Notes Differential Equations

8/6/2019 Notes Differential Equations

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Differential Equations 1

Table of Contents

1.1 Classification of differential equations.............................................. 2

1.2 Verifying a solution to a differential equation ................................... 3

1.3 Solving differential equations ............................................................ 4

1.4 Type I.................................................................................................. 5

1.5 Type II ................................................................................................ 6

1.6 Type III ............................................................................................... 9

1.7

Growth and decay problems............................................................. 10

1.8 The law of natural decay .................................................................. 12

1.9 Newton's law of cooling.................................................................. 14

1.10 Electrical problems ......................................................................... 16

1.11 Integrating factor............................................................................. 18

1.12 Homogeneous first order................................................................. 20

1.13 Second order differential equations ................................................ 22

1.14 Second order constant coefficients ................................................. 231.15 Second order constant coefficients non-homogeneous................... 29

Written and typed

by R.Rozen 2008


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1.1 Classification of differential equations

A differential equation ( often abbreviated to simply d.e. ) is merely an equation

involving the differential co-efficients. It is of the form

2

2( , , , ,.....) 0dy d yf x y

dx dx=

It contains the unknown functiony the dependant variable,x the independant variable and

the various derivatives.

We will consider only ORDINARY differential equations that is where we have

functions of only one variable ( )y f x= . We can further classify differential equationsaccording to their order of degree.

The ORDER of a differential equations is the order of the highest derivative present.

The DEGREE of a differential equations is the degree of the highest power of the highest

derivative.

A LINEAR differential equations is one which is linear in y and various derivatives.

Some examples of differential equations are

(a)dy

a bxdx

= + (b)2

2

20

d yx

dx+ =

(c) 2x tx x t + = (d)2

20

d y

dx=

(e)

3

3 5 0dy dy

x ydx dx

+ + =

(f)

3 2 1tD x x= +

Note that (a) and (e) are first order, while (b),(c) and (d) are second order, (f) is thirdorder, (e) has a degree of 3, all others have a degree of one, while all are linear except for

(e) and (f). Notice that there are many different notations for derivatives,2

2

d xx

dt= and

22

2t

d xD x

dt=

Differential equations are extremely important in the study of mathematics and appear in

almost every branch of science.


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Example

If ( )cos 4y Ax x= satisfies the differential equation ( )2

216 16sin 4

d yy x

dx+ =

find the value ofA

Solution

Given ( )cos 4y Ax x= we need to finddy

dx, differentiating using the product rule

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

2

2

2

2

cos 4 4 sin 4 differentiating again

4 sin 4 4 sin 4 16 cos 4 substituting

16 8 sin 4 16 cos 4 16 cos 4 8 sin 4 16sin 4

so it follows from equating coefficients that 2

dyA x Ax x

dx

d yA x A x Ax x

dx

d yy A x Ax x Ax x A x x

dx

A

=

=

+ = + = =

=

1.3 Solving differential equations

The solution of a differential equation can be complicated and is usually obtained by theprocess of integration. Since the integration process produces an arbitrary constant of

integration, the solution of differential equation are classified as follows,

A GENERAL SOLUTION is one which contains arbitrary constants and satisfies the

differential equation.

A PARTICULAR SOLUTION is one which satisfies the differential equation and some

other initial value condition ( or boundary value ) which enable the constants ofintegration to be found.

In general the number of arbitrary constants to be found is equal to the order of thedifferential equation.

Throughout this course we will study only special types of differential equations.


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First Order Differential Equations

1.4 Type I

This is of the form

0 0

0 0

( ) ( ) integrating wrt gives

( ) this is the general solution

to find , use when this is then the particular solution.

dyf x y x y x

dx

y f x dx C

C x x y y

= =

= +

= =

Example

Expressy in terms ofx given 6 0 (1) 2dy

x ydx

+ = =

Solution

Given 6 0 (1) 2dy

x ydx

+ = = it follows that 6dy

xdx

= integrating once wrtx gives

2

2

6

3 this is the general solution

to find the value of use (1) 2 this means that when 1 2 substituting

2 3 so 5

3 5 is the particular solution

y x dx

x C

C y x y

C C

y x

=

= +

= = =

= + =

= +

Example

Expressy in terms ofx given2

1(0) 0

(3 2 )

dyy

dx x= =

Solution

This is of Type I2

1

(3 2 )

dy

dx x=

integrating wrtx gives

( )2

3 2

dxy

x=

to find this integral let


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12

2 2 11 12 2

1 16 6

3 2 so 2 or so

1

2

1this is the general solution

2(3 2 )

to find use 0 when 0

0 so

1 1 3 (3 2 ) 2

2(3 2 ) 6 6(3 2 ) 6(3 2 )

