Setting the coefficients of the polynomial on the left hand side equal to zero: the coefficients of the polynomial on the

right hand side. We see that the coefficient of xr−1 is zero at both r1 = 0 and r2 = −12 so a0 is a free is not necessarily

zero. We also get our recurrence relation from setting the coefficient of xn+r−1 to zero.

2(n+ r − 1)(n+ r − 2)an−1 − 2(n+ r)(n+ r − 1)an + 3(n+ r − 1)an−1 − 3(n+ r)an − an−1 = 0

Solving for an

an = an−11− 3(n+ r − 1)− 2(n+ r − 1)(n+ r − 2)

−2(n+ r)(n+ r − 1)− 3(n+ r)

We will find two linearly independent solutions by substituting our two values of r into the recurrence relation. For

r1 = 0


an = an−11− 3(n− 1)− 2(n− 1)(n− 2)

−2n(n− 1)− 3n

Which simplifies to:

an =

(2n− 3

2n+ 1

)an−1

To obtain the first 6 terms in the solution we will find the first 3 terms in y1 using the above recurrence relation

created by using r1 = 0 and the first 3 terms in y2 using the using a recurrence relation created by using r2 = −12 . The

first 3 terms in this recurrence relation will have coefficients a0, a1 and a2. They are:

a1 =−1

3a0 a2 =

1

5a1 =

1

5

−1

3a0 =

−1

15a0

Using r1 = 0 the first 3 terms in y1 are:

y1 = a0 + a1x+ a2x2

y1 = a0

(1− x

3− x2

15

)Since this is a second order differential equation we would expect the form of the solution to be:

y = C1y1 + C2y2

In our answer C1 is the constant a0.

To find y2 we insert r2 = −12 into the recurrence relation:

an =1− 3(n− 3

2 )− 2(n− 32 )(n− 5

2 )

−2(n− 12 )(n− 3

2 )− 3(n− 12 )

After a bit of simplifying we arrive at our recurrence relation:

an =

(2n− 1

n

)an−1

The first three terms in this recurrence relation will have coefficients a0, a1 and a2. They are:

a1 = a0 a2 =3

2a1 =

3

2a0

y2 = a0x−12 + a1x

12 + a2x

32

y2 = a0x−12 + a0x

12 +

3

2a0x

32

y2 = a0

(x−12 + x

12 +

3

2x

32

)An Example:

Solve the following differential equation by method of Forbenious.


4xy′′ + 2y′ + y = 0

In standard form our equation is:

y′′ +1

2xy′ +

1

4xy = 0

p0 = limx→0

x1

2x=

1

2

q0 = limx→0

x21

4x= 0

Our indical equation becomes:

r(r − 1) +1

2r = 0

Which has roots r1 = 12 and r2 = 0

Assume a solution to the differential equation of the form:

y =

∞∑n=0

anxn+r

Therefore:

y′ =

∞∑n=0


∞∑n=0

(n+ r)(n+ r − 1)anxn+r−2

Inserting the series for y, y’ and y” into our differential equation yields the following:

∞∑n=0

4(n+ r)(n+ r − 1)anxn+r−1 +

∞∑n=0

2(n+ r)anxn+r−1 +

∞∑n=0

anxn+r = 0

Shifting the last series on the LHS of the above equation yields:

∞∑n=0

4(n+ r)(n+ r − 1)anxn+r−1 +

∞∑n=0

2(n+ r)anxn+r−1 +

∞∑n=1

an−1xn+r−1 = 0

Adding the n = 0 terms in the first two series and then combining the rest into one series yields:

(4r(r − 1)a0 + 2ra0)xr−1 +

∞∑n=1

[4(n+ r)(n+ r − 1)an + 2(n+ r)an + an−1]xn+r−1 = 0

This gives the recurrence relation:

4(n+ r)(n+ r − 1)an + 2(n+ r)an + an−1 = 0


an =−an−1

(2n+ 2r)(2n+ 2r − 1)


For r1 = 12 our recurrence relation becomes:

an =−an−1

(2n+ 1)(2n)

Substituting values of n into our recurrence relation yields:

a1 =−a03 · 2

a2 =−a15 · 4

=a0

5 · 4 · 3 · 2

a3 =−a27 · 6

=−a0

7 · 6 · 5 · 4 · 3 · 2The solution to this recurrence relation is:

an =(−1)n · a0(2n+ 1)!

Making the solution:

y1 =

∞∑n=0

(−1)n

(2n+ 1)!xn+1/2 = sin(

√x)

For r2 = 0 our recurrence relation becomes:

an =−an−1

(2n)(2n− 1)

Substituting values of n into our recurrence relation yields:

a1 =−a02 · 1

a2 =−a14 · 3

=a0

4 · 3 · 2

a3 =−a26 · 5

=−a0

6 · 5 · 4 · 3 · 2The solution to this recurrence relation is:

an =(−1)n · a0

(2n)!

Making the solution:

y2 =

∞∑n=0

(−1)n

(2n)!xn = cos(

√x)

453.

Determain if x = 0 is a regular or iregular singular point. If x = 0 is a regular singular point find the indicial equation

and its roots.

y′′ +

(e2x − 2x− 1

xex − x

)y′ + +

(sin(4x)

(x+ sin(x))2

)y = 0

454.


and its roots.


y′′ + x sin

(2

x

)y′ + +

(arctan(x)

x

)y = 0

455.


and its roots.

y′′ +

(1

x− 1

sin(x)

)y′ + +

(arcsin(x)

ln(x+ 1)

)y = 0

456.

Solve by method of Frobenious. You should be able to recognise a pattern and identify the power series as a known

function.

2xy′′ + (x+ 1)y′ + y = 0

457.

Solve by method of Frobenious. You should be able to recognise a pattern and identify the power series as a known

function.

y′′ +2

xy′ + y = 0

458.


x2y′′ − xy′ + (1− x)y = 0

459.


(x2 − 2x)y′′ + (2− x2)y′ + (2x− 2)y = 0

460.


(6x2 + 2x)y′′ + (x+ 1)y′ − y = 0

461.


xy′′ + y′ + xy = 0

462.


xy′′ + y′ − 4y = 0

463.


3.3. THE GAMMA FUNCTION 117

xy′′ + 2y′ + xy = 0

464. Show that one solution to

x2y′′ = (a2 − a+ bx)y

is

y = xa(

1 +bx

1! · 2a+

(bx)2

2! · (2a)(2a+ 1)+

(bx)3

3! · (2a)(2a+ 1)(2a+ 2)+ ...

)

3.3 The Gamma Function

As you have seen many of our series solutions involve factorials. The problem factorials is that they are only defined for non

negative integers. The gamma function extend the idea of factorials to all positive real numbers. The Gamma Function

is given by the improper integral:

Γ(x) =

∫ ∞0

e−ttx−1dt x > 0

The first important property that we will prove of the gamma function shows its similarities to the factorial function:

Thm:

Γ(x+ 1) = xΓ(x)

Proof:

Γ(x+ 1) =

∫ ∞0

e−ttxdt

Using integration by parts

u = tx dv = e−tdt du = xtx−1dt v = −e−t

Γ(x+ 1) = limb→∞

−txe−t∣∣∣∣b0

+ x

∫ b

0

tx−1e−tdt

The first term is zero at both zero and infinity so we hvae

Γ(x+ 1) = x

∫ ∞0

e−ttx−1dt = xΓ(x)

This theorem demonstrates the relationship with the factorial function. The next result is a direct corollary to the

above theorem and is the most well known property of the gamma function.

If n is a positive integer

Γ(n) = (n− 1)!

Example:


Calculate Γ( 12 )

Γ

(1

2

)=

∫ ∞0

e−tt−12 dt =

∫ ∞0

e−t√tdt

Substituting

u =√t 2du =

dt√t

when t = 0 u = 0 as t→∞ u→∞

Γ

(1

2

)= 2

∫ ∞0

e−u2

du

Although we cannot find an antiderivative for e−u2

we can evaluate this integral with a clever trick

Γ

(1

2

)= 2

∫ ∞0

e−u2

du = 2

∫ ∞0

e−v2

dv

So (Γ

(1

2

))2

=

(2

∫ ∞0

e−u2

du

)·(

2

∫ ∞0

e−v2

dv

)= 4

∫ ∞0

∫ ∞0

e−u2−v2dudv

Converting to polar coordinates (Γ

(1

2

))2

= 4

∫ π2

0

∫ ∞0

e−r2

rdrdθ

(Γ

(1

2

))2

= −2

∫ π2

0

limb→∞

e−r2

∣∣∣∣b0

dθ = 2

∫ π2

0

dθ = π

So

Γ

(1

2

)=√π

465.

Find

Γ

(3

2

)466.

Show

limx→0+

Γ(x) =∞

467.

Use the principle of mathematical induction to prove:

Γ

(n+

1

2

)=

(2n)!√π

4nn!n = 0, 1, 2, ...

468.

3.4. BESSEL’S EQUATION 119

Show that the definition of the Gamma function is equivalent to Euler’s original definition:

Γ(x) =

∫ 1

0

(ln

(1

t

))x−1dt

469.

Another important function related to the Gamma function is Euler’s Psi function: the derivative of the logarithm of

the Gamma function.

