Ordinary Differential Equations · 2020-03-19 · The solution of an ordinary differential equation satisfying a set of conditions is a particular solution. Example Consider the differential

UNENE Math Refresher Course

Ordinary Differential Equations

1

Ordinary Differential Equations — Introduction © Wei-Chau Xie

Example – Vibration of a Mass-Spring System

Consider a mass-spring system as shown

m

x(t)

F(t)k

m

x, x, x

F(t)kxm

x(t)

F(t)k

x(t) is the displacement of mass m from its equilibrium position.

The spring force is kx.

Newton’s Second Law requires ma =∑

F :

m · x(t) = F(t) − kx =⇒ mx + kx = F(t)

This equation is a differential equation.

2


Definition

A differential equation is an equation that contains derivatives of dependent

variables with respect to one or more independent variables.

A differential equation is called an ordinary differential equation (ODE)

if there is only one independent variable and only ordinary derivatives (not

partial derivatives) are involved.

Let x be an independent variable and y be a dependent variable. In general, an

ODE is of the form

F(

x, y,dy

dx, . . . ,

dny

dxn

)

= 0

The order of a differential equation is the order of the highest derivative

appearing in the differential equation.

If y and its various derivatives y′, y′′, . . . appear linearly in the equation, it

is a linear differential equation; otherwise, it is nonlinear.

3


Definition – Linear Ordinary Differential Equations

The general form of an nth-order linear ordinary differential equation is

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · · + a1(x)

dy

dx+ a0(x) y = f(x)

If a0(x), a1(x), . . . , an(x) are constants, the ordinary differential equation is said

to have constant coefficients. Otherwise it is said to have variable coefficients.

Example

.d2y

dx2+ 2 y = sin 3x Second-order linear with constant coefficients

.

(dy

dx

)2+ 2 y = sin 3x First-order nonlinear

. x2 d2y

dx2+ 2x

dy

dx− y = ln x Second-order linear with variable coefficients

4


Definition – Partial Differential Equations

If the dependent variable u is a function of more than one independent variable,

say x1, x2, . . . , xm, an equation involving the variables x1, x2, . . . , xm, u and

various partial derivatives of u with respect to x1, x2, . . . , xm is called a partial

differential equation (PDE).

Example

.∂2u

∂x2+

∂2u

∂y2= f(x, y) Poisson’s equation in two-dimensions

.∂2u

∂x2=

1

c

∂u

∂tHeat equation in one-dimension

5


Definition – General Solution

The general solution of an ordinary differential equation contains all possible

solutions. In general, an nth-order ordinary differential equation will contain n

arbitrary constants in its general solution.

The solution of an ordinary differential equation satisfying a set of conditions is

a particular solution.

Example

Consider the differential equation y ′′ + y = 0.

It can be shown that

y = C1 sin x + C2 cos x

is a solution for any constants C1 and C2 =⇒ general solution

y1 = sin x is a particular solution (C1 = 1, C2 = 0)

y2 = cos x is a particular solution (C1 = 0, C2 = 1)

y3 = 2 sin x−3 cos x is a particular solution (C1 = 2, C2 = −3)6


Definition – Initial Value Problem

If a differential equation is required to satisfy conditions on the dependent

variable and its derivatives specified at one value of the independent variable,

these conditions are called initial conditions (IC) and the problem is called an

initial value problem (IVP).

Definition – Boundary Value Problem

If a differential equation is required to satisfy conditions on the dependent

variable and possibly its derivatives specified at two or more values of the

independent variable, these conditions are called boundary conditions (BC) and

the problem is called a boundary value problem (BVP).

7


Example – Initial Value Problem

Consider the differential equation y′′ + y = 0.

If we impose the initial conditions y(0)= 2, y′(0) = −1, determine the solution

of this IVP.

We know that y(x) = C1 sin x + C2 cos x is the general solution.

. y′(x) = C1 cos x − C2 sin x

Applying the initial conditions:

. y(0) = C1 sin 0 + C2 cos 0 = C2 = 2

. y′(0) = C1 cos 0 − C2 sin 0 = C1 = −1

The solution of this IVP is

y(x) = − sin x + 2 cos x

8


Example – Boundary Value Problem

Consider the differential equation y′′ + y = 0.

If we impose the boundary conditions y(0)= 2, y( π4 ) = −1, determine the

solution of this BVP.

We know that y(x) = C1 sin x + C2 cos x is the general solution.

Applying the boundary conditions:

. y(0) = C1 sin 0 + C2 cos 0 = C2 = 2

. y( π4 ) = C1 sin π

4 + C2 cos π4 = 1√

2(C1 + C2) = −1 =⇒ C1 = −2−

√2

The solution of this BVP is

y(x) = −(2+√

2) sin x + 2 cos x.

9

First-Order Differential Equations — Method of Separation of Variables © Wei-Chau Xie

The Method of Separation of Variables

Consider the first-order DE:dy

dx= f(x) φ( y)

Case 1: If φ( y)=/ 0, divide the DE by φ( y):

1

φ( y)dy

︸︷︷︸

= f(x) dx︸︷︷︸

Contains y only Contains x only

General solution can be obtained by integrating both sides∫

1

φ( y)dy =

∫

f(x) dx + C C = arbitrary constant

Case 2: If φ( y) = 0 =⇒ y = y0 is a root of the equation

. =⇒ y = y0 is a solution.

10


Example

Solvedy

dx= 2x y

Case 1: y = 0 is a solution.

Case 2: y =/ 0, divide the DE by y and integrate both sides

.

∫1

ydy =

∫

2x dx + C

. ln∣∣ y

∣∣ = x2 + C =⇒

∣∣ y

∣∣ = ex2+C =⇒ y = ±eC · ex2

Let A = ±eC. The general solution is

y = Aex2or y = Cex2

C = arbitrary real constant

Case 1: y = 0 is included in the general solutionwith C = 0.

11


Example

Solve the IVPdy

dx=

6x5 + 2x + 1

cos y + e y, y(0) = π

The DE is variable separable

.

∫

(cos y + e y) dy =∫

(6x5 + 2x + 1) dx + C

. ∴ sin y + e y = x6 + x2 + x + C General solution

Using the initial condition: x = 0, y = π :

sin π + eπ = C =⇒ C = eπ

The solution of the IVP is

sin y + e y = x6 + x2 + x + eπ Particular solution

12


Example

Solvedr

dϕ=

sin ϕ + e2r sin ϕ

3er + er cos ϕ

dr

dϕ=

sin ϕ(1 + e2r)

er(3 + cos ϕ)Variable separable

∴

∫er

1 + e2rdr =

∫sin ϕ

cos ϕ + 3dϕ + C

∫1

1 + e2rd(er) = −

∫1

cos ϕ+3d(cos ϕ+3) + C

∫1

1 + x2dx = tan−1 x

∴ tan−1er = − ln(cos ϕ+3) + C General solution

13

First-Order Differential Equations — Linear First-Order Equations © Wei-Chau Xie

Linear First-Order Equations

Linear first-order equations are of the formdy

dx+ P(x)· y = Q(x)

The roles of x and y may be exchanged to give linear first-order DE of the form

dx

dy+ P( y)·x = Q( y)

Linear First-Order Equations

dy

dx+ P(x)· y = Q(x) =⇒ y = e

−∫

P(x)dx[ ∫

Q(x)e

∫

P(x)dxdx + C]

dx

dy+ P( y)·x = Q( y) =⇒ x = e

−∫

P( y)dy[ ∫

Q( y)e

∫

P( y)dydy + C]

14


Example

Solvedy

dx= 5 y + 3e5x

Rewrite the DE in the standard formdy

dx+ P(x)· y = Q(x):

.dy

dx− 5 · y = 3e5x, P(x) = −5, Q(x) = 3e5x

.