3(3

duu x dx du

dx

y u dx u du u C C u

y Cx

C x y

C C

x xy

x x x

xy

= = =

= = = + = +

= +

= =

= + =

= = =

=

is the particular solution2 )x

1.5 Type II

This is of the form

0 0( ) ( )

this is solved by first inverting both sides

1

( )

integrating wrt gives

this is the general solution( )

Ofcourse we should try to re-arrange to make the subject.

to find

dyf y y x y

dx

dx

dy f y

y

dyx C

f y

y

= =

=

= +

0 0, use when this is then the particular solution.C x x y y= =

Example

Expressy in terms ofx given 3 0 (0) 5dy

y ydx

+ = =


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Solution

This is of Type II, rewriting as 3 then inverting both sides givesdy

ydx

=

1integrating wrt gives

3

1

3

1ln

3

dxy

dy y

dyx

y

x y C

=

=

= +

At this point we have two alternatives to follow

(i) First find the value ofCthen transpose to make y the subject(ii) First transpose to makey the subject then find the value ofC

Method I

[ ]

1ln first find then rearrange

3

Now 5 when 0 so that

1 10 ln5 so that ln5 ( do not evaluate ) substituting gives

3 3

1 1ln ln 53 3

1ln 5 ln

3

1 5ln using log laws

3

53 ln

x y C C

y x

C C

x y

x y

xy

xy

= +

= =

= + =

= +

=

=

= 3

3

3

5or inverting again gives

y

5

so 5

x

x

x

e

ye

y e

=

=

=


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Method II

3 3 3

1ln rearranging first then find

3

1

ln3

ln 3( )

ln 3 3

ln 3

where 3 since is a constant so is

in index form gives .

where since is a constant

B x B x x

B

x y C C

y C x

y C x

y C x

y B x

B C C B

y e e e Ae

A e B

= +

= =

=

=

=

= = =

=

3

so is

Now to find when 5 0 it follows that 5

so 5 as beforex

A

A y x A

y e

= = =

=

Example

Expressy in terms ofx given 2(3 2 ) (0) 0dy

y ydx

= =

Solution

2

2

2 1

22

2 11 12 2

1(3 2 ) (0) 0 inverting integrating wrt

(3 2 )

where 3 2 Now 2 so(3 2 )

1 1to find

2 2(3 2 )

when 0 0 so substitut

dy dxy y y

dx dy y

dy dux u dy u y dy du

y dy

x u du u C C C C u y

x y

= = =

= = = = =

= = + = + = +

= =

1 1ing 0 so

6 6

1 1transposing to make the subject

2(3 2 ) 6

1 1

6 2(3 2 )

6 1 1 3inverting 3 2

6 2(3 2 ) 6 1

3 3(6 1) 3 182 3

6 1 6 1 6 1

9

6 1

C C

x yy

xy

xy

y x

x xy

x x x

xy

x

= + =

=

+ =

+ = = +

+ = = =

+ + +

=+


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1.6 Type III

Variables separable

This is of the form

0 0

0 0

( ) ( ) ( )

this is solved by separating the variables

( ) this is the general solution( )

If possible we should try to re-arrange to make y the subject.

to find , use when

dy f x g y y x ydx

dyf x dx C

g y

C x x y y

= =

= +

= =

this is then the particular solution.

Example

Expressy in terms ofx given 26dy y x ydx + =

Solution

rewriting as 2 26 (6 1)dy

x y y y xdx

= = this is now of Type III, with variables separable

3 3

2

3

2 2

(6 1)

ln 2 or

wherex x C x x C

dyx dx

y

y x x C

y e Ae A e +

=

= +

= = =

as the general solution.