ψ(x) =d

dxln

(Γ(x)

)=

Γ′(x)

Γ(x)

Show the following property of the Psi function

ψ(x+ 1) =1

x+ ψ(x)

Use the above results to also show for positive integers n

ψ(n+ 1) = ψ(1) +n∑k=1

1

k

3.4 Bessel’s Equation

We will now consider a differential equation of great importance in applied mathematics. The Bessel Equation of order

v is:

x2y′′ + xy′ + (x2 − v2)y = 0

or in standard form:

y′′ +1

xy′ +

(1− v2

x2

)y = 0

We see that x = 0 is a singular point of the Bessel Equation. Since both

P0 = limx→0

x1

x= 1 And Q0 = lim

x→0x2(

1− v2

x2

)= −v2

exist x = 0 is a regular singular point and our indical equation is

r(r − 1) + r − v2 = 0 Or r2 − v2 = 0

Which has roots

r = ±v

If the difference between the roots: 2n is not an integer then the method of Frobenius gives two linearly independent

solutions:

Jv =

∞∑n=0

(−1)nx2n+v

22n+vn!Γ(1 + v + n)

and


J−v =

∞∑n=0

(−1)nx2n−v

22n−vn!Γ(1− v + n)

An Example:

Solve the Bessel equation of order 12

x2y′′ + xy′ +

(x2 − 1

4

)= 0

In standard form we have

y′′ +1

xy′ +

(1− 1

4x2

)y = 0

We see that x = 0 is a singular point. To show it is a regular singular point we compute:

P0 = limx→0

x1

x= 1 Q0 = lim

x→0x2(

1− 1

4x2

)=−1

4

Since both limits exist we see that x = 0 is a regular singular point and we can use the method of Frobenius to find a

series solution about x = 0. The indical equation and roots are:

r(r − 1) + r − 1

4= 0 r = ±1

2

y =

∞∑n=0

anxn+r y′ =

∞∑n=0

(n+ r)anxn+r−1 y =

∞∑n=0

(n+ r)(n+ r − 1)anxn+r−2

Inserting the series for y, y′ and y′′ into the original differential equation gives:

∞∑n=0

(n+ r)(n+ r − 1)anxn+r +

∞∑n=0

(n+ r)anxn+r +

∞∑n=0

anxn+r+2 −

∞∑n=0

1

4anx

n+r = 0

Shifting the third series so that it too has an exponent of n+ r

∞∑n=0

(n+ r)(n+ r − 1)anxn+r +

∞∑n=0

(n+ r)anxn+r +

∞∑n=2

an−2xn+r −

∞∑n=0

1

4anx

n+r = 0

Adding the first and second terms in the first, second and fourth series will produce four series all with an index

starting at n = 2 that can be simplified into a single series.

(r(r− 1) + r− 1

4

)a0x

r +

((1 + r)r+ (1 + r)− 1

4

)a1x

1+r +

∞∑n=2

(((n+ r)(n+ r− 1) + (n+ r)− 1

4

)an + an−2

)xn+r = 0

The coefficient of a0xr is identical to the indical equation and is zero at both values of r making a0 not zero whereas

the coefficient of a1x1+r is not zero at both values of r making a1 = 0. Our recurrence relation is:(

(n+ r)(n+ r − 1) + (n+ r)− 1

4

)an + an−2 = 0

3.4. BESSEL’S EQUATION 121

an =−an−2

(n+ r)2 − 14

For the larger root r = 12 our recurrence relation simplifies to:

an =−an−2n(n+ 1)

Producing a sequence of terms in an

a2 =−a02 · 3

=−a03!

a3 =−a14 · 3

= 0

Since a5 depends on a3 it too must be zero. In fact an = 0 for all odd values of n

a4 =−a25 · 4

=a0

5 · 4 · 3 · 2=a05!

a6 =−a47 · 6

=−a07!

Recognizing the pattern we see

a2n =(−1)na0(2n+ 1)!

And the first solution is

y1 =

∞∑n=0

(−1)na0(2n+ 1)!

x2n+12 =

∞∑n=0

(−1)na0(2n+ 1)!

x2n+1

√x

=1√x

∞∑n=0

(−1)na0(2n+ 1)!

x2n+1

Recognizing the series gives:

y1 = J 12(x) = a0

cos(x)√x

Since r1 − r2 is an integer the second linearly independent solution would be quite difficult to find using series, but

we were able to find a closed form expression for y1 = cos(x)√x

so we can use the method of reduction of order form chapter

2 to find the second linearly independent solution.

y′′ +1

xy′ +

(1− 1

4x2

)y = 0

v =

∫e−

∫1xdx

cos2(x)x

dx =

∫sec2(x)dx = tan(x)

Our second linearly independent solution is

y2 = y1v =cos(x)√

xtan(x)

y2 = J−12

(x) =sin(x)√

x

And the homogenous solution is:

y = C1cos(x)√

x+ C2

sin(x)√x


A few important properties of the solutions to the Bessel equation are:

d

dx

(xvJv(x)

)= xvJv−1(x)

d

dx

(x−vJv(x)

)= −x−vJv+1(x)

Jv+1(x) =2v

xJv(x)− Jv−1(x) Jv+1(x) = Jv−1(x)− 2J ′v(x)

470.

The Bessel functions of order n + 12 for integer n are related to the spherical Bessel functions. Use one of the above

properties of the Bessel function and the results of the example to calculate J 32

and J 52

471.

Show the orthogonality of the Bessel functions by showing∫ 1

0

xJ−12J 1

2dx = 0

472.

Show that the substitution z =√xy transforms the Bessel equation to normal form:

z′′ +

(1 +

1− 4v2

4x2

)z = 0

473.

The parametric Bessel equation is:

x2y′′ + xy′ + (λ2x2 − v2)y = 0 x > 0

Show

y = C1Jv(λx) + C2J−v(λx) v /∈ Z

is the solution

474.

Use the results from the previous problem and to find the solution to

x2y′′ + xy′ +

(25x2 − 1

4

)y = 0 x > 0

475.

Use ideas from the previous two problems and the fact ı2x2 = −x2 to solve

x2y′′ + xy′ −(x2 +

1

4

)y = 0 x > 0

476.

Use the change of variable y = x−12 v(x) to solve

x2y′′ + 2xy′ + λ2x2y = 0

Chapter 4

Laplace Transform

4.1 Calculating Laplace and Inverse Laplace Transforms

The Laplace Transform of a function f(t) will be a function of s given by the improper integral:

L(f) =

∫ ∞0

f(t)e−stdt

provided the integral converges. Functions that outgrow ekt, for constant k, do not have a Laplace Transform since

the integral diverges. So f(t) = etn

will not have a Laplace Transform for n > 1. Let us now try to find the Laplace

Transform of f(t) = eat.

L(eat) =

∫ ∞0

eate−stdt =

∫ ∞0

e(a−s)tdt

= limb→∞

e(a−s)t

a− s

∣∣∣∣b0

= limb→∞

e(a−s)b

a− s− 1

a− s=

1

s− aFor s > a.

One key property of the Laplace Transform is that it is a Linear Transformation, meaning:

L(af(t) + bg(t)) = aL(f(t)) + bL(g(t))

To derive the Laplace Transform of sin(t) and cos(t) we will use the linearity of the Laplace Transform and: Euler’s

Identity

eıat = cos(at) + ı sin(at)

Making

L(eıat) = L(cos(at)) + ıL(sin(at))

L(eıat) =

∫ ∞0

eıate−stdt =

∫ ∞0

e(ıa−s)tdt

123

124 CHAPTER 4. LAPLACE TRANSFORM

= limb→∞

e(ıa−s)t

ıa− s

∣∣∣∣b0

= limb→∞

e(ıa−s)b

ıa− s− 1

ıa− s=

1

s− ıa

Multiplying the result by the complex conjugant

1

s− ıa· s+ ıa

s+ ıa=

s+ ıa

s2 + a2

Separating into real and imaginary parts gives

L(eıat) = L(cos(at)) + ıL(sin(at)) =s

s2 + a2+ ı

a

s2 + a2

So

L(cos(at)) =s

s2 + a2L(sin(at)) =

a

s2 + a2

How would one compute L(tnf(t))?

L(f(t)) =

∫ ∞0

f(t)e−stdt

Differentiating with respect to s

d

dsL(f(t)) =

∫ ∞0

d

dsf(t)e−stdt =

∫ ∞0

−tf(t)e−stdt = L(−tf(t))

Differentiating again with respect to s

d2

d2sL(f(t)) =

∫ ∞0

d

ds− tf(t)e−stdt =

∫ ∞0

t2f(t)e−stdt = L(t2f(t))

In general

dn

dnsL(f(t)) = L((−1)nttf(t))

Due to linearity we can factor out the (−1)n and arrive at

L(ttf(t)) = (−1)ndn

dnsL(f(t))

Here is a basic table of laplace transforms:

4.1. CALCULATING LAPLACE AND INVERSE LAPLACE TRANSFORMS 125

f(t) F(s) = L(f(t))eat 1

s−asin(at) a

s2+a2

cos(at) ss2+a2

tn n ∈ N n!sn+1

eattn n ∈ N n!(s−a)n+1

eat − ebt a−b(s−a)(s−b)

eat sin(bt) b(s−a)2+b2

eat cos(bt) s−a(s−a)2+b2

t sin(at) 2as(s2+a2)2

t cos(at) s2−a2(s2+a2)2

sinh(at) as2−a2

cosh(at) ss2−a2

tr r ∈ R r > −1 Γ(r+1)sr+1√

t√π

2s32

1√t

√π√s

tn−12 n ∈ N 1·3·5...(2n−1)

√π

2nsn+12

sin(at) cosh(at)− cos(at) sinh(at) 4a3

s4+4aa

sin(at) sinh(at) 2a2ss4+4a4

sinh(at)− sin(at) 2a3

s4−a4

cosh(at)− cos(at) 2a2ss4−a4

sin(at)− at cos(at) 2a3

(s2+a2)2

sin(at) + at cos(at) 2as2

(s2+a2)2

t2 sin(at) −2a(a2−3s2)(s2+a2)3

t2 cos(at) 2s(s2−3a2)(s2+a2)3

eatg(t) G(s− a) G(s) = L(g(t))

tng(t) (−1)n dn

dsn

(L(g(t))

)Notice there is not a Laplace Transform for any function that outgrows f(x) = eat. This is because the improper

integral from the definition will diverge.