∫

P(x) dx =∫

−5 dx = −5x, e

∫

P(x)dx = e−5x

.

∫

Q(x)e

∫

P(x)dx dx =∫

3e5x · e−5x dx = 3x

. y = e−∫

P(x)dx[ ∫

Q(x)e

∫

P(x)dx dx + C]

= e5x(3x + C)

15


Example

Solve y′ =1

x − 3 y

.dy

dx=

1

x − 3 y=⇒

dx

dy= x − 3 y

.dx

dy− 1 · x = −3 y, P( y) = −1, Q( y) = −3 y

.

∫

P( y) dy =∫

−1 · dy = − y, e

∫

P( y)dy = e−y

.

∫

Q( y)e

∫

P( y)dydy =∫

−3 y · e−y dy = −3− y − 1

(−1)2e−y = 3( y + 1)e−y

. x = e−∫

P( y)dy[ ∫

Q( y)e

∫

P( y)dydy + C]

. = e y[3( y + 1)e−y + C] = 3( y + 1) + Ce y

16

Applications of First-Order Differential Equations — Heating and Cooling © Wei-Chau Xie

Newton’s Law of Cooling

The rate of change in the temperature T(t), dT/dt, of a body in a medium of

temperature Tm is proportional to the temperature difference between the body

and the medium, i.e.,

dT

dt= −k(T − Tm),

where k>0 is a constant of proportionality.

17


Example – Newton’s Law of Cooling

A thermometer is removed from a room where the temperature is 20◦C to the

outside where the temperature is −10◦C. After 1 minute, the thermometer

reads 10◦C.

1. What is the reading of the thermometer at t = 2 minutes?

2. How long will it take for the thermometer to read −5◦C?

By Newton’s Law of CoolingdT

dt= −k(T − Tm)

T: Temperature of the thermometer at time t

Tm: Outside temperature, Tm = −10◦C

The DE can be written as

.dT

dt+ k·T = kTm, P(t) = k, Q(t) = kTm Linear first-order

18


.

∫

P(t) dt = kt, e

∫

P(t)dt = ekt,

∫

Q(t)e

∫

P(t)dtdt =∫

kTm · ekt dt = Tm ekt

. T = e−∫

P(t)dt[ ∫

Q(t) e

∫

P(t)dtdx + C]

= e−kt(Tm ekt + C)

. ∴ T = −10 + C e−kt

At t = 0, T = 20◦C: 20 = −10 + Ce−k·0 =⇒ C = 30

At t = 1, T = 10◦C: 10 = −10 + 30e−k·1 =⇒ e−k = 23

=⇒ k = − ln23

. ∴ T(t) = −10 + 30·et· ln 23 = −10 + 30·

(23

)t

At t = 2: T(2) = −10 + 30·(2

3

)2 = 103

= 3.33◦C

When T = −5: −5 = −10 + 30·(2

3

)t=⇒

(23

)t = 16

=⇒ t· ln23

= ln16

t =ln

16

ln23

= 4.4 min

19

Applications of First-Order Differential Equations — Radioactive Decay Law © Wei-Chau Xie

Example – Radioactive Decay Law

Suppose that, in a sample of radioactive materials, there are N(t) nuclei which

have not decayed at time t.

During the time interval from t to t+1t, the number of radioactive nuclei

which will decay is proportional to N(t) and 1t, i.e.

N(t) − N(t+1t) = λ·N(t)·1t

where λ is the constant of proportionality called the decay constant.

Divide by 1t and take the limit 1t→0

−N(t+1t) − N(t)

1t= λN(t) =⇒ −

dN

dt= λN

The differential equation is variable separable∫

1

NdN =

∫

−λ dt + C =⇒ ln N = −λt + C =⇒ N(t) = C e−λt

20

Applications of First-Order Differential Equations — Radioactive Decay Law © Wei-Chau Xie

Initial condition: t = 0, N(0) = N0 =⇒ C = N0

N(t) = N0 e−λt Radioactive Decay Law

Half-Life t1/2 is the length of time it takes for the radioactivity of a

radioisotope to decrease by a factor of two, i.e.

N(t1/2) = N0 e−λt1/2 = 1

2N0

e−λt1/2 = 1

2=⇒ − λt1/2 = ln

12

= − ln 2

∴ t1/2 =ln 2

λor λ =

ln 2

t1/2

21

Applications of First-Order Differential Equations — Motion of a Particle © Wei-Chau Xie

Newton’s Second Law

The product of the mass of an object and its acceleration is equal to the sum of

forces applied on the object, i.e., ma =∑

F.

Example

A motor boat is travelling at a speed v0 and the motor is turned off at time

t = 0. The resistance from air is proportional to the velocity, i.e., Rair = −αv,

and the resistance from water is proportional to the square of the velocity, i.e.,

Rwater = −βv2, in which α and β are positive constants. Determine the velocity

of the boat for t>0.

αv

x,v,a

βv2

22

Applications of First-Order Differential Equations — Motion of a Particle © Wei-Chau Xie

Apply Newton’s Second Law to the boat:

ma =∑

F : mdv

dt= −α v − β v2 = −β v (v + k), k =

α

β

.

∫dv

v (v + k)= −

β

m

∫

dt + C Variable separable

.1

k

∫ ( 1

v−

1

v + k

)

dv = −β

mt + C =⇒

1

k[ ln v − ln(v + k)] = −

β

mt + C

.1

kln

v

v + k= −

β

mt + C =⇒

v

v + k= C e

−kβm t

At t = 0, v = v0 =⇒v0

v0 + k= C · e0

∴ C =v0

v0 + k=

v0

v0 +α

β

=β v0

α + β v0

v(t) =C k e

−αtm

1 − C e−αt

m

=

β v0

α + β v0

·α

β· e

−αtm

1 −β v0

α + β v0

· e−αt

m

=α v0 e

−αtm

(α+βv0) − βv0 e−αt

m

23

Applications of First-Order Differential Equations — Mixture Problem © Wei-Chau Xie

Example – Mixture Problem

Initially 50 kg of salt is dissolved in a large tank holding 1000 L of water. A brine

solution with a concentration of 0.2 kg/L is pumped into the tank at the rate of

10 L/min, and the well-stirred solution is pumped out at the same rate.

1. Determine the amount of salt in the tank at any time.

2. How much salt is present after 50 minutes?

3. How much salt is present after a long long time?

OutflowQout , cout

InflowQin , cin

Volume = 1000

Amount of salt = s(t)

Concentration c(t)=s(t)

1000

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s(t)=Amount of salt in the tank at time t =⇒ Salt concentration c(t) =s(t)

1000

Inflow

Flow rate Qin = 10 (L/min)

Salt concentration cin = 0.2 (kg/L)

Outflow

Flow rate Qout = 10 (L/min)

Salt concentration cout = c(t) =s(t)

1000(kg/L)

Consider time interval 1t:

. 1s = Change of amount of salt in 1t

. = (Amount of salt in inflow in 1t) − (Amount of salt in outflow in 1t)

. = cin Qin 1t − cout Qout 1t

. = 0.2 × 10 × 1t −s

1000× 10 × 1t

25


Divide by 1t and take the limit 1→0 =⇒ds

dt= 2 −

s

100

.ds

dt+

1

100·s = 2, P(t) =

1

100, Q(t) = 2 Linear first-order

.