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Applications of first order differential equations

1.7 Growth and decay problems

If ( )N N t = represents the population number at a certain time t, then the law of naturalgrowth states that the rate of increase of population is proportional to the current

population at that time. This leads to the equation

so thatdN dN

N kN dt dt

=

where kis a positive constant. Assuming that the initial population number is

0 0then the particular solution of this differential equation isktN N N e=

In the absence of such factors as wars, famine etc, this equation has been found to model

population growth, but only over a limited time frame, as the model predicts that as time

increases the population number increases without bounds, that is as t N .The constant kcan be interpreted as the excess birth rate over the death rate. We have

consideredNto be a continuous variable but is in fact a discrete quantity, this is of littleconsequence. It is easy to verify the given solution or integrate to solve the differentialequation similar to previous examples. The model is called exponential growth.

Example

The population of a certain city increases at a rate proportional to the current population.

In 1970 the population was half a million and in 1980 the population was one million,

express the population numberNin terms oftthe time in years after 1970.i) What is the predicted population in 2000?

ii) When does the population reach five million?

( )N t

time

N


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Solution

IfNrepresents the population number tyears after 1970 then

0 0

0

the particular solution is where and are constants

to be found, since in 1970 0, 500, 000 (0) 500, 000to find the value of in 1980, 10 (10) 1,000,000

( ) 500,000 so

kt

kt

dNkN N N e N k

dt

t N N N k t N

N t e

= =

= = = == =

=10

10

0.0693

30 0.0693

substituting gives

1,000,000 500,000

2

1ln(2) 0.0693 thus

10

( ) 500, 000

i) in the year 2000, 30 so (30) 500, 000 4, 000, 000there are 4 million people in 2000

ii) now to fi

k

k

t

X

e

e

k

N t e

t N e

=

=

=

=

= = =

0.0693

0.063

nd when 5 million

5,000,000 500,000

10

0.0693 ln(10)

1ln(10) 33.22 this works out to be in 2003 late February.

0.0693

t

t

t N

e

e

t

t

=

=

=

=

=

the model predicts that the population will double every 10 years.

million


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1.8 The law of natural decay

In the mass of a radioactive substance, atoms disintegrate spontaneously, although the

process is not a continuous one but averaged over the large number of atoms in a

specimen, it is found that the time rate of decay is proportional to the mass of the

radioactive substance present at that time.If we let ( )m m t= be the mass of the radioactive substance at a time t,

0

0

then so that the particular solution is

where (0) is the initial mass present.

ktdm dmm km m m edt dt

m m

= =

=

This is the law of exponential decay, as , 0t m , an infinite time is required for allof the radioactive material to disintegrate. For this reason the rate of disintegration isoften measured in terms of the half-life, that is the time required for half of the original

mass to disintegrate, to find the half-life T, we need to find 1 02? whent m m= =

Substituting gives1

0 0 02

12

1

dividing by gives

or 2 taking logs gives ln(2) so that

ln(2)

kT

kT kT

k

m m e m

e e kT

T

== = =

=

Notice that this half-life formula does not depend upon the initial mass 0m and is thus

independent of the time when the particular observations have begun. Half-lives forradioactive substances can range from milliseconds to centuries.

m(t)

m0

time


13/37


Example

The rate of disintegration of a radioactive substance is proportional to the amount of the

substance remaining at that time. If it takes three years for 20% to disintegrate, find the

half-life of the substance.

Solution

0 0

0

3

0 0

3

Let ( ) denote the mass present after years

so that where (0) is the initial mass present

Now when 3 20% disintegrates so 80% remains then 0.8

0.8

0.8

3 l

kt

k

k

m m t t

dmkm m m e m m

dt

t m m

m m e

e

k

=

= = =

= =

=

=

=13

1

n(0.8)

ln(0.8) 0.0744

ln(2)so the half life ln(2) 9.32 years

0.0744k

k

T

=

= = =

Example

The rate of decay of a radioactive substance is proportional to the amount of the

substance present at that time. Initially 50 milligrams of a radioactive substance is present

and after one hour it is observed that 10% has disintegrated. Find the amount remainingafter a further two hours.