An Example: Use the table to find the Laplace Transform of:

f(t) = 4t sin(3t) + t2e6t

Using the table we see that the Laplace Transform of 4t sin(3t) is 4 2·3s(s2+32)2 and the Laplace Transform of t2e6t is

2(s−6)3 . So:

L(f(t)) =24s

(s2 + 9)2+

2

(s− 6)3


An Example: Find the inverse Laplace Transform of:

F (s) =s3 − 8s2 + 23s− 7

(s2 − 4s+ 20)(s− 3)2

Looking at the table of Laplace Transforms we notice that there is not an inverse Laplace Transform for a function

with a repeated linear factor multiplied by a quadratic factor so we must use partial fraction decomposition to separate

this function into a sum of functions that do have an inverse Laplace Transform. By partial fractions

F (s) =s− 3

s2 − 4s+ 20+

1

(s− 3)2

Looking at the table we notice all terms with a quadratic denominator are written in the completed square form.

So we will complete the square on the quadratic denominator and do nothing to the second term since we can find the

inverse Laplace Transform it as it is:

F (s) =s− 3

(s− 2)2 + 16+

1

(s− 3)2

To find the inverse Laplace of a term with a denominator: (s− 2)2 + 42 we need to have either an s− 2 or a 4 in the

numerator so I will rewrite the s− 3 term as s− 2− 1 and separate the fraction into two fractions:

F (s) =s− 2

(s− 2)2 + 42− 1

(s− 2)2 + 42+

1

(s− 3)2

I will now multiply the second term by 44 to make it fit the table:

F (s) =s− 2

(s− 2)2 + 42− 1

4

(4

(s− 2)2 + 42

)+

1

(s− 3)2

Using the table to find the inverse Laplace Transform gives:

f(x) = e2t cos(4t)− 1

4e2t sin(4t) + te3t

477. Find the Laplace Transform of:

f(x) = t3 − 3t cos(4t)


f(t) = t2 sinh(3t) + e2t sin(3t)


f(t) = t4e5t − sin(2t) sinh(2t)

480.

Use trig identities and the table in this section to find the Laplace transform of:

f(t) = sin(At+B) g(t) = cos(At+B)

481.

Use trig identities and the table in this section to find the Laplace transform of:


f(t) = sin2(At) g(t) = cos2(At)

482. Use the definition to find the Laplace Transform of:

f(t) =

1 t < 2

t 2 ≤ t < 6

t2 t ≥ 6


f(t) = t32 − t 5

2

484. Find the Inverse Laplace Transform of:

F (s) =3s

s2 + 4+

12

s2 − 10s+ 34


F (s) =500 + 40s

(s2 + 25)2


F (s) =4s3 + 4s2 − 27s− 18

s4 − 3s3 + 2s2


F (s) =3s4 + s3 − 5s2 − 10s− 25

s5 + 2s4 + 5s3


F (s) =9s2 − 30s+ 49

(s− 3)2(s2 + 1)


F (s) =3s3 − 6s2 + 39s+ 54

(s2 − 4s+ 13)(s2 + 9)

490.

Find the Inverse Laplace Transform of:

F (s) =

∞∑n=1

1

sn

You should be able to recognize the power series.

491.


F (s) =

∞∑n=0

(−1)n

s2n+2



492.


F (s) =

∞∑n=0

(−1)n(2n)!

s2n+2


493.

Use the series expansion for ln(1 + t) to find

L(ln(1 + t))

494.

Show L(et2

) does not exist.

495.

Show L(

1t2

)does not exist.

496.

Use

cosh(t) =et + e−t

2sinh(t) =

et − e−t

2

to derive formulas for

L{cosh(at+ b)} L{sinh(at+ b)}

and

L{cosh2(t)} L{sinh2(t)}

Then find the inverse Laplace transform of

L{cosh2(t)} − L{sinh2(t)}

What can you conclude about the value of

cosh2(t)− sinh2(t)

497.

Calculate:

L(t2 · y′)

498.

Calculate:

L(teat cos(t))

499.

Calculate:


L(teat sin(t))

500.

Let

y =

n∑i=0

(1 + t)n

and

F (s) = L(y)

Show that the coefficients of both 1sn and 1

sn+1 in F (s) are both n!

501.

Use the following property:

L(f(t)

t

)=

∫ ∞s

F (t)dt

to find

L(

sinh(t)

t

)502.

Use a power series expansion to show:

L−1(e−

1s

√s

)=

cos(2√t)√

πt

503.

Recall the Gamma function from the last chapter is given by

Γ(t) =

∫ ∞0

ut−1e−udu

Show that

L(tr) =Γ(r + 1)

sr+1r > −1

504.

Show

L{y · y′} =1

2

(sL{y2} − y2(0)

)505.

Another interesting inverse Laplace Transform is:

L−1(F (s)

)=−1

xL−1

(F ′(s)

)Use this transform to calculate the inverse Laplace Transform of


F (s) = arctan

(1

s

)506.

Use the previous problem to find the inverse Laplace Transform of

F (s) = ln(1 + s2)

In the special case where you are trying to find the inverse Laplace Transform of a rational function

F (s) =P (s)

Q(s)

with the degree of P less than the degree of Q, if Q has n distinct, possibly complex, roots ri, the inverse Laplace

Transform is given by

L−1(F (s)

)=

n∑i=1

P (ri)

Q′(ri)

507.

Use the previous above results to find the inverse Laplace Transform of

F (s) =(s− 1)(s− 3)(s− 5)

(s− 2)(s− 4)(s− 6)

508.

Use the previous above results to find the inverse Laplace Transform of

F (s) =s2

(s2 + 1)(s− 5)

4.2 Solving Initial Value Problems

What is the use of Laplace Transforms in differential equations? The answer comes from the next derivation of the

Laplace Transform of y′(t):

L(y′(t)) =

∫ ∞0

y′(t)e−stdt using integration by parts

u = e−st dv = y′(t)dt du = −se−stdt v = y(t)

L(y′(t)) =

∫ ∞0

y′(t)e−stdt = limb→∞

e−sty(t)

∣∣∣∣b0

−∫ b

0

−sy(t)e−stdt

= limb→∞

e−sby(b)− y(0) + sL(y(t))

4.2. SOLVING INITIAL VALUE PROBLEMS 131

= sL(y(t))− y(0)

In a similar way we can find the Laplace Transform of y′′, y′′′ ...

L(y′(t)) = sL(y(t))− y(0)

L(y′′(t)) = s2L(y(t))− sy(0)− y′(0)

L(y′′′(t)) = s3L(y(t))− s2y(0)− sy′(0)− y′′(0)

In general

L(y(n)(t)) = snL(y(t))− sn−1y(0)− sn−2y′(0)− sn−3y′′(0)− ...− y(n−1)(0)

From the above equations it should be clear that you must have initial conditions evaluated at x = 0 to use Laplace

Transforms to solve a differential equation.

The procedure to solve a differential equation using Laplace Transforms is: First, using the above tables take the

Laplace Transform of both sides of the differential equation; Second, Solve for L(y(t)) as a function of s; Third, using the

above tables calculate the inverse Laplace of both sides of the equation.

A Example:

Solve:

y′′ + 4y = 20e t cos(t) y(0) = 7 y′(0) = 10

Taking the Laplace transform of both sides gives:

s2L(y)− 7s− 10 + 4L(y) = 20s− 1

(s− 1)2 + 1

Solving for L(y):

L(y)(s2 + 4) = 7s+ 10 + 20s− 1

(s− 1)2 + 1

L(y) =7s+ 10

s2 + 4+ 20

s− 1

((s− 1)2 + 1)(s2 + 4)

Applying Partial Fractions to the second fraction on the right hand side and adding the results to the first fraction

yields

L(y) =4s− 2

s2 − 2s+ 2+

3s+ 4

s2 + 4

Completing the square on the first fraction

L(y) =4s− 2

(s− 1)2 + 1+

3s+ 4

s2 + 4


Manipulating the right hand side into our table of Laplace Transforms

L(y) = 4s− 1

(s− 1)2 + 1+ 2

1

(s− 1)2 + 1+ 3

s

s2 + 4+ 2

2

s2 + 4

Taking the inverse Laplace of each side

y = 4et cos(t) + 2et sin(t) + 3 cos(2t) + 2 sin(2t)

An Example: Solve:

y′′ − 2y′ + y = 4t2e2t y(0) = 1 y′(0) = 3

Taking the Laplace transform of both sides gives:

s2L(y)− s− 3− 2

(sL(y)− 1

)+ L(y) = 4

(2

(s− 1)3

)Solving for L(y):

L(y)(s2 − 2s+ 1) = 4

(2

(s− 1)3

)+ s+ 1

Solving for L(y)

L(y) = 4

(2

(s− 1)5

)+

s+ 1

(s− 1)2

Manipulating the right hand side to fit the table

L(y) = 4

(2

(s− 1)5

)+

s+ 1

(s− 1)2=

8

4!