∫

P(t) dt =t

100,

∫

Q(t)e

∫

P(t)dtdt =∫

2 · et

100 dt = 200 et

100

. s(t) = e−∫

P(t)dt[ ∫

Q(t) e

∫

P(t)dtdx + C]

. = e− t

100(

200 et

100 + C)

= 200 + C e− t

100

C is determine from the initial condition: t = 0, s = 50

50 = 200 + C e0 =⇒ C = −150 =⇒ s(t) = 200 − 150 e− t

100

At t = 50 min: s(50) = 200 − 150 e− 50

100 = 109.02 kg

When t → ∞: limt→∞

s(t) = limt→∞

(

200 − 150 e− t

100)

= 200 kg

26

Applications of First-Order Differential Equations — Tank Problem © Wei-Chau Xie

Example – Tank Problem

The moderator in a CANDU unit has been poisoned with gadolinium nitrate

Gd(NO3)3 due to activation of shut-down system two (SDS-2). Referring to

the schematic diagram, it is given that the volume of heavy water D2O in

the calandria is V = 220, 000 L, with the initial concentration of gadolinium

nitrate being C0 = 12 mg/L, the flow rate through the ion exchange columns is

Q = 2, 300 L/min, and the removal efficiency is α = 95%. Determine how long

it will take to get the gadolinium nitrate concentration below 0.01 mg/L in order

to begin plant startup activities.

Initial Gd(NO3)3

concentration

C0=12 mg/L

Flow rate

Q=2,300 L/min

Calandria

V=220,000 LIon Exchanger

Removal efficience α

27


At time t, the Gd(NO3)3 concentration is C(t).

Amount of Gd(NO3)3 in calandria is C(t)·V.

During time interval 1t,

Outflow = Q1t

Amount of Gd(NO3)3 removed = C(t)·Q1t·α

At time t+1t, the Gd(NO3)3 concentration is C(t+1t).

Amount of Gd(NO3)3 in calandria is C(t+1t)·V.

∴ (Amount of Gd(NO3)3 at t) − (Amount of Gd(NO3)3 at t+1t)

= Amount of Gd(NO3)3 removed

∴ C(t)·V − C(t+1t)·V = C(t)·Q1t·α

Divide by 1t and take the limit 1t→0

−VC(t+1t) − C(t)

1t= αQ C(t) =⇒ V

dC(t)

dt= −αQ C(t)

28


∫1

CdC = −

∫1

τdt + D, τ =

V

αQVariable separable

ln C = −t

τ+ D =⇒ C(t) = D e−t/τ

When t = 0, C(0) = C0 =⇒ D = C0

∴ C(t) = C0 e−t/τ =⇒ t = −τ lnC(t)

C0

= −V

αQln

C(t)

C0

When C(t) = 0.01 mg/L

t = −220, 000

0.95×2, 300ln

0.01

12= 713.87 min = 11 hr 54 min

29

Applications of First-Order Differential Equations — Chemical Reaction © Wei-Chau Xie

Law of Mass Action

The Law of Mass Action states that the rate of reaction is directly proportional

to the product of the masses of reactants.

Example – Chemical Reaction

A chemical reaction may be expressed as

A + B → C

in which M grams of chemical A reacts with N grams of chemical B to

generate M+N grams of compound C.

Suppose we add a grams of chemical A to b grams of chemical B.

At time t, x(t) grams of compound C is produced, in whichM

M+Nx(t) grams

of chemical A andN

M+Nx(t) grams of chemical B are used.

The amounts of chemicals A and B left are, respectively,

a −M

M+Nx(t), b −

N

M+Nx(t)

30


By the Law of Mass Action, the rate of reaction satisfies

.dx

dt= k

(

a −M

M+Nx)(

b −N

M+Nx)

. = k ·M

M+N

[a (M+N)

M− x

]

·N

M+N

[b (M+N)

N− x

]

= k (x − α) (x − β)

where k = k ·M

M+N·

N

M+N, α =

a (M+N)

M, β =

b (M+N)

N

. ∴

∫dx

(x−α)(x−β)= k dt + C =⇒

1

α−β

∫ ( 1

x−α−

1

x−β

)

dx = k t + C

.1

α−βln

∣∣∣∣

x−α

x−β

∣∣∣∣= k t + C =⇒

x−α

x−β= C ek t k = k (α−β)

C is determined by the initial condition t = 0, x(0) = 0:

0−α

0−β= C e0 =⇒ C =

α

β=⇒

x−α

x−β=

α

βek t

∴ x(t) =αβ (1 − e−k t)

α − β e−k t

31


Numerical Example – Chemical Reaction

Suppose M = 1, N = 4, a = 50, b = 32. If 30 grams of compound C are

formed in 10 minutes

1. Determine the amount of compound C at any time.

2. How much of compound C is produced in 15 minutes?

3. What happens after a long long time?

. x(10) = 30, α =a (M+N)

M=

50 (1+4)

1= 250

. β =b (M+N)

N=

32 (1+4)

4= 40

. x(t) =αβ (1 − e−k t)

α − β e−k t=

250·40 (1 − e−k t)

250 − 40 e−k t=

1000 (1 − e−k t)

25 − 4 e−k t

At t = 10, x(10) =1000 (1 − e−10 k)

25 − 4 e−10 k= 30 =⇒ k = 1

10ln

8825

32


x(t) =1000

(

1 − e− t

10 ln8825

)

25 − 4 e− t

10 ln8825

At time t = 15: x(15) =1000

(

1 − e−3

2 ln8825

)

25 − 4 e−3

2 ln8825

= 34.79

When t → ∞: limt→∞

x(t) =1000

(

1 − e− t

10 ln8825

)

25 − 4 e− t

10 ln8825

=1000

25= 40

After a long long time, 40 grams of compound C is produced, which uses 8

grams of chemical A and all 32 grams of chemical B. Since there is no chemical

B left, the chemical reaction stops.

33

Linear Ordinary Differential Equations © Wei-Chau Xie

Linear Ordinary Differential Equations

In general, an nth-order linear ODE is of the form

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · · + a1(x)

dy

dx+ a0(x) y = F(x)

If a0, a1, . . . , an are constants, the ODE has constant coefficients.

Using the D-operator, the general nth-order linear ODE can be rewritten as

[

an(x)Dn + an−1(x)Dn−1 + · · · + a1(x)D + a0(x)]

︸︷︷︸y = F(x), Dy ≡

d y

dx

φ(D)

. ∴ φ(D) y = F(x)

34

Linear Ordinary Differential Equations © Wei-Chau Xie

Procedure for Finding the General Solution

φ(D) y = F(x), φ(D) = an Dn + an−1 Dn−1 + · · · + a1 D + a0

1. Find a particular solution yP(x) of the differential equation:

φ(D) yP = F(x) Particular solution

where the subscript P stands for particular solution.

2. Find the general solution yC(x) of the complementary differential equation:

φ(D) yC = 0 Complementary solution

where the subscript C stands for complementary solution.

3. Add yP(x) and yC(x) to obtain the general solution y(x):

y(x) = yC(x) + yP(x) General solution

35

Linear Ordinary Differential Equations — Complementary Solutions © Wei-Chau Xie

Procedure for Finding the Complementary Solution

For the nth-order linear differential equation

φ(D) y = F(x), φ(D) = anDn+an−1Dn−1+ · · · +a1D+a0

set the right-hand side to zero to obtain the complementary DE

φ(D) y = 0

Replace D by λ to obtain the characteristic equation

φ(λ) = 0 or an λn + an−1 λn−1 + · · · + a1 λ + a0 = 0

Solve the characteristic equation, which is an algebraic equation, to find the

characteristic numbers (roots) λ1, λ2, . . . , λn.

Write the complementary solution yC using the characteristic numbers.

36


Forms of Complementary Solution

Having obtained the characteristic numbers, the complementary solution can be

easily written.