Solution

0 0

Let ( ) denote the mass present after years

so that where (0) 50 mg is the initial mass present

Now when 1 10% is lost so 45 mg remains

45 50

45

5050

ln 0.45

kt

k

k

m m t t

dmkm m m e m m

dt

t m

e

e

k

=

= = = =

= =

=

=

=

0.1053

0.1053 3

1053

so ( ) 50

Now after a further two hours, means when 3

(3) 50 36.457

so 36.46 mg remains

t

X

m m t e

t

m e

= =

=

= =


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1.9 Newton's law of cooling

Newton's Law of cooling states that the time rate of change of the temperature of a body

is proportional to the difference between the temperature of the body and the surrounding

medium. Let Tdenote the temperature of a body after a time t, and Tm denote the

temperature of the surrounding medium, then Newton's Law of cooling can be written as

( ) ( )

( )

0

0 0

where is a constant of proportionality.

If we let be the difference in the temperature, so that then

the general solution of this is where

0

m m

m

kt

dT dT T T or k T T k

dt dt

T T

dk e

dt

T

=

=

= =

= = 0and is the initial temperature of the body.mT T

Example

A metal ball is heated to a temperature of 200oCand is then placed in a room, which is

maintained at a constant temperature of 30oC. After five minutes the temperature of the

ball has dropped to 150oC, assuming Newton's law of cooling,

i) find the temperature of the ball after a further 10 minutes

ii) how long before the temperature of the ball reaches 40oC

Solution

( )

0

0

the surrounding room temperature is constant at 30

Let 30, then the general solution is

We need two sets of conditions, to find the two constants, now

0, 200

o

m m

kt

o

dTk T T T C

dt

dT k e

dt

t T C

= =

= = =

= = 0

5

5

0.0696

so 170 and when

5, 150 so 120 substituting gives

120 170

120

170

125 ln17

1 12ln 0.0696

5 17

so 170

o

k

k

t

t T C

e

e

k

k

e

=

= = =

=

=

=

=

=


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0.0696 15

i) after a further ten minutes means we are required to find when 15

(15) 170 59.79 so the temperature of the ball is 30 or 89.79X o

T t

e T C

=

= = = +

0.0696

0.0696

ii) we are required to find the value of when 40 that is 10

10 170

10

170

100.0696 ln

170

1 10ln 40.67

0.0696 170

after 40.67 minutes the temperature of the ball reache

t

t

t T

e

e

t

t

= =

=

=

=

= =

os 40 C

Example

A cold can of beer is taken from the refrigerator ( which is kept at a constant temperatureof 3oC) and placed in a warm summer's room at a temperature of 30oC. If after two

minutes the temperature of the can is 4oC, find its temperature after a further three

minutes.

Solution

0

0 0

2

2

( ) the surrounding room temperature is constant at 30

Let 30, then the general solution is

0, 3 so 27 and when

2, 4 so 26

26 27

26

27

2

o

m m

kt

o

o

k

k

dTk T T T C

dtd

T k edt

t T C

t T C

e

e

k

= =

= = =

= = =

= = =

=

=

=

0.0189

26ln

27

1 26ln 0.0189

2 27

so 27 t

k

e

=

=

0.0189 5

after a further three minutes means we are required to find when 5

(5) 27 24.57 so the temperature of the can is 30 or 5.43X o

T t

e T C

=

= = = +


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1.10 Electrical problems

Example

The charge Q coulombs at a time tseconds in a capacitor of capacitance CFarads when

discharging through a resistance ofR ohms, satisfies the differential equation0

dQRC Q

dt+ =

i) Assuming an initial charge of 0Q solve the differential equation to obtain the

charge Q at any time tafter discharging commences.

ii) A circuit contains a resistance of 400k and a capacitance of 7.3 F and after225 milliseconds, the charge falls to 7.0 Coulombs, find the initial charge.

iii) After how long is the charge half its initial value?

Solution

i) 0 transposing gives

Now and are constants, the variables here are and , inverting both sides gives

1 1separating the variables and integrating gives

1

dQ dQRC Q RC Qdt dt

R C Q t

dt

RC dQ Q

dQdt

RC

+ = =

=

=

0

0 0

00

so that

ln( ) where is the contant of integration, to find use when 0

0 ln( ) so that ln( ) substituting gives

ln( ) ln( ) using log laws ln in

Q

tQ K K K Q Q t

RC

Q K K Q

Qt tQ Q

RC RC Q

= + = =

= + =

= + =

00

0

0

0.077

0

index form

inverting both sides so that ( )

ii) given 7.3 and 400 , the product 2.92,

to find use 0.225 when 7.0 substituting gives

7.0 so t

t t t

RC RC RC Q Q

e e Q Q t Q eQ Q

C F R k RC

Q t s Q C

Q e

= = = =

= = =

= =

= 0.077 0.3401

2

0.341 12 2

hat 7.0 7.56 so ( ) 7.56 for 0

iii) to find ? when 3.78so 0.34 ln( )