(4!

(s− 1)5

)+

(s− 1) + 2

(s− 1)2

L(y) =1

3

(4!

(s− 1)5

)+

1

s− 1+ 2

(1

(s− 1)2

)Calculating the inverse Laplace Transform

y =1

3t4et + 2tet + et

The Laplace Transform can also be used to solve linear differential equations without constant coefficients. To do so

we would like to have nice formula for:

L(f(t)y) L(f(t)y′) L(f(t)y′′)

Unfortunately, these formulas either do not exist or are well beyond the scope of this book, so instead we will find

formulas for the specific case f(t) = t. Lets now calculate L(ty)

We know

L(tf(t)) = − d

dsL(f(t))

So


L(ty) = − d

dsL(y)

And

L(ty′) = − d

dsL(y′) = − d

ds

(sL(y)− y(0)

)= −L(y)− s d

dsL(y)

And

L(ty′′) = − d

dsL(y′′) = − d

ds

(s2L(y)− sy(0)− y′(0)

)= −2sL(y)− s2 d

dsL(y) + y(0)

Using the notation L′(y) = ddsL(y) our formulas become:

L(ty) = −L′(y) L(ty′) = −L(y)− sL′(y) L(ty′′) = −2sL(y)− s2L′(y) + y(0)

A Example:

Solve:

ty′′ + 2(t− 1)y′ − 2y = 0 y(0) = 0

Taking the Laplace Transform of each side gives:

−2sL(y)− s2L′(y) + y(0)− 2L(y)− 2sL′(y)− sL(y) + y(0)− 2L(y) = 0

−2sL(y)− s2L′(y)− 2L(y)− 2sL(y)− sL′(y)− 2L(y) = 0


(s2 + 2s)L′(y) + (4s+ 4)L(y) = 0

This equation is now separable

L′(y)

L(y)= − 4s+ 4

s(s+ 2)

Applying partial fraction to the right hand side gives

L′(y)

L(y)= −2

s− 2

s+ 2

Integrating gives

ln

(L(y)

)= −2 ln(s)− 2 ln(s+ 2) + C

Treating C as ln(K) and using some log rules we get:

ln(s2(s+ 2)2L(y)) = ln(K)


So

L(y) =K

s2(s+ 2)2

Applying partial fraction to the right hand side gives

L(y) = K

(−1

4· 1

s+

1

4· 1

s2+

1

4· 1

s+ 2+

1

4· 1

(s+ 2)2

)Absorbing the 1

4 into the constant and calculating the inverse Laplace Transform gives:

y = K

(− 1 + t+ e−2t + te−2t

)We see the first initial condition y(0) = 0 gives us no information about the constant K but our second initial condition

y(1) = 2e−2 does. Applying this condition gives

2e−2 = K

(− 1 + 1 + e−2 + e−2

)So K = 1 and the solution is

y =

(− 1 + t+ e−2t + te−2t

)

509. Solve:

y′ + y = et y(0) = 1

510. Solve:

y′ + 2y = sin(2t) y(0) = 1

511. Solve:

y′′ + 6y′ + 5y = 12et y(0) = −1 y′(0) = 7

512. Solve:

y′′ − 2y′ + y = 6tet y(0) = 1 y′(0) = 2

513. Solve:

y′′ − 2y′ − 3y = 8− 16t− 12t2 y(0) = 2 y′(0) = 2

514. Solve:

y′′ + y′ − 2y = 3et y(0) = 5 y′(0) = 8

515. Solve:


y′′ + y = et(3 cos(t)− sin(t)) y(0) = 3 y′(0) = 5

516. Solve:

y′′ − 2y′ + y = (t+ 2)e2t y(0) = 1 y′(0) = 0

517. Solve:

y′′ − 4y′ + 5y = te2t y(0) = 1 y′(0) = 4

518. Solve:

y′′ − 3y′ + 2y = (1− 2t)et y(0) = 3 y′(0) = 5

519. Solve:

y′′ − 4y′ + 3y = −et(3 sin(t) + cos(t)) y(0) = 1 y′(0) = 4

520. Solve:

y′′ + y = 2 sin(t) + 4t cos(t) y(0) = 3 y′(0) = 2

521.

Solve:

y′′ + 4y = sin(2t) y(0) = 10 y′(0) = 0

522. Solve:

y′′ − 7y′ + 10y = 9 cos(t) + 7 sin(t) y(0) = 5 y′(0) = −4

523. Solve:

y′′ + 4y = 4t2 − 4t+ 10 y(0) = 0 y′(0) = 3

524. Solve:

y′′ + 4y′ + 5y = e2t sin(t) y(0) = 1 y′(0) = −6

525. Solve:

y′′ − 2y′ + y = 18tet y(0) = 1 y′(0) = 3

526. Solve:

y′′ − 5y′ + 6y = et(4 sin(t)− 2 cos(t)) y(0) = 3 y′(0) = 3

527. Solve:


y′′ + y = e2t(4t3 + 12t2 + 6t) y(0) = 1 y′(0) = 1

528. Solve:

y′′ − y′ − 2y = e2t(t2 + 6t+ 2) y(0) = 2 y′(0) = 1

529. Solve:

y′′ − 2y′ + 10y = −12 cos(3t) + t sin(t) + sin(t)− 6t cos(t) y(0) = 1 y′(0) = 13

530. Solve:

y′′′ − y′′ + y′ − y = −2et y(0) = 2 y′(0) = 4 y′′(0) = 4

531. Solve:

y′′′ − 6y′′ + 11y′ − 6y = et + e2t + e3t y(0) = 0 y′(0) = 0 y′′(0) = 0

532. Solve:

y′′′ + y′′ + 3y′ − 5y = 16e−t y(0) = 0 y′(0) = 2 y′′(0) = −4

533. Solve:

y′′′ − 2y′′ + 4y′ − 8y = 8e2t cos(2t)− 16e2t sin(2t) y(0) = 4 y′(0) = 8 y′′(0) = 0

534. Solve:

y′′′ − 8y′′ + 21y′ − 18y = 2e3t y(0) = 2 y′(0) = 5 y′′(0) = 15

535. Solve:

y′′ − y =

∞∑n=0

ent y(0) = 0 y′(0) = 0

536. Solve:

ty′′ − y′ = t2 y(0) = 0 y(1) = 2

537. Solve:

ty′′ + y′ − 2ty = et y(0) = 1 y(1) = 2

538. Solve:

ty′′ + 2ty′ + 2y = 0 y(0) = 0 y(1) = 2

539. Solve:

ty′′ + (t+ 2)y′ + y = −1 y(0) = 0 y(1) = 2


540.

Find the Laplace Transform of the solution to Bessel order 0.

ty′′ + y′ + ty = 0

541.

Use the formula

L(y′′(t)) = s2L(y(t))− sy(0)− y′(0)

To find the Laplace transform of

y(t) = sin(at)

542.

Convert the Cauchy Euler equation to the variable t and then use Laplace transforms to solve it.

x2d2y

dx2+ x

dy

dx− 4y = sin(ln(x)) y(1) = 1 y′(1) = 3

543.

Derive a formula for L(t2y′′(t)) and use it along with the formula that we created for L(ty′(t)) to try to solve the

Cauchy Euler Equation

at2y′′ + bty′ + cy = 0 y(0) = k1 y′(0) = k2

544.

Solve

y′ + 2y +

∫ ∞0

y(u)du = et y(0) = 1 y′(0) = 0

Hint

L{∫ ∞

0

y(u)du

}=L{y}s

545.

Use

L{∫ ∞

0

y(u)du

}=L{y}s

to find the inverse Laplace Transform of 1s(s−1)


4.3 Unit Step Function

So far we have learned how to use Laplace Transforms to solve differential equations involving continuous functions. We

will now learn how to solve differential equations involving piecewise functions. First a definition.

The Unit Step Function is defined as:

uc(t) =

{0 t < c

1 t ≥ c

To express a the piecewise function:

g(t) =

f1 t < c1

f2 c1 < t ≤ c2f3 c2 < t

we write it in terms of the unit step function

g(t) = f1 + uc1(t)

(f2 − f1

)+ uc2(t)

(f3 − f2

)It can be shown that:

L(uc(t)f()) = e−csL(f(t+ c)) L−1(e−csF (s)) = uc(t)f(t− c)

An Example: Solve:

y′′ + y = g(t) y(0) = 0 y′(0) = 1

g(t) =

{t t < 2

4 2 < t

Start by writing the equation using the unit step function:

y′′ + y = t+ (4− t)u2(t)

Take the Laplace Transform of both sides:

s2L(y)− 1 + L(y) =1

s2+ e−2sL(4− (t+ 2))

L(y)(s2 + 1) =1

s2+ e−2s

(2

s− 1

s2

)+ 1

L(y) =1

s2(s2 + 1)+

1

(s2 + 1)+ e−2s

(2

s(s2 + 1)− 1

s2(s2 + 1)

)Applying partial fraction to the first term and on the terms multiplied by e−2s gives:

L(y) =1

s2− 1

s2 + 1+

1

(s2 + 1)+ e−2s

(−2s

s2 + 1+

1

s2 + 1+

2

s− 1

s2

)

L(y) =1

s2+ e−2s

(−2s

s2 + 1+

1

s2 + 1+

2

s− 1

s2

)

4.3. UNIT STEP FUNCTION 139

Calculating the inverse Laplace Transform:

y = t+ u2(t)

(− 2 cos(t) + sin(t) + 2− t

)∣∣∣∣t−2

y = t+ u2(t)

(− 2 cos(t− 2) + sin(t− 2) + 2− (t− 2)

)

y = t+ u2(t)

(− 2 cos(t− 2) + sin(t− 2)− t

)

546. Express the piecewise function using the unit step function and then find its Laplace Transform:

f(t) =

{t t < 4

e3t t > 4

547. Express the piecewise function using the unit step function and then find its Laplace Transform:

f(x) =

sin(2t) t < 3

te4t 3 < t < 6

cosh(2t) 6 < t


F (s) =e−5s

(s+ 4)


F (s) =e−3s(s− 5)

(s+ 1)(s+ 2)


F (s) = e−s4s2 − 17s+ 17

(s− 1)(s− 2)(s− 3)

551.