λ = a =⇒ yC = C eax

λ = α± iβ =⇒ yC = eαx(A cos βx + B sin βx)

If the characteristic equation φ(λ)=0 has a p-fold root λ = a

λ = a, a, . . . , a =⇒ yC =(

C0 + C1x + · · · + Cp−1x p−1)

eax

︸︷︷︸︸︷︷︸

p times polynomial of degree p−1

If φ(λ)=0 has a pair of complex roots λ = α±iβ of p-fold

λ = α±iβ, . . . , α±iβ =⇒ yC = eαx[(

A0 + A1x + · · · + Ap−1x p−1)

cos βx

+(

B0 + B1x + · · · + Bp−1x p−1)

sin βx ]︸︷︷︸︸︷︷︸

p times polynomials of degree p−1

37


Example

Solve y′′ − 9 y = 0

Using the D-operator, (D2 − 9) y = 0 =⇒ φ(D) = D2 − 9

The characteristic equation is φ(λ) = 0 =⇒ λ2 − 9 = 0 =⇒ λ = ±3

The complementary solution is yC = C1 e−3x + C2 e3x

Example

Solve (D2 + 2 D + 2) y = 0

The characteristic equation is λ2 + 4λ + 20 = 0

λ =−4 ±

√

42 − 4×1×20

2×1=

−4 ±√

−64

2= −2 ± i 4 i =

√−1

The complementary solution is yC = e−2x(A cos 4x + B sin 4x)

38


Example

Solve (D2 − 4D + 4) y = 0.

The characteristic equation is λ2−4λ+4 = 0 =⇒ (λ−2)2 = 0 =⇒ λ = 2, 2

The complementary solution is y(x) = (C0 + C1x) e2x

Example

Given the characteristic numbers of a DE, write the complementary solution.

1. λ = 2, 2, 2, 0, 0

yC = (C0 + C1 x + C2 x2) e2x + (D0 + D1 x)

2. λ = 1 ± i√

3, −2, −2, 0, −1

yC = ex(A cos√

3x + B sin√

3x) + (C0 + C1 x) e−2x + D0 + D1 e−x

3. λ = 1, 1, −1, ± i, ± i

yC = (C0 + C1 x) ex + D e−x + [(A0 + A1 x) cos x + (B0 + B1 x) sin x]

39

Linear Ordinary Differential Equations — Particular Solutions © Wei-Chau Xie

Method of Undetermined Coefficients

The method of undetermined coefficients assumes a form for a particular

solution, with coefficients to be determined, according to the form of the

right-hand side of the DE.

Corresponding to RHS F(x) Assumed Form of yP

(1) Polynomial of degree k Polynomial of degree k

(2) eαx C eαx

(3) sin βx, cos βx A cos βx + B sin βx

(4) eαx(

a0+a1x+ · · · +akxk)

cos βx, eαx[(

A0+A1x+ · · · +Akxk)

cos βx

eαx(

b0+b1x+ · · · +bkxk)

sin βx +(

B0+B1x+ · · · +Bkxk)

sin βx]︸︷︷︸︸︷︷︸

Polynomial of degree k Polynomial of degree k

The coefficients are determined by substituting the assumed particular

solution into the DE.

40


Example

Find yP for y′′ + 4 y′ − 2 y = 2x2 − 3x + 6.

RHS function F(x) = 2x2 − 3x + 6 Polynomial of degree 2

Assume yP = C2 x2 + C1 x + C0 Polynomial of degree 2

Substitute into the DE

yP′′ + 4 yP

′ − 2 yP = 2C2 + 4(2C2 x + C1) − 2(C2 x2 + C1 x + C0)

= −2C2 x2 + (8C2 − 2C1)x + (2C2 + 4C1 − 2C0)

= 2x2 − 3x + 6

Equate coefficients of corresponding terms

−2C2 = 2 =⇒ C2 = −1

8C2 − 2C1 = −3 =⇒ C1 = 12(8C2 + 3) = − 5

2

2C2 + 4C1 − 2C0 = 6 =⇒ C0 = C2 + 2C1 − 3 = −9

∴ yP = −x2 − 52

x − 9 Particular solution

41


Example

Solve (D2 − 3D + 2) y = 4 e3x.

Characteristic equation λ2−3λ+2 = 0 =⇒ (λ−1)(λ−2) = 0 =⇒ λ = 1, 2

yC = C1 ex + C2 e2x

Corresponding to F(x) = 4 e3x, assume yP = C e3x


(D2 − 3D + 2) yP = 9C e3x − 3 · 3C e3x + 2 · C e3x = 2C e3x = 4 e3x

∴ 2C = 4 =⇒ C = 2 =⇒ yP = 2 e3x

∴ y = yC + yP = C1 ex + C2 e2x + 2 e3x General solution

42


Example

Find yP for y′′ − y′ + y = 2 sin 3x.

Corresponding to F(x) = 2 sin 3x, assume yP = A cos 3x + B sin 3x


yP′′ − yP

′ + yP = (−9A cos 3x − 9B sin 3x) − (−3A sin 3x + 3B cos 3x)

+ (A cos 3x + B sin 3x)

= (−8A − 3B) cos 3x + (3A − 8B) sin 3x

= 2 sin 3x


−8A − 3B = 0

3A − 8B = 2

=⇒

A = − 1673

B = 673

∴ yP = − 1673

cos 3x + 673

sin 3x Particular solution

43


Example

Solve (D2 − 2D − 3) y = 4x − 5 + 6x e2x.

Characteristic equation λ2−2λ−3 = 0 =⇒ (λ+1)(λ−3) = 0 =⇒ λ = −1, 3

yC = C1 e−x + C2 e3x

Corresponding to F1(x) = 4x − 5, assume yP1 = A1 x + A0

Corresponding to F2(x) = 6x e2x, assume yP2 = (B1 x + B0) e2x

yP = yP1 + yP2 = A1 x + A0 + (B1 x + B0) e2x

yP′ = A1 + B1 e2x + 2(B1 x + B0) e2x, yP

′′ = 4B1 e2x + 4(B1 x + B0) e2x


(D2−2D−3) yP = [4B1e2x+4(B1 x+B0)e2x]−2[A1+B1e2x+2(B1 x+B0)e2x]

−3[A1 x+A0+(B1 x+B0)e2x]

= −3A1x − (2A1+3A0) + [(2B1−3B0) − 3B1 x]e2x

= 4x − 5 + 6x e2x

44



−3A1 = 4

2A1+3A0 = 5

2B1−3B0 = 0

−3B1 = 6

=⇒

A1 = −43

A0 = 13

(5 − 2A1) = 239

B0 = 23

B1 = −43

B1 = −2

∴ y = yC + yP

= C1 e−x + C2 e3x − 43

x + 239

−(

2x + 43

)

e2x General solution

45


Exceptions in the Method of Undetermined Coefficients

In general, if any of the normally assumed terms of a particular solution occurs

in the complementary solution, one must multiply these assumed terms by a

power of x which is sufficiently high, but not higher, so that none of these

assumed terms occur in the complementary solution.

Example

Given the complementary solution yC and the right-hand side F(x) of the DE, specify

the form of yP using the method of undetermined coefficients.

(1) yC = c1ex + c2e2x + (d0 + d1x)e−x, F(x) = 3ex + 2e3x − x e−x

F(x) Normally Assumed Form for yP Contained in yC Modification

3ex C1ex c1ex x·C1ex

2e3x C2e3x ---------- ----------

−x e−x (D0 + D1 x)e−x (d0 + d1x)e−x x2 ·(D0 + D1 x)e−x

∴ yP = x·C1ex + C2e3x + x2 ·(D0 + D1 x)e−x

46


(2) yC = c0 + c1 x + c2 x2 + a cos 2x + b sin 2x, F(x) = 2 − x + sin x − cos 2x


2 − x C0 + C1 x c0 + c1 x + c2 x2 x3 ·(C0 + C1 x)

sin x A1 cos x + B1 sin x ---------- ----------

− cos 2x A2 cos 2x+B2 sin 2x a cos 2x + b sin 2x x·(A2 cos 2x+B2 sin 2x)

∴ yP = x3 ·(C0 + C1 x) + A1 cos x + B1 sin x + x·(A2 cos 2x + B2 sin 2x)

(3) yC = (c0 + c1 x)ex + a cos x + b sin x, F(x) = ex + e−x − x sin x


ex C1 ex (c0 + c1 x)ex x2 ·C1 ex

e−x C2 e−x ---------- ----------

−x sin x (A0+A1x) cos x a cos x + b sin x x·[(A0+A1x) cos x

+(B0+B1x) sin x +(B0+B1x) sin x]

∴ yP = x2 ·C1 ex + C2 e−x + x·[(A0+A1x) cos x + (B0+B1x) sin x]

47


Example

Solve y′′ − 2 y′ + y = ex.