1ln(2) 2.023s

0.34

t

o

t

Q e C Q Q t e t

t Q Qe t

t

= = = =

= = == =

= =


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Example

The basic equation governing the amount of current i Amperes in anRL series circuitconsisting of a resistanceR ohms and an inductanceL Henries connected to a voltage

source ofEvolts after a time tseconds satisfies the differential equationdi

L Ri E dt

+ =

Assuming the initial current is zero, find the current at any time t.

Solution

transposing gives

Now , and are constants while and are the variables, inverting

1 1 1separating the variables gives integrating gives

1ln

di diL Ri E L E Ri

dt dt

R L E i t

dt didt

L di E Ri L E Ri

tE R

L R

+ = =

= =

=

( )

[ ]

where is the constant of integration,

1 1to find use 0 when 0 substituting gives 0 ln( ) so that ln( )

1 1 1 1ln( ) ln( ) ln( ) ln( ) ln by log laws

transpos

i C C

C i t E C C E R R

t EE Ri E E E Ri

L R R R R E Ri

+

= = = + =

= + = =

ing ln so that inverting to make the subject

so

1

( ) 1 typical values are 20 V 4 an

Rt

L

Rt Rt

L L

Rt Rt L L

Rt

L

Rt E E e i

L E Ri E Ri

E Rie E Ri Ee

E

Ri E Ee E e

Ei i t e E R

R

= =

= =

= =

= = = =

d 2 HL =


18/37


1.11 Integrating factor

This method is applicable to first-order differential equations of the form

( ) ( ) ( ) ( )where and are given functions ofdy

p x y q x p x q x xdx

+ = .

If we can multiply both sides of the above equation by a suitable function ( )r x giving

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( )( ) ( ) ( )

( )

now if then by the product rule

this is now a first-order separable differential equation which can be solved

directly, then substitute back for to obtain

dyr x r x p x y r x q x

dx

r x p x r x

dr x y r x q x

dx

r x

+ =

=

=

( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )( ) ( )

the unknown function

Now must satisfy or

that is log

so that where

now the constant can be omitted as it is a multiplying

e

p x dx C p x dx c

y y x

r x r x p x r xdr dr

r x p x r x p x dx C dx r

r x e Ae A e

A

+

=

=

= = = +

= = =

( )( )

factor,

the function is called the integrating factor.p x dx

r x e=

Example Expressy in terms ofx given ( )24 3 0 2xdy

y e ydx

= =

Solution

( )

( ) ( )

( )

( )

2

2

4 4

4 2 4 6

4 6 6 4

4 3 0 2

here 4 and 3

so the differential equation can be written as

3 3 so that

1

3 multiplying both sides by giv2

x

x

dx x

x x x x

x x x x

dyy e y

dx

p x q x e

r x e e

dye e e e

dx

ye e dx e C e

= =

= =

= =

= =

= = +

( )

( )

2 4

4 2

es

1to find use

2

1 50 2 so that

2 2

15

2

x x

x x

y e Ce C

y C C

y e e

= +

= = + =

=

Often we may need to use the results


19/37


log

log log

log

e

ne e

e

x

n x x n

a n x a n

e x

e e x

e e x+

=

= =

=

Example

Expressy in terms ofx given

( )42 4 1 3dy

x y x ydx

= =

Solution

( )

( ) ( )

( )

( )

2

4

3

3

2

2log log 2

2

2 3 2

2

2 4 1 3

we need to divide by and write the

differential equation in the form2

4 now in this form

2and 4

1

4 . 4

4

e e

dxx xx

dyx y x y

dx

x

dyy x

dx x

p x q x xx

r x e e e xx

dyx x x x

dx

yxdx

x

= =

=

= =

= = = = =

= =

= =

( )