Solve:

y′ − 4y = 1 + uπ(t)t sin(t) y(0) = 1

552.

Solve:

y′′ + 3y′ + 2y = g(t) y(0) = 2 y′(0) = −1

g(t) =

{t2 t < 3

1 3 < t

553.

Solve:


y′′ + y = g(t) y(0) = 2 y′(0) = −1

g(t) =

{t t < 1

t2 t > 1

554.

Solve:

y′′ + 5y′ + 6y = g(t) y(0) = −1 y′(0) = 0

g(t) =

0 t < 1

4t 1 < t < 5

8 5 < t

555.

Solve:

y′′ − 2y′ + y = g(t) y(0) = 1 y′(0) = 3

g(t) =

2t t < 2

t2 2 < t < 5

1 5 < t

556.

Solve:

y′′ + 8y′ + 17y = g(t) y(0) = 1 y′(0) = 0

g(t) =

t t < 1

t2 1 < t < 5

t 5 < t

557.

Solve:

y′′ + 4y = g(t) y(0) = 0 y′(0) = 0

g(t) =

t t < 1

t2 1 < t < 2

1 2 < t

558.

Solve:

y′′ + y = g(t) y(0) = 0 y′(0) = 1

4.4. CONVOLUTION 141

g(t) =

{cos(2t) t < π

2

sin(2t) π2 < t

559.

Solve:

y′′ + 4y = g(t) y(0) = −3 y′(0) = 1

g(t) =

{| sin(t)| t < 2π

0 t > 2π

560.

Show: if

Y (s) =

∞∑n=0

e−ns(

1

s2+n

s

)Then

L−1(Y

)=

t t < 1

2t 1 < t < 2

3t 2 < t < 3...

nt n− 1 < t < n

561.

Express the floor function as a sum of unit step functions and find its Laplace Transform.

4.4 Convolution

If you are asked to find the inverse Laplace Transform of F (s)G(s) and you know the inverse Laplace Transform of both

F (s) and G(s) what is the relationship between the Laplace Transform of f(t)g(t) and Laplace Transform of f(t) and

g(t)? The answer is a new operation called Convolution.

The Convolution of f(t) and g(t) is denoted f ∗ g and is given by the integral:

f ∗ g =

∫ t

0

f(t− v)g(v)dv

For example the Convolution of t2 and t3 is:

t2 ∗ t3 =

∫ t

0

(t− v)2v3dv =

∫ t

0

(t2 − 2tv + v2)v3dv =

∫ t

0

(t2v3 − 2tv4 + v5)dv =t2v4

4− 2tv5

5+v6

6

∣∣∣∣t0

=t6

4− 2t6

5+t6

6=t6

60

Some basic properties of Convolution are:


f ∗ g = g ∗ f f ∗ (g + h) = f ∗ g + f ∗ h

(f ∗ g) ∗ h = f ∗ (g ∗ h) f ∗ 0 = 0

It is also true that

f ∗ 1 6= f

since

f ∗ 1 =

∫ t

0

f(t− v)dv 6= f

What makes Convolution useful in differential equations are the following properties involving the Laplace Transform:

L(f ∗ g) = L(f) · L(g) (L)−1(F (s)G(s)) = ((L)−1(F (s))) ∗ ((L)−1(G(s)))

To show the first property:

L(f ∗ g) = L(f) · L(g)

holds we let

F (s) = L(f) =

∫ ∞0

f(t)e−stdt G(s) = L(f) =

∫ ∞0

g(y)e−sydy

L(f ∗ g) =

∫ ∞0

e−st(∫ t

0

f(t− v)g(v)dv

)dt =

∫ ∞0

e−st(∫ ∞

0

ut−v(t)f(t− v)g(v)dv

)dt

Remember ut−v(t) will be zero if v > t

Reversing the order if integration gives

L(f ∗ g) =

∫ ∞0

g(v)

(∫ ∞0

e−stut−v(t)f(t− v)dt

)dv

The integral in parentheses is equal to e−svF (s) so we have:

L(f ∗ g) =

∫ ∞0

g(v)e−svF (s)dv = F (s)

∫ ∞0

g(v)e−svdv

So

L(f ∗ g) = F (s) ·G(s)

An Example: Solve:

y′ − 2

∫ t

0

et−vy(v)dv = t y(0) = 2

Rewriting the integral as convolution


y′ − 2et ∗ y = t y(0) = 2

Taking the Laplace Transform of each side gives:

sL(y)− 2− 2

(1

s− 1L(y)

)=

1

s2

Solving for L(y)

(s− 1)sL(y)− 2L(y) =s− 1

s2+ 2(s− 1)

L(y)(s2 − s− 2) =s− 1

s2+ 2(s− 1)

L(y) =s− 1

s2(s2 − s− 2)+

2(s− 1)

s2 − s− 2

L(y) =s− 1

s2(s− 2)(s+ 1)+

2(s− 1)

(s− 2)(s+ 1)

After partial fractions we have:

L(y) =2

s+ 1+

3

4

1

s− 2− 3

4

1

s+

1

2

1

s2

Taking the inverse Laplace Transform

y = 2e−t +3

4e2t − 3

4+

1

2t

Convolution can also be used to find the output response y(t) of a physical system for some input function f(t).

Consider the physical system governing the motion of a mass connected to a spring:

my′′ + by′ + ky = f(t)

For Simplicity lets assume

y(0) = y′(0) = 0

Taking the Laplace transform of both sides of the differential equation gives

ms2L(y) + bsL(y) + kL(y) = L(f(t))

L(y) =L(f(t))

ms2 + bs+ k

The function

W (s) =1

ms2 + bs+ k

is called the Transfer Function for the physical system.


L(y) = L(f(t))W (s)

The function

w(t) = L−1(W (s)

)is called the Weight Function of the system.

Through convolution we see the solution to the differentia equations is

y(t) =

∫ t

0

w(u)f(t− u)du

This equation reduces solving the same mass spring system for different input functions into solving a definite integral

for each input function.

562.

Find:

t ∗ et

563.

Find:

sin(t) ∗ sin(2t)

564.

If y(0) = 0 show:

2 ∗ (y(t) · y′(t)) = y2(t)

565.

Find the Laplace Transform of:

f(t) =

∫ t

0

(t− v)2e5vdv

566.

Find the Laplace Transform of:

f(t) =

∫ t

0

cos(t− v) · sin(4v)dv

567.

Find the inverse Laplace Transform of the following function by using Convolution:

F (s) =1

s2(s2 + 1)

568.

Find the inverse Laplace Transform of the following function by using Convolution:


F (s) =s

(s2 + 1)2

569.

Solve:

y′ + 8

∫ t

0

(t− v)y(v)dv = t y(0) = 0

570.

Solve:

y′ + y −∫ t

0

sin(t− v)y(v)dv = − sin(t) y(0) = 1

571.

Solve:

y′ = 1− sin(t)

∫ t

0

ty(v)dv y(0) = 0

572.

Solve:

y′ + et ∗ y + e−t ∗ y = 0 y(0) = 1

573.

Solve:

y′ + t ∗ y′ = 0 y(0) = 1

574.

Solve:

y′′ + y = 2t ∗ et y(0) = y′(0) = 0

575.

Solve:

y′′ − y = 2t ∗ sin(t) y(0) = 4 y′(0) = 2

576.

Solve using convolution. Leave your answer in terms of an integral involving g(t).

y′′ + y = g(t) y(0) = 1 y′(0) = 1

577.

Use convolution to evaluate the following integral∫ t

0

(t− u)5u8du


578.

Use convolution to evaluate the following integral

∫ t

0

(t− u)numdu

579.

If f and g have the following properties

f(t) ∗ f(t) = tf(t) f(0) = 4

Find f(t)

580.

Without using the definition of convolution calculate

f(t) = u1(t)t ∗ u2(t)t2

By calculating the Laplace Transform of f(t) simplifying and then calculating the inverse Laplace transform to find

f(t)

581.

If

f(t) ∗ f ′(t) =t2

2f(0) = f ′(0) = 0

Find f(t)

582.

Prove The following with mathematical induction

n-times︷︸︸︷t ∗ t ∗ t ∗ ... ∗ t =

t2n−1

(2n− 1)!

583.

Show the following property holds

t ∗ tn =tn+2

(n+ 1)(n+ 2)

584.