Characteristic equation λ2 − 2λ + 1 = 0 =⇒ (λ − 1)2 = 0 =⇒ λ = 1, 1

yC = (C0 + C1 x)ex

Corresponding to F(x) = ex, the normally assumed form for yP is Cex.

Since Cex is contained in yC, assume yP = x2 · Cex.

yP′ = 2Cxex + Cx2 ex, yP

′′ = 2Cex + 4Cxex + Cx2 ex.


yP′′ − 2 yP

′ + yP = (2Cex + 4Cxex + Cx2 ex) − 2(2Cxex + Cx2 ex) + Cx2 ex

= 2Cex = ex

2C = 1 =⇒ C = 12

=⇒ yP = 12

x2 ex

∴ y = yC + yP = (C0 + C1 x)ex + 12

x2 ex General solution

48

Applications of Linear Differential Equations © Wei-Chau Xie

Vibration of a Single Degree-of-Freedom System

F(t)m

x(t)

k

c m

x0(t) x(t)

k

c

Consider a single degree-of-freedom system consisting of a mass m, a spring

of stiffness k, and a viscous damper with damping coefficient c.

The motion is described by the horizontal displacement x(t) of mass m from

its equilibrium position.

The system is called single degree-of-freedom (DOF) because it needs one

variable to describe its motion.

49


1. Single DOF System under Externally Applied Force

F(t)

F(t)F(t)

m

x(t)

k

c

m

x, x, x

cx

kx

m

x(t)

k

c

Consider mass m. It moves towards the right by a displacement x(t).

The spring is stretched by x(t) =⇒ the spring pulls the mass back towards

the left and the spring force is kx.

The damping force is similar to the spring force but is proportional to the

velocity =⇒ the damping force is cx towards the left.


F :

m · x(t) = F(t) − kx − cx =⇒ mx + cx + kx = F(t)

50


2. Single DOF System under Base Excitation

m

x0(t) x(t)

k

c

m

x0(t)x(t)

k

c m

x, x, x

c(x−x0)

k(x−x0)

The support moves towards the right by a displacement x0(t).

Applications of this model include a structure under earthquake excitation or

a robotic arm under base movement.

Consider mass m. It moves towards the right by a displacement x(t).

Mass moves towards the right =⇒ the spring pulls it back towards the left.

The stretch of the spring is x(t)−x0(t) =⇒ the spring force is k(x−x0).

The damping force is similar to the spring force but is proportional to the

velocity =⇒ the damping force is c(x−x0) towards the left.

51


m

x0(t) x(t)

k

c

m

x0(t)x(t)

k

c m

x, x, x

c(x−x0)

k(x−x0)


F :

m · x(t) = −k(x−x0) − c(x−x0)

Let y(t) = x(t)−x0(t) be the relative displacement of the mass.

The equation of motion becomes

my + cy + ky = −mx0

where x0(t) is the base acceleration.

The loading on the mass created from base excitation is F(t) = −mx0(t),

which is proportional to the mass and the base acceleration.

52


Equation of motion of a single degree-of-freedom system

mx + cx + kx = F(t)

Divide the equation by m

x +c

mx +

k

mx =

F(t)

m, ω2

0 =k

m, 2ζω0 =

c

m

ω0 natural circular frequency of the system

ζ nondimensional damping coefficient

Equation of motion in standard form x + 2ζω0 x + ω20 x =

F(t)

m

Solution x(t) has two parts: complementary solution and particular solution.

Complementary solution xC(t) corresponds to F(t) = 0. The system is in

free (not forced) vibration, and the solution is the response of free vibration.

Particular solution xP(t) corresponds to the right-hand side of the DE or

the forcing term F(t) and is the response of forced vibration.

53


Free Vibration—Complementary Solution

Complementary DE x + 2ζω0 x + ω20 x = 0

Characteristic equation λ2 + 2ζω0λ + ω20 = 0 =⇒ λ = ω0

(

−ζ ±√

ζ 2−1)

Underdamped System 06ζ <1

Most engineering structures fall in this category with ζ usually less than 10%.

λ = ω0

(

−ζ ± i√

1−ζ 2)

= −ζω0 ± iωd

where ωd = ω0

√

1−ζ 2 is the damped natural circular frequency.

xC(t) = e−ζω0t(A cos ωd t + B sin ωd t)

A and B are determined from the initial conditions x(0) = x0 and x(0) = v0:

A = x0, B =v0+ζω0x0

ωd

∴ xC(t) = e−ζω0t(

x0 cos ωd t +v0+ζω0x0

ωd

sin ωd t)

, 06ζ <1.

54


Special Case: Undamped System ζ = 0, ωd = ω0

The response becomes

. xC(t) = x0 cos ω0t +v0

ω0

sin ω0t

. =√

x20+

( v0

ω0

)2{

x0√

x20+(v0/ω0)

2cos ω0t +

v0/ω0√

x20+(v0/ω0)

2sin ω0t

}

. ︸︷︷︸︸︷︷︸︸︷︷︸

. a cos ϕ sin ϕ

. = a cos(ω0t−ϕ) Harmonic function: amplitude a, phase angle ϕ

x0

acos(ω0t−ϕ)

xC(t)

t

Slope=v0 at t=0

Amplitude a

−a

a

Period T= 2πω0

ϕω0

55


Harmonic solution xC(t) = a cos(ω0t−ϕ)

Period T =2π

ω0

ω0 is called the natural circular frequency of the system, unit rad/sec.

The natural frequency is f =ω0

2π=

1

T, unit

cycles

secor Hz.

The maximum displacement is

max xC(t) = a =√

x20+

( v0

ω0

)2= amplitude of the motion

56


Underdamped Free Vibration

xC(t) = ae−ζω0t cos(ωd t−ϕ) a =√

x20 +

(v0+ζω0 x0ωd

)2, ϕ = tan−1

(v0+ζω0 x0ωd x0

)

cos(ωdt−ϕ)

ae−ζω0tcos(ωdt−ϕ)

xC(t)

t

t

−a

−1

1

−ae−ζω0t (Envelope)

Harmonic oscillatory response with exponentially decaying amplitude

ae−ζω0t

a

Td= 2πωd

ϕωd

57


Forced Vibration—Particular Solution

For an underdamped system with 0<ζ <1, the complementary solution or

the response of free vibration is

xC(t) = e−ζω0t(

x0 cos ωd t +v0+ζω0x0

ωdsin ωd t

)

The response decays exponentially and approaches zero as t→∞.

The complementary solution is called the transient solution.

Because its value becomes negligible after some time, its effect is small and is

not important in practice.

The particular solution xP(t) is associated with the right-hand side of the DE

and hence corresponds to forced vibration.

The particular solution is called the steady-state solution, because it is the

solution that persists when time is large.