2

4 2

4 2

2

2 to find use

1 3 2 so that 1

2

x C

y x Cx C

y C C

y x x

+

= +

= = + =

= +


20/37


1.12 Homogeneous first order

The first order differential equation of the form ( ),dy

f x ydx

= can be reduced to a first

order separable type if ( ),y

f x y g x

= that is if we lety

v x= or

( )

. then differentiating using the product rule

in this case the differential equation becomes

we can then solve this using the variables separable techniques

and back sub

y v x

dy dvv x

dx dx

dvv x g v

dx

=

= +

+ =

( )stitute for , to obtain the solution for

Sometimes it may not be possible to obtain an explicit solution.

v y y x=

Example Solve ( )22 1 2dyx y y ydx

= =

Solution

( )

( )

( )

2

2

2

2 2

2

2 1 2

1 22 dividing by , provided that 0 gives let .

1 2 2

2 cancelling ' gives 2 now separating the variables

dyx y y y

dx

y ydy dyx y y x x y v x

dx dx x

dy dvv x v vx v v x

dx dxdv dv

x v x x s vdx dx

dv

v

= =

+= + = =

= + = + = +

= =

=

( ) ( )

52

2

12

1

2

to find use2

1 51 2 so that 2 2 1 hence

2 2

2

2 5 4

dx

x Cv

yv

x x C

xy Cx C

y C C C

x xy

x x

= +

= =

+

= +

= = + = = +

= =


21/37


Example

Expressy in terms ofx given ( )2 2 22 1 2dy

x x y ydx

= + =

Solution

( )

( )

( )

2 2 2

2 22

2

2 2 2

2

22 2

2

2 1 2

dividing by gives2

let .

1

2 2

11 1 2separating the variables gives

2 2 22

1

2 2 2log to f

1 1ey

x

dyx x y y

dx

dy x yx

dx x

y x v

dy dv x y vv x

dx dx x

vdv v v vx v

dxdv dx

xv

xC C C x

v x y

= + =

+=

=

+ += + = =

+ += = =

=

+ = + = + =

( )

( )

( )

ind use 1 2

2log 1 0 so that 2

1 2

22 log transposing to make the subject

2 log 2 inverting

1

2 log 2

2

log 2

log 2 22

log 2 log 2

log 4

log 2

e

e

e

e

e

e

e e

e

e

C y

C C

xx y

x y

x xx y

x y

x x

xx y

x

x x xxy x

x x

x xy

x

=

+ = = =

+ =

=

=

=

= =

=


22/37


1.13 Second order differential equations

Type IV, this is of the form2

2( )

d yf x

dx= this type of differential equation can easily be solved by direct integration

once, integrating both sides gives

1( )dy

f x dx C dx

= + this is now of the form Type I, and can be solved by integratingagain, however two sets of initial conditions must be given to find the two arbitrary

constants of integration.

Example

Given2

236 0 '(1) 2 (1) 3

d yx y y

dx+ = = = expressy in terms ofx

Solution

2

2

2

1

1

1 1

2

transposing gives 36 integrating wrt gives

36 18

to find the value of '(1) 2 means 1 when 2 substituting gives

2 18 so that 20, now

18 20 i

d yx x

dx

dyxdx x C

dx

dyC y x

dx

C C

dyx

dx

=

= = +

= = =

= + =

= +

2 3

2

2

2 2

3

ntegrating again wrt gives

( 18 20) 6 20

to find the value of (1) 3 means 1 when 3 substituting gives

3 6 20 so that 11

6 20 11

x

y x dx x x C

C y x y

C C

y x x

= + = + +

= = =

= + + =

= +


23/37


1.14 Second order constant coefficients

In this section we solve second order differential equations with constant coefficients of

the form2

20

d y dya b cy

dx dx+ + = where a, b and c are real constants. Since this is a

second order differential equation it must have two linearly independent solutions and

assuming a solution exists it must be of the form mxy Ae= whereA and m are constantsto be found. Now if

22

2then differentiating and differentiating againmx mx mx

dy d yy Ae mAe m Ae

dx dx= = =

substituting into2

20

d y dya b cy

dx dx+ + = gives

( )

2

2

2

0 taking out the common factor of

0 since 0 since 0 we want a non-trivial solution

it follows that 0 this equation is called the auxilary e

mx mx mx mx

mx mx

am Ae bmAe cAe Ae

Ae am bm c Ae y

am bm c

+ + =

+ + =

+ + = quation.

Now to