If f(t) = (1 + t)2

f(t) ∗ g(t) = f(t) · g(t) g(0) = e−1

Find g(t)

585.

Show

f ∗ g′ = f ′ ∗ g

4.5. DELTA FUNCTION 147

4.5 Delta Function

The Dirac Delta Function δ(x) is defined as

δ(t) =

{0 t 6= 0

∞ t = 0

And has the property: ∫ ∞−∞

f(t)δ(t)dt = f(0)

L(δ(t− c)) = e−cs

The delta function shows up in science when considering the impulse of a force over a short interval. If a force F (t)

on the time interval t0 to t1 then the impulse due to the force is:

Impulse =

∫ t1

t0

F (t)dt

By Newton’s second law ∫ t1

t0

F (t)dt =

∫ t1

t0

mdv

dtdt = mv(t1)−mv(t0)

where m is the mass and v is the velocity. Since an objects momentum is the product of mass and velocity we see

that the impulse is equal to the change in momentum.

An Example: Solve:

y′′ + y = δ(t− π) y(0) = 0 y′(0) = 0

Taking the Laplace Transform of both sides gives:

s2L(y) + L(y) = e−πs

L(y)(s2 + 1) = e−πs

L(y) = e−πs1

s2 + 1

Finding the inverse Laplace Transform:

y = uπ(t) sin(t)

∣∣∣∣t−π

y = uπ(t) sin(t− π)

After a trig identity

y = −uπ(t) sin(t)


586.

Find: ∫ 3

0

et2

δ(t− 1)dx

587.

Find the value of k so that ∫ 1

0

sin2(π(t− k))δ

(t− 1

2

)dt =

3

4

588.

Find

L{δ(sin(πt))}

589. Solve

y′ + y = δ(t− 1) y(0) = 2

590. Solve

y′′ − 4y′ + 4y = e2t + δ(t− 2) y(0) = 1 y′(0) = 3

591. Solve

y′′ + 2y′ + 2y = δ(t− π) y(0) = 1 y′(0) = 1

592. Solve

y′′ + 2y′ − 3y = δ(t− 1)− δ(t− 2) y(0) = 2 y′(0) = −2

593. Solve

y′′ − y′ − 2y = 3δ(t− 1) + et y(0) = 0 y′(0) = 3

594. Solve

y′′ − 3y′ + 2y = δ

(t− 1

)− δ(t− 2

)y(0) = 2 y′(0) = 3

595. Solve

y′′ + y = δ

(t− π

2

)− δ(t− 3π

2

)y(0) = 0 y′(0) = 0

596. Solve

y′′ − 4y′ + 13y = tu2(t) + δ

(t− 2

)y(0) = 1 y′(0) = 1

597. Solve

y′ − y = t ∗ et + δ

(t− 1

)y(0) = 1

598. Solve

y′′ + y =

∞∑n=0

δ

(t− nπ

)y(0) = 0 y′(0) = 0

599.

Show

L{δ(t− n) ∗ δ(t−m)} = L{δ(t−m− n)}

Chapter 5

First Order Systems of Differential

Equations

In this chapter we will study systems of differential equation of the form:

dx

dt= F (t, x, y)

dy

dt= G(t, x, y)

Where the dependent variables x and y are linked together by the independent variable t.

5.1 Homogenous Linear Systems

The theory of first order linear systems is very similar to the theory of second order equation that we studied in chapter

2. Consider the following system of equations:

dx

dt= 4x− y;

dy

dt= 2x+ y

This system can be transformed into a second order equation by solving for y in the first equation and substituting it

into the second:

y = 4x− dx

dt

d

dt

(4x− dx

dt

)= 2x+ 4x− dx

dt

This produces the second order equation:

4dx

dt− d2x

dt2= 2x+ 4x− dx

dt

Which simplifies to

d2x

dt2− 5

dx

dt+ 6x = 0

This has the characteristic equation:

149

150 CHAPTER 5. FIRST ORDER SYSTEMS OF DIFFERENTIAL EQUATIONS

r2 − 5r + 6 = 0 (r − 3)(r − 2) = 0

Giving the solution:

x1 = e2t x2 = e3t

Substituting these equations into y = 4x− dxdt gives two solution for y

y1 = 2e2t y2 = e3t

Making the general solution:

x = C1e3t + C2e

2t y = C1e3t + 2C2e

2t

In chapter 2 we learned that if the Wronskian of the two solutions to a second order equation is nonzero on an interval

then the two solutions are linearly independent and form a Fundamental Solution Set. For systems of equations the

Wronskian is:

W (t) =

∣∣∣∣∣x1 x2

y1 y2

∣∣∣∣∣ = x1y2 − x2y1

It can be shown that if the homogenous system:

dx

dt= a1(t)x(t) + b1y(t)

dy

dt= a2(t)x(t) + b2y(t)

has solutions:

x = x1(t) y = y1(t) and x = x2(t) y = y2(t)

and the Wronskian is non zero on the interval [a, b] then:

x = x1(t) y = y1(t) and x = x2(t) y = y2(t)

is the general solution to the system of differential equations on the interval [a, b]. We see that the Wronskian for the

last problem we solved is:

W (t) =

∣∣∣∣∣e3t e2t

e3t 2e2t

∣∣∣∣∣ = e5t 6= 0

So the general solution is indeed:

x = C1e3t + C2e

2t y = C1e3t + 2C2e

2t

You can also solve the same system of differential equations using matrices. Let us first write the system of equations

as a matrix equation of the form:

Ax = x’ where x =

[x

y

]

5.1. HOMOGENOUS LINEAR SYSTEMS 151

dx

dt= 4x− y;

dy

dt= 2x+ y

The system in matrix form is: [4 −1

2 1

][x

y

]=

[x′

y′

]As we earlier showed the system of equations can be reduced to a second order equation with constant coefficients

whose solutions were of the form: ert. It is reasonable to assume the above matrix equation will have a solution of the

form:

x(t) =

[x

y

]= ertu

Where r is a constant and u is a constant nonzero vector. Substituting x(t) = ertu into Ax = x’ gives:

rertu = Aertu

Dividing by the nonzero factor ert and rearranging the terms gives:(A− rI

)u = 0

The values of r and u that satisfy the above equation are the Eigenvalues and Eigenvectors of the matrix A. To find

the eigenvalues r of a matrix we take the determinant of both sides to the above equation:∣∣∣∣(A− rI)

u

∣∣∣∣ = |0| = 0

Since u is nonzero the solutions: r, to the above equation come from the solutions to:∣∣∣∣A− rI∣∣∣∣ = 0

This equation is a polynomial with the variable r. This equation is called the Characteristic Equation whose roots

are the eigenvalues of the matrix A. Back to the problem we were solving. Our characteristic equation is:∣∣∣∣∣ 4− r −1

2 1− r

∣∣∣∣∣ = (4− r)(1− r) + 2 = 0

Factoring the characteristic equation and solving gives:

(r − 2)(r − 3) = 0 r1 = 2 r2 = 3

NOTE: this is the same characteristic equation we got in our first solution to this problem.

Now that we have the eigenvalues for A we need the eigenvectors u. To find the eigenvector for each eigenvalue we

must solve the following equation for u. (A− rI

)u = 0

For r1 = 2 we have:


[2 −1

2 −1

][x

y

]= 0

We see the solution to this matrix equation is y = 2x. Taking x to be 1 we get our eigenvector:

u1 =

[1

2

]For r2 = 3 we have: [

1 −1

2 −2

][x

y

]= 0

We see the solution to this matrix equation is y = x. Taking x to be 1 we get our eigenvector:

u2 =

[1

1

]Making the solution to the differential equation:[

x

y

]= C1e

2t

[1

2

]+ C2e

3t

[1

1

]Making:

x = C1e2t + C2e

3t y = 2C1e2t + C2e

3t

The same solution we obtained earlier.

The matrix with column i given by eritui is called the Fundamental Matrix. The fundamental matrix for our problem

is:

X(t) =

[e2t e3t

2e2t e3t

]

Some times the matrix has complex eigenvalues: r = α± βı and eigenvectors a± bı the solution becomes:

x(t) = C1

(eαt cos(βt)a− eαt sin(βt)b

)+ C2

(eαt cos(βt)b + eαt sin(βt)a

)Example:

Solve the system of differential equations:

x′ = −x− 2y y′ = 8x− y

Writing the system as a matrix equation gives:[−1 −2

8 −1

][x

y

]=

[x′

y′

]Find the characteristic equation:


∣∣∣∣∣ −1− r −2

8 −1− r

∣∣∣∣∣ = (−1− r)2 + 16 = 0

Which has roots:

r = −1± 4ı

Using r = −1 + 4ı the eigenvector is: [−4ı −2

8 −4ı

][x

y

]= 0

This has the solution:

y = −2xı taking x = 1 the eigenvector is

u1 =

[1

−2ı

]=

[1

0

]+ ı

[0

−2

]We could find the eigenvector corresponding to r = −1− 4ı in the same we found the first eigenvector but we do not

need to. It is true that if u1 = a + ıb is a eigenvector for eigenvalue r1 = α + βı then u2 = a − ıb is a eigenvector for

r1 = α− βı. This means the eigenvector for r = −1− 4ı is

u2 =

[1

2ı

]=

[1

0

]+ ı

[0

2

]Making the solution:

x(t) = C1

(e−t cos(4t)

[1

0

]− e−t sin(4t)

[0

2

])+ C2

(e−t cos(4t)

[0

2

]+ e−t sin(4t)

[1

0

])

There are cases where the algebraic multiplicity of an eigenvalue is greater than its geometric multiplicity. This can

occur when you have repeated eigenvalues. We see this in the next example

Example:

Solve the system of differential equations:

x′ = x+ y y′ = y

Writing the system as a matrix equation gives:[x′

y′

]=

[1 1

0 1

][x

y

]Find the characteristic equation: ∣∣∣∣∣ 1− r 1

0 1− r

∣∣∣∣∣ = (1− r)2 = 0


Which has the repeated root

r = 1

and only one eigenvector

v1 =

[0

1

]Your options at this point are either find the a generalized eigenvector or convert the system into a second order

equation.