58


Suppose F(t) = F0 sin �t. The particular solution is

.xP(t) =F0

m

1√

(ω20−�2)2+(2ζω0�)2

sin(�t−ϕ), ϕ = tan−1 2ζω0�

ω20−�2

Let r =�

ω0

=Excitation frequency

Undamped natural frequency= Frequency ratio

The amplitude of the response of forced vibration is

.∣∣xP(t)

∣∣

max=

F0

m

1√

(ω20−�2)2+(2ζω0�)2

=F0

k

1√

(1−r2)2+(2ζ r)2, ω2

0 =k

m

If dynamic effect is not considered one obtains

mx + cx + kx = F(t) = F0 sin �t =⇒ xstatic =F0

k

This is the static displacement of the system under the static force F0.

59


∣∣xP(t)

∣∣

max

xstatic

=1

√

(1−r2)2+(2ζ r)2= Dynamic Magnification Factor

0

1

2

3

4

5

6

7

0.5 1 1.5 2 2.5

ζ=0

ζ=0.1

ζ=0.2

ζ=0.3

�ω0

r=

Dyn

amic

Mag

nif

icat

ion

Fac

tor

(DM

F)

60


When r → 0 (excitation frequency � → 0), DMF → 1.

Dynamic excitation approaches a static force. The amplitude of dynamic

response approaches the static displacement.

When r → ∞ (excitation frequency � ≫ ω0), DMF → 0.

The dynamic response approaches zero.

Maximum value of DMF occurs whend(DMF)

dr= 0 =⇒ r ≈ 1, if ζ ≪1

Maximum value of DMF occurs when r ≈ 1 (� ≈ ω0)

DMFmax ≈ DMF∣∣

r=1=

1√

(1−r2)2+(2ζ r)2

∣∣∣∣

r=1

=1

2ζ

The smaller the damping coefficient, the larger the amplitudes of dynamic

response.

When ζ = 0 and � = ω0, a response of unbound amplitude occurs and the

system is in resonance.

61


Resonance

When ζ = 0 and � = ω0, the equation of motion becomes

xP + ω20 xP =

F0

msin ω0t =⇒ xP(t) = −

F0

2mω0

t cos ω0t

t

xP(t)

Excitation Frequency � = Natural Frequency of the System ω0

=⇒ Resonance

Amplitude of response of the system increases linearly with time.

62

The Laplace Transform © Wei-Chau Xie

The Laplace Transform

The Laplace transform is one of the most important integral transforms.

Applying the Laplace transform to a linear ODE with constant coefficients

converts it into a linear algebraic equation.

The solution of the DE can then be obtained by determining the inverse Laplace

transform.

The method of Laplace transform is particularly advantageous in solving linear

ODE with the RHS functions involving discontinuous and impulse functions.

Definition — Laplace Transform

Let f(t), t>0, be a given function. The Laplace transform F(s) of function f(t)

is defined by

F(s) = L{

f(t)}

=∫ ∞

0e−st f(t)dt, s>0

63


Example

Determine the Laplace transform of f(t) = 1.

F(s) = L{

1}

=∫ ∞

0e−st · 1 dt = −

1

se−st

∣∣∣

∞

t=0=

1

s, s>0

Example

Determine the Laplace transforms of f(t) = sin ωt and f(t) = cos ωt.

. L{

eiωt}

=∫ ∞

0e−st · eiωtdt =

∫ ∞

0e−(s−iω)tdt = −

1

s−iωe−(s−iω)t

∣∣∣

∞

t=0

. =1

s−iω=

s + iω

(s−iω)(s+iω)=

s + iω

s2 + ω2, s>0

. ∵ cos ωt = Re(eiωt), sin ωt = Im(eiωt) eiωt = cos ωt+ i sin ωt

. ∴ L{

cos ωt}

= Re[L{

eiωt}

] =s

s2 + ω2

. L{

sin ωt}

= Im[L{

eiωt}

] =ω

s2 + ω2

64


Properties of the Laplace Transform

1.. Laplace transform L { · } is a linear operator

2.. Laplace Transform of Derivatives

L{

f (n)(t)}

= snF(s) − sn−1 f(0) − sn−2 f ′(0) − · · · − s f (n−2)(0) − f (n−1)(0)

3.. Property of Shifting

L{

eat f(t)}

= F(s−a) = L{

f(t)}∣∣∣

s→(s−a)

4.. Property of Differentiation

L{

tn f(t)}

= (−1)nF(n)(s) = (−1)n dnF(s)

dsn, n = 1, 2, . . .

65


Example

Determine the Laplace transform of f(t) = eat.

Solution 1: F(s) = L{

eat}

=∫ ∞

0e−st · eatdt Definition

. =∫ ∞

0e−(s−a)tdt = −

1

s−ae−(s−a)t

∣∣∣

∞

t=0=

1

s−a, s>a

Solution 2: F(s) = L{

eat · 1}

= L{

1}∣∣

s→(s−a)Property of Shifting

. =1

s

∣∣∣

s→(s−a)=

1

s−a

66


Example

Determine the Laplace transform of f(t) = tn, n=0, 1, . . . .

Using the formula of Laplace transform of derivative

. L{

f ′(t)}

= s L{

f(t)}

− f(0)f(t) = tn

==⇒ L{

n tn−1}

= s L{

tn}

− 0n

L{

tn}

=n

sL

{

tn−1}

. n = 1 : L{

t}

=1

sL

{

1}

=1

s2, L

{

1}

=1

s

. n = 2 : L{

t2}

=2

sL

{

t}

=2

s·

1

s2=

2·1s3

. n = 3 : L{

t3}

=3

sL

{

t2}

=3

s·

2·1s3

=3·2·1

s4

. · · · · · ·

In general, L{

tn}

=n!

sn+1, n = 0, 1, . . . 0!=1

67

The Heaviside Step Function © Wei-Chau Xie

The Heaviside Step Function

The Heaviside step function is defined by

H(t−a) ={

0, t<a

1, t>aa is a real number

t

a

H(t−a)

1

0

. L{

H(t−a)}

=1

se−as, s>0

. L{

f(t−a) H(t−a)}

= e−asF(s), s > a > 0

68


Example

Evaluate (a) L{

(t−2)3 H(t−2)}

(b) L{

sin t H(t−2π)}

(a) L{

(t−2)3 H(t−2)}

= e−2sL

{

t3}

= e−2s 3!s4

=6

s4e−2s

(b) L{

sin t H(t−2π)}

= L{

sin(t−2π) H(t−2π)}

. = e−2πsL

{

sin t}

=e−2πs

s2 + 1

69


The Heaviside step function is very useful in dealing with functions with

discontinuities or piecewise smooth functions.

f(t) ={

0, t< t0

f1(t), t>t0

= f1(t)H(t−t0)

tt0

f (t)

f1(t)

f(t) ={

f2(t), t< t0

0, t>t0

= f2(t)[1−H(t−t0)]

tt0

f (t)

f2(t)

f(t) =

0, t<a

g(t), a< t<b

0, t>b

= g(t)[H(t−a) − H(t−b)]

ta b

f (t)

g(t)

70

Impulse Functions © Wei-Chau Xie

Impulse Functions

Consider an impulse function f(t) over a time interval t0< t< t0+ε with constant

amplitude I/ε as shown in Figure (a).

tt0 t0+ε1ε1

f (t)

ε1

I

(a) (b) (c)

Area=I

tt0 t0+ε2ε2

f (t)

ε2

IArea=I

tt0

f (t)

Area=I

ε→0

The area under the function f(t), or the impulse, is I.

f(t) =I

ε

{

H(t−t0) − H[t−(t0+ε)]}

When ε decreases as in Figures (a) and (b)

The width of the time interval over which the impulse is defined decreases.

The amplitude I/ε of the function increases.

The area under the function remains constant.

71

The Dirac Delta Function © Wei-Chau Xie

Dirac Delta Function

In the case of I = 1, the limiting function as ε→0,

limε→0

f(t) = limε→0

H(t−t0) − H[t−(t0+ε)]ε

≡ δ(t−t0)

is called the unit impulse function or the Dirac delta function.

Properties of the Dirac Delta Function

ta

δ(t−a)

Area=1

1. δ(t−a) = 0, if t 6=a.

2. δ(t−a) → +∞, as t→a.

3.

∫ a+α

a−α

δ(t−a)dt = 1, α>0.

4. Shifting Property If g(t) is any function,∫ a+α

a−α

g(t) δ(t−a)dt = g(a), α>0

72

The Dirac Delta Function © Wei-Chau Xie

5.

∫ t

−∞δ(t−a)dt = H(t−a) =

{

0, t<a

1, t>a=⇒

dH(t−a)

dt= δ(t−a)

6. L{

δ(t−a)}

=∫ ∞

0e−st · δ(t−a)dt = e−as, a>0 Shifting property

7. L{

f(t) δ(t−a)}

=∫ ∞

0e−st· f(t)δ(t−a)dt = e−as f(a), a>0

δ(t−a) is not a function in the usual sense.

It is known as a generalized function.

73

The Inverse Laplace Transform © Wei-Chau Xie

The Inverse Laplace Transform

Given the Laplace transform F(s) of function f(t), F(s)=L{

f(t)}

, the inverse

Laplace transform is f(t)=L−1

{

F(s)}

.

Properties of the Inverse Laplace Transform

1. L{

·}

is a linear operator =⇒ L−1

{

·}

is a linear operator

2. L{

eat f(t)}

= F(s−a) =⇒ L−1

{

F(s−a)}

= eatL

−1{

F(s)}

= eat f(t)

3. L{

tn f(t)}

= (−1)nF(n)(s) =⇒ L−1

{

F(n)(s)}

= (−1)n tn f(t)

4. L{

f(t−a)H(t−a)}

= e−as F(s) =⇒ L−1

{

e−as F(s)}

= f(t−a) H(t−a)

74


Example

Evaluate L−1

{

F(s)}

, where F(s) =2s

s2+4+

1

s3−

4

s−1.

. L−1

{

F(s)}

= L−1

{2s

s2+4

}

+ L−1

{1

s3

}

− L−1

{4

s−1

}

. = 2 L−1

{s

s2+22

}

+ 12

L−1

{2!s3

}

− 4 L−1

{1

s−1

}

. = 2 cos 2t + 12

t2 − 4 et

75


Example

Evaluate L−1

{s

s2 + 6s + 11+

1

(s−3)3

}

.

.L −1

{s

s2 + 6s + 11+

1

(s−3)3

}

= L−1

{(s+3) − 3

(s+3)2 + 2

}

+ 12

L−1

{2

(s−3)3

}

. = L−1

{s+3

(s+3)2 + 2

}

− 3 L−1

{1

(s+3)2 + 2

}

+ 12

L−1

{2

(s−3)3

}

L−1

{

F(s−a)}

= eatL

−1{

F(s)}

. = e−3tL

−1

{s

s2 + (√

2)2

}

−3e−3t

√2

L−1

{ √2

s2 + (√

2)2

}

+ 12

e3tL

−1

{2!s3

}

. = e−3t cos√

2t − 3√2

e−3t sin√

2t + 12

e3t t2

76


Example

Evaluate L−1

{e−πs

s2+9

}

.

. ∵ L−1

{1

s2+9

}

= 13

L−1

{3

s2+32

}

= 13

sin 3t

. ∴ L−1

{e−πs

s2+9

}

= 13

sin 3(t−π) H(t−π)

L−1

{

e−as F(s)}

= f(t−a) H(t−a)

. = − 13

sin 3t H(t−π)

77


Theorem — Convolution Integral

If L−1

{

F(s)}

= f(t) and L−1

{

G(s)}

= g(t), then

L−1

{

F(s)G(s)}

=(

f ∗ g)

(t) =∫ t

0f(u)g(t−u)du =

∫ t

0g(u) f(t−u)du

in which the integral is known as a Convolution Integral.

Example

Evaluate L−1

{1

s2(s2+1)

}

.

. ∵ L−1

{1

s2

}

= t, L−1

{1

s2+1

}

= sin t

. ∴ L−1

{1

s2(s2+1)

}

= L−1

{1

s2·

1

s2+1

}

= t ∗ sin t =∫ t

0(t−u) sin u du

. = −∫ t

0(t−u) d(cos u) = −(t−u) cos u

∣∣∣

t

u=0+

∫ t

0cos u du = t − sin t

78

Solving Differential Equations Using the Laplace Transform © Wei-Chau Xie

Solving Differential Equations Using the Laplace Transform

Consider an nth-order linear ODE with constant coefficients

an y (n)(t) + an−1 y (n−1)(t) + · · · + a1 y ′(t) + a0 y(t) = f(t)

L{

y(t)}

= Y(s)

L{

y ′(t)}

= sY(s) − y(0)

L{

y ′′(t)}

= s2 Y(s)) − s y(0) − y ′(0)

· · · · · ·L

{

y(n)(t)}

= sn Y(s) − sn−1 y(0) − sn−2 y ′(0) − · · · − s y(n−2)(0) − y(n−1)(0)

Taking the Laplace transform on both sides of the equation leads to an

algebraic equation for Y(s)

an

[

snY(s) −n∑

i=1

sn−i y(i−1)(0)

]

+ an−1

[

sn−1Y(s) −n−1∑

i=1

sn−1−i y(i−1)(0)

]

+ · · ·

+ a1[sY(s) − y(0)] + a0Y(s) = L{

f(t)}

79


Solving for Y(s) yields

Y(s) =L

{

f(t)}

+n∑

k=1

k∑

i=1

ak y(i−1)(0)sk−i

n∑

i=0

ai si

Taking the inverse transform y(t)=L−1

{

Y(s)}

gives the solution of the DE.

80


Example

Solve the initial value problem (IVP) y ′′ + 4 y = 4t, y(0) = 1, y ′(0) = 5.

Let Y(s)=L{

y(t)}

. Taking the Laplace transform of both sides of the DE yields

. [s2 Y(s) − s y(0) − y ′(0)] + 4Y(s) = 4 ·1

s2

. (s2 + 4)Y(s) =4

s2+ s + 5

Solving for Y(s) leads to

. Y(s) =4

s2(s2+4)+

s

s2+4+

5

s2+4

. =( 1

s2−

1

s2+4

)

+s

s2+4+

5

s2+4=

1

s2+

s

s2+4+

4

s2+4

Taking the inverse Laplace transform gives

. y(t) = L−1

{

Y(s)}

= L−1

{1

s2

}

+ L−1

{s

s2+22

}

+ 2 L−1

{2

s2+22

}

. = t + cos 2t + 2 sin 2t

81


Example

Solve the IVP y ′′ + 2 y ′ + 5 y = f(t), t > 0, y(0) = 1, y ′(0) = 3, where

f(t) ={

1, 0 6 t < 2

0, t > 2

Using the Heaviside step function, f(t) can be written as

. f(t) = 1 − H(t−2) =⇒ L{

f(t)}

= L{

1}

− L{

H(t−2)}

=1

s−

e−2s

s

Taking the Laplace transform of both sides of the DE yields

. [s2 Y(s) − s y(0) − y ′(0)] + 2[sY(s) − y(0)] + 5Y(s) =1−e−2s

s

. (s2 + 2s + 5)Y(s) =1−e−2s

s+ s + 5

Solving for Y(s) leads to

.Y(s) =1−e−2s

s(s2+2s+5)+

s+5

s2+2s+5=

1−e−2s

5

{ 1

s−

s+2

s2+2s+5

}

+(s+1) + 4

(s+1)2+4

. =1−e−2s

5

{ 1

s−

(s+1)