First converting to a second order equation. Solving the first equation for y and differentiating gives

y = x′ − x y′ = x′′ − x′

Substituting this into the second equation gives

x′′ − x′ = x′ − x

x′′ − 2x′ + x = 0

Forming the characteristic equation and finding the roots

r2 − 2r + 1 = 0 r = 1

The solution is

x = C1et + C2te

t

y = x′ − x = C1et + C2(t+ 1)et − (C1e

t + C2tet)

y = C2et

[x

y

]=

[C1e

t + C2tet

C2et

]= C1e

t

[1

0

]+ C2e

t

[0

1

]+ C2te

t

[1

0

]Or you can find a generalized eigenvector by solving

(A− rI)v2 = v1

[0 1

0 0

][x

y

]=

[1

0

]Making

v2 =

[0

1

]


And the general solution will have the form

x = C1ertv1 + C2te

rtv1 + C2ertv2

Making our solution

x = C1et

[1

0

]+ C2te

t

[1

0

]+ C2e

t

[0

1

]600.

Solve:

x′ = 2x+ y y′ = −x+ 4y

USN

601.

Solve:

x′ = 2x+ 4y y′ = −5x− 2y

C

602.

Solve:

x′ = −x− 2y y′ = 13x+ y

C

603.

Solve:

x′ = −7x− 2y y′ = 5x− y

SSp

604.

Solve:

x′ = x+ 2y y′ = 21x+ 2y

SP

605.

Solve:

x′ = x+ 2y y′ = x

SP

606.

Solve:


x′ = 2x+ 3y y′ = −5x− 2y

SN

607.

Solve:

x′ = x− 4y y′ = 2x+ 5y

USp

608.

Solve:

x′ = x+ 3y y′ = 12x+ y

609.

Solve:

x′ = x+ 2y + 2z y′ = 2x+ 3y z′ = 2x+ 3y

610.

Solve:

x′ = 2x− 4y y′ = 2x− 2y

611.

Solve:

x′ = −2x− 5y y′ = x+ 2y

612.

Solve:

x′ = 4x− y y′ = x+ 2y

613.

Solve:

x′ = −x+ 5y y′ = −x+ 3y

614.

Solve:

x′ = −6x+ 4y y′ = −5x+ 3y

615.

Convert the following differential equation governing the motion of a mass attached to a spring into a system of

equations

5.2. NON HOMOGENOUS LINEAR EQUATIONS 157

my′′ + by′ + ky = 0

616.

Solve: [2 −3

1 −2

][x

y

]=

[x′

y′

]617.

Solve: [−1 2

−1 −3

][x

y

]=

[x′

y′

]618.

Use the definition of the Wronskian in this section to prove Abel’s Theorem for systems: If

{y1, y2}

are solutions to

x′ = Ax

Then

W (y1, y2) = Ce∫ tt0tr(A)du

619.

Given a Linear System

x′ = Ax

with fundamental matrix X(t). The matrix exponential eAt can be found with the following formula:

eAt = X(t)X−1(0)

Find the matrix exponential for x′ = Ax with

A =

1 2 −1

1 0 1

4 −4 5

5.2 Non Homogenous Linear Equations

In this section we will develop ways to solve the non homogenous system:

x′(t) = Ax(t) + f(t)


As in chapter 2 these nonhomogeneous systems will have a homogenous solution: xh(t), and a particular solution: xp(t)

associated with them. And as in chapter 2 we will find the nonhomogeneous solution using the methods of Undetermined

Coefficients and Variation of Parameters.

Undetermined Coefficients

The method of Undetermined Coefficients can be applied to the above nonhomogeneous linear system if f(t) is a

polynomial, exponential function, sines and cosines and any product of these functions. The table in section 2.5 is still

useful in picking the form of the particular solution: xp(t).

An Example:

Solve: [x′

y′

]=

[4 −1

2 1

][x

y

]+

[−1− 5t

−7t

]From the previous section we know the homogenous solution is

xh =

[x

y

]= C1e

2t

[1

2

]+ C2e

3t

[1

1

]Seeing that the f(t) is a two row column vector containing two linear equations we will assume a particular solution

also be a row column vector containing two linear equations.

xp =

[At+B

Ct+D

]Differentiating gives

x’p =

[A

C

]Substituting x and x’ into the differential equation gives:[

A

C

]=

[4 −1

2 1

][At+B

Ct+D

]+

[−1− 5t

−7t

]After matrix multiplication we see the top row produces the equation:

A = 4At+ 4B − Ct−D − 1− 5t

And the Second Row produces the equation:

C = 2At+ 2B + Ct+D − 7t

Equating corresponding coefficients yields the following system of equations


A+D − 4B = −1 4A− C = 5 2B +D − C = 0 2A+ C = 7

This system has the solution

A = 2 B = 1 C = 3 D = 1

Making the particular solution

xp =

[2t+ 1

3t+ 1

]As in chapter 2 the general solution is the sum of the homogenous solution and the particular solution

x =

[x

y

]= C1e

2t

[1

2

]+ C2e

3t

[1

1

]+

[2t+ 1

3t+ 1

]620.

Solve using Undetermined Coefficients[x′

y′

]=

[3 2

3 4

][x

y

]+

[−5t2 + t− 1

−7t2 − 3t+ 2

]621.


y′

]=

[2 2

1 3

][x

y

]+

[(−2t− 1)e2t

−2te2t

]622.


y′

]=

[1 3

5 3

][x

y

]+

[−6e2t

−2e2t − 5et

]623.

Consider the nonhomogeneous system [x′

y′

]=

[1 1

0 1

][x

y

]+

[et

et

]The homogenous solution was found to be

x = C1et

[1

0

]+ C2te

t

[1

0

]+ C2e

t

[0

1

]Show that you cannot find a particular solution of the form etv or tetv. Then find a particular solution of the form

tetu1 + etu2

Variation of Parameters


The method of variation of parameters can be applied to most nonhomogeneous linear systems. Let X(t) be the

fundamental matrix for the homogenous system:

x′(t) = Ax An×n

This makes the general solution to the system X(t)c where c is a n× 1 column vector of constants.

To find a solution to the nonhomogeneous system

x′(t) = Ax(t) + f(t)

we look for a particular solution of the form:

xp(t) = X(t)v(t)

Where v(t) is a n× 1 column vector of functions.

Differentiating xp and substituting it and xp in to the nonhomogeneous equation gives

x′p = X(t)v′(t) + X′(t)v(t)

X(t)v′(t) + X′(t)v(t) = AX(t)v(t) + f(t)

Since X′(t) = AX(t) the above equation becomes

X(t)v′(t) + X′(t)v(t) = X′(t)v(t) + f(t)

So

X(t)v′(t) = f(t)

v′(t) = X−1(t)f(t)

v(t) =

∫X−1(t)f(t)dt

xp = X(t)

∫X−1(t)f(t)dt

An Example:

Solve [x′

y′

]=

[1 2

10 2

][x

y

]+

[9et

18e2t

]We first find the solution to the homogeneous system:[

x′

y′

]=

[1 2

10 2

][x

y

]This homogeneous system has a solution


[x

y

]= C1e

6t

[2

5

]+ C2e

−3t

[1

−2

]and a Fundamental Matrix

X(t) =

[2e6t e−3t

5e6t −2e−3t

]whose inverse is

X−1(t) =1

9

[2e−6t e−6t

5e3t −2e3t

]The particular solution is

xp =

[2e6t e−3t

5e6t −2e−3t

]∫1

9

[2e−6t e−6t

5e3t −2e3t

][9et

18e2t

]dt

xp =

[2e6t e−3t

5e6t −2e−3t

]∫ [2e−5t + 2e−4t

5e4t − 4e5t

]dt

xp =

[920e

t − 95e

2t

−92 e

t − 910e

2t

]And the general solution is

x = C1e6t

[2

5

]+ C2e

−3t

[1

−2

]+

9

20

[et − 4e2t

−10et − 2e2t

]

Cauchy Euler Equation

The Cauchy Euler equation for first order systems is an equation of the form

tx′ = Ax

This system can be converted to a system of differential equations with constant coefficients with the following

substitution

x = trv

Making

x′ = rtr−1v So tx′ = rtrv

Substituting x and tx′ into the Cauchy Euler equation yields


rtrv = Atrv

rv = Av

So we see r is the eigenvalue and v is the eigenvector of matrix A

An Example:

Solve

t

[x′

y′

]=

[4 −1

2 1

][x

y

]We have already shown that the eigenvalues and eigenvectors of[

4 −1

2 1

]are

r1 = 2 v1 =

[1

2

]and r2 = 3 v1 =

[1

1

]Making the Solution [

x

y

]= C1t

2

[1

2

]+ C2

[1

1

]624.

Solve using Variation of Parameters[x′

y′

]=

[2 2

5 −1

][x

y

]+

[−2t2 − 2t− 1

t2 − 3t+ 1

]625.

Solve using Variation of Parameters [x′

y′

]=

[2 8

2 2

][x

y

]+

[2e2t

e6t

]626.

Solve using Variation of Parameters [x′

y′

]=

[1 4

4 1

][x

y

]+

[−3et

−3tet

]627.

Solve the Cauchy-Euler equation.

t

[x′

y′

]=

[2 −1

3 −2

][x

y

]+

[t−1

1

]628.

5.3. LOCALLY LINEAR SYSTEMS 163


t

[x′

y′

]=

[1 2

10 2

][x

y

]+

[t2

t

]629.


t

[x′

y′

]=

[1 1

15 3

][x

y

]+

[t2

t

]630.


t

[x′

y′

]=

[5 −1

3 1

][x

y

]+

[t2

t

]

5.3 Locally Linear Systems

We will now discuss nonlinear systems around their critical or equilibrium points. The general system of equations we

will be studying are systems of the form:

x′ = F (x, y) y′ = G(x, y)

The Critical or Equilibrium Points of this system are the values (x0, y0) such that

x′ = F (x0, y0) = 0 and y′ = G(x0, y0) = 0

If both x0 > 0 and y0 > 0 then the critical point is defined to be a Positive Equilibrium Point and will be discussed

when we consider the competing species problem.

The system is said to be Locally Linear around a critical or equilibrium point if both F and G have continuous first

and second order partial derivatives at the critical point (x0, y0).

For the general system:

x′ = F (x, y) y′ = G(x, y)

with equilibrium point (x0, y0) we can form the locally linear system in matrix form:[Fx(x0, y0) Fy(x0, y0)

Gx(x0, y0) Gy(x0, y0)

][x

y

]=

[x′

y′

]With the matrix:

J =

[Fx(x0, y0) Fy(x0, y0)

Gx(x0, y0) Gy(x0, y0)

]called the Jacobian Matrix

An Example


Find the positive equilibrium solution to the system and discuss the stability near this point.

dx

dt= x

(4− 2y

)dy

dt= y

(− 2 + x

)

We see that the positive equilibrium solutions come from the equations :

4− 2y = 0 − 2 + x = 0

Giving the positive equilibrium solution (2, 2). The Jacobian Matrix for this problem is:

J =

[4− 2y −2x

y x− 2y

]

The Jacobian Matrix at the equilibrium point (2, 2) is:

J =

[0 −4

2 −2

]

Producing the locally linear system:

[0 −4

2 −2

][x

y

]=

[x′

y′

]

Now we need the eigenvalues and eigenvector of our Jacobian Matrix evaluated at the equilibrium points. The

characteristic equation is:

J =

∣∣∣∣∣ 0− r −4

2 −2− r

∣∣∣∣∣ = −r(−2− r) + 8 = 0 r2 + 2r + 8 = 0

Which has roots:

r1,2 = −1±√

7ı

I have no desire to find the eigenvectors so, instead of the two eigenvectors I present you with one rabbit with two

pancakes on its head.

5.3. LOCALLY LINEAR SYSTEMS 165

631.

Find all equilibrium solutions and discuss the stability of each point and interpret the results of the competing species

modle.

dx

dt= x

(1− x− y

)dy

dt= y

(2− y − 3x

)Show that if the initial condition is on the seperatrix: the line connection the origin to the positive equilibrium

solution the the constant corresponding to the positive eigenvalue is zero and the solution will tend towards the positive

equilibrium solution.

632.


modle.

dx

dt= x

(1− x− y

)dy

dt= y

(3− 2x− 4y

)633.


modle.

dx

dt= x

(2− x− y

)dy

dt= y

(3− y − 2x

)634.


modle.


dx

dt= x

(2− 2x− y

)dy

dt= y

(3− 2x− 2y

)635.


modle

dx

dt= x

(3− x− y

)dy

dt= y

(4− x− 2y

)636.


modle

dx

dt= x

(2− x− 1

2y

)dy

dt= y

(3− x− y

)637.


modle

x′ = x(2− x− y) y′ = y(3− 2y − x)

638.


modle

x′ = x(2− x− 2y) y′ = y(7− 2y − 6x)

639.


modle

x′ = x(4− 2x− 2y) y′ = y(5− y − 4x)

640.


modle

x′ = x(4− x− 2y) y′ = y(3− y − x)

641.

Find the positive equilibrium solution to the system and discuss the stability near this point.

dx

dt= x

(1− x− y

)dy

dt= y

(3− x− 2y

)642.

Find the positive equilibrium solution to the competing species system of equations and discuss the stability near this

point.

5.4. LINEAR SYSTEMS AND THE LAPLACE TRANSFORM 167

dx

dt= x

(2− x

3− 4y

x+ 4

)dy

dt= y

(x

x+ 4− 1

2

)643.

Find the equilibrium solution to the system and sketch phase plane diagram.

dx

dy= −y(y − 2)

dy

dt= (x− 2)(y − 2)

644.

Solve the system of equations by converting it into a first order differential equation

dx

dy= xy − x3

y

dy

dt= 3x2 − y2

645.

Solve the system of equations by converting it into a first order differential equation

dx

dt= x

dy

dt= 2y − x3y2

5.4 Linear Systems and the Laplace Transform

We can also use Laplace Transforms to solve linear systems with initial conditions. The procedure will be quite similar to

the procedure used to solve initial value problems in chapter 4. Given a system of equations we will start by taking the

Laplace Transform of each side of the equation, solving for the Laplace Transform of both x(t) and y(t) and then using

a table to find the inverse.

An Example: Solve:

dx

dt= 2y + 4t x(0) = 4

dy

dt= 4x− 2y − 4t− 2 y(0) = −5

Taking the Laplace Transform of both sides of the differential equation involving x′ gives:

L(x′) = L(y) + L(4t)

Using a table we get:

sL(x)− x(0) = 2L(y) +4

s2

sL(x)− 4 = 2L(y) +4

s2

Taking the Laplace Transform of both sides of the differential equation involving y′ gives:

L(y′) = 4L(x)− 2L(y)− L(4t)− L(2)

sL(y)− y(0) = 4L(x)− 2L(y)− 4

s2− 2

s


sL(y) + 5 = 4L(x)− 2L(y)− 4

s2− 2

s

We now have two equations with two desired variables: L(x) and L(y) that we will need to solve for. In our first

equation I will solve for L(x) and insert it into the second.

sL(x)− 4 = 2L(y) +4

s2

L(x) =2

sL(y) +

4

s3+

4

s

Inserting L(x) into the second of our two equations gives:

sL(y) + 5 = 4L(x)− 2L(y)− 4

s2− 2

s

sL(y) + 5 = 4

(2

sL(y) +

4

s3+

4

s

)− 2L(y)− 4

s2− 2

s

Multiplying both sides by s3 and simplifying the result gives:

L(y)(s4 + 2s3 − 8s2) = 16− 4s+ 14s2 − 5s3

L(y) =16− 4s+ 14s2 − 5s3

(s4 + 2s3 − 8s2)

Applying partial fractions to the right hand side gives:

L(y) =−6

(x+ 4)+

1

(s− 2)− 2

s2

Using a table to find the inverse Laplace gives:

y = −6e−4t + e2t − 2t

Now that we have y we can substitute its equation into the original equation for x′:

dx

dt= 2y + 4t

becomes:

dx

dt= 2(−6e−4t + e2t − 2t) + 4t

dx

dt= −12e−4t + 2e2t

Integrating gives:

x = 3e−4t + e2t + C

Applying the initial condition shows C = 0 and our solutions are:

x = 3e−4t + e2t y = −6e−4t + e2t − 2t

5.4. LINEAR SYSTEMS AND THE LAPLACE TRANSFORM 169

646.

Solve:

x′ = 2y + t x(0) = 0 y′ = x+ y y(0) = 0

647.

Solve:

x′ = x+ y x(0) = 0 y′ = x+ y − tet y(0) = 1

648.

Solve:

x′ = y − x+ 4t2 + 1 x(0) = 1 y′ = x− y + 2t2 + 2t− 1 y(0) = 0

649.

Solve:

x′ = 2x+ y − t− 1 x(0) = 1 y′ = y − t y(0) = 1

650.

Solve:

x′ = x+ y − t sin(t) + t cos(t) x(0) = 0 y′ = x− t sin(t) + cos(t) y(0) = 0

651.

Solve:

x′ = 2x− y + 1− 2t x(0) = 1 y′ = 2y − x+ t y(0) = 1

652.

Solve:

x′ = x− y − t3 + 4t2 + 1 x(0) = 0 y′ = 2x− y − 2t3 + t2 + t+ 1 y(0) = 0

653.

Solve:

x′ = x+ y + 2tet x(0) = 0 y′ = x′ − t2et − 3tet y(0) = 1

654.

Solve:

x′ = x+ y x(0) = 0 y′ = x′ − x y(0) = 1

655.

Solve:

x′ = y − 3t x(0) = 0 y′ = x′ + 6x− 6t3 − 3t2 + 2 y(0) = 1

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1 Di erential Equations - University of Texas at Dallasx+1 to transform the equation into a separable equation and then solve it dy dx = y+ 3 x+ 1 + y 3x x+ 1 2 34. Use the substitution