(s+1)2+22−

1

(s+1)2+22

}

+(s+1)

(s+1)2+22+

2 · 2

(s+1)2+22

82


. ∵ L−1

{1

s−

(s+1)

(s+1)2+22−

1

2·

2

(s+1)2+22

}

= t − e−t(

cos 2t − 12

sin 2t)

L−1

{

F(s−a)}

= eatL

−1{

F(s)}

. ∴ L−1

{

e−2s

[1

s−

(s+1)

(s+1)2+22−

1

2·

2

(s+1)2+22

]}

. ={

(t−2) − e−(t−2) [ cos 2(t−2) − 12

sin 2(t−2)]}

H(t−2)

L−1

{

e−as F(s)}

= f(t−a) H(t−a)


. y(t) = L−1

{

Y(s)}

. = 15

{

[t − e−t(

cos 2t − 12

sin 2t)

]

. −{

(t−2) − e−(t−2) [ cos 2(t−2) − 12

sin 2(t−2)]}

H(t−2)

}

. + e−t cos 2t + 2e−t sin 2t

83


Example

Solve the IVP y ′′ + y = 4 δ(t−2π), y(0) = 1, y ′(0) = 0.

Let Y(s)=L{

y(t)}


. [s2 Y(s) − s y(0) − y ′(0)] + Y(s) = 4 L{

δ(t−2π)}

. (s2 + 1)Y(s) = 4 e−2πs + s

Solving for Y(s) leads to Y(s) =4 e−2πs

s2+1+

s

s2+1


. y(t) = L−1

{

Y(s)}

= L−1

{4 e−2πs

s2+1

}

+ L−1

{s

s2+1

}

. = 4 sin(t−2π) H(t−2π) + cos t

. L−1

{

e−as F(s)}

= f(t−a) H(t−a)

84


Example

Solve the IVP y ′′ − 2 y ′ + 5 y = 2 δ(t−3), y(0) = 0, y ′(0) = 1.

Let Y(s)=L{

y(t)}


. [s2 Y(s) − s y(0) − y ′(0)] − 2[sY(s) − y(0)] + 5Y(s) = 2 L{

δ(t−3)}

. (s2 − 2s + 5)Y(s) = 2 e−3s + 1 =⇒ Y(s) =2 e−3s + 1

(s−1)2+22


. ∵ L−1

{2

(s−1)2+22

}

= et sin 2t L−1

{

F(s−a)}

= eat f(t)

. ∴ y(t) = L−1

{

Y(s)}

= L−1

{

e−3s 2

(s−1)2+22

}

+1

2L

−1

{2

(s−1)2+22

}

. = et−3 sin 2(t−3) H(t−3) + 12

et sin 2t

L−1

{

e−as F(s)}

= f(t−a) H(t−a)

85


Example

Find y(t) where y(t) = t3 +∫ t

0sin(t−u) y(u) du.

Let Y(s)=L{

y(t)}

. Take the Laplace transform of both sides of the equation

. L{

y(t)}

= L{

t3}

+ L

{ ∫ t

0sin(t−u) y(u) du

}

Using the convolution integral

. L

{ ∫ t

0sin(t−u) y(u) du

}

= L{

sin t}

L{

y(t)}

=1

s2+1Y(s)

. ∴ Y(s) =3!s4

+1

s2+1Y(s) =⇒

(

1 −1

s2+1

)

Y(s) =6

s4

Solving for Y(s) leads to Y(s) =s2+1

s2·

6

s4=

6

s4+

6

s6

. ∴ y(t) = L−1

{

Y(s)}

= L−1

{3!s4

}

+ 120

L−1

{5!s6

}

= t3 + 120

t5

86


Example

Find y(t) where y(t) = 3t2 − e−t −∫ t

0et−u y(u) du.

Let Y(s)=L{

y(t)}

. Take the Laplace transform of both sides of the equation

. L{

y(t)}

= 3 L{

t2}

− L{

e−t}

− L

{ ∫ t

0et−u y(u) du

}

. ∵ L

{ ∫ t

0et−u y(u) du

}

= L{

et}

L{

y(t)}

=1

s−1Y(s) Convolution integral

. ∴ Y(s) = 3·2!s3

−1

s+1−

1

s−1Y(s) =⇒

(

1 +1

s−1

)

Y(s) =6

s3−

1

s+1

. ∴ Y(s) =s−1

s

( 6

s3−

1

s+1

)

=6

s3−

6

s4−

1

s+1+

1

s(s+1)

. =6

s3−

6

s4−

1

s+1+

( 1

s−

1

s+1

)

. ∴ y(t) = L−1

{

Y(s)}

= 3L−1

{2!s3

}

− L−1

{3!s4

}

+ L−1

{1

s

}

− 2L−1

{1

s+1

}

. = 3t2 − t3 + 1 − 2e−t

87


Example

Determine the current i(t) in the series RLC circuit shown when R = 2 �,

L = 0.1 H, C = 0.1 F, and V(t) = 120t−120t H(t−1).

V(t)

R

C

L

i V(t)

0

120

120 t

t 1

For t 6 0, V(t) = 0. The capacitor behaves as an open circuit =⇒ i(0) = 0.

For t > 0, apply Kirchhoff ’s Voltage Law

. −V(t) + Ri + Ldi

dt+

1

C

∫ t

−∞i(u)du = 0

. 0.1di

dt+ 2i + 10

∫ t

−∞i(u)du = 120t − 120t H(t−1)

88


Let I(s)=L{

i(t)}

.

.L

{ ∫ t

−∞i(u)du

}

= L

{ ∫ t

−∞i(u) · 1du

}

. = L{

1}

L{

i(t)}

=1

sI(s) Convolution integral

.L{

t H(t−1)}

= L{

[(t−1)+1] H(t−1)}

. = L{

(t−1) H(t−1)}

+ L{

H(t−1)}

=e−s

s2+

e−s

s

Take the Laplace transform of both sides of the equation

.0.1 [s I(s) − i(0)] + 2 I(s) + 10·1

sI(s) = 120·

1

s2− 120

( e−s

s2+

e−s

s

)

.

(

0.1 s + 2 +10

s

)

I(s) = 120( 1−e−s

s2−

e−s

s

)

. ∵ 0.1 s + 2 +10

s=

0.1 s2 + 2 s + 10

s=

s2 + 20 s + 100

10 s

89


. ∴ I(s) =1200 s

s2 + 20 s + 100

( 1−e−s

s2−

e−s

s

)

. = 1200

{1 − e−s

s (s+10)2−

e−s

(s+10)2

}

. ∵ L−1

{1

s

}

= 1, L−1

{1

(s+10)2

}

= e−10t t

. ∴ L−1

{1

s (s+10)2

}

=∫ t

0u e−10u · 1 du Convolution integral

. =−10u − 1

(−10)2e−10u

∣∣∣

t

0= −

10t + 1

100e−10t +

1

100

. L−1

{e−s

s (s+10)2

}

={

−10(t−1) + 1

100e−10(t−1) +

1

100

}

H(t−1)

. L−1

{e−s

(s+10)2

}

= e−10(t−1)(t−1) H(t−1)

90


Taking the inverse Laplace transform gives the solution to the DE

.i(t) = L−1

{

I(s)}

. = 1200

{

−10t+1

100e−10t +

1

100−

[

−10(t−1)+1

100e−10(t−1) +

1

100

]

H(t−1)

. − (t−1) e−10(t−1) H(t−1)

}

. = 12{

1 − (10t+1) e−10t − H(t−1) − [90(t−1) + 1] e−10(t−1) H(t−1)

}

91

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Ordinary Differential Equations · 2020-03-19 · The solution of an ordinary differential equation satisfying a set of conditions is a particular solution. Example Consider the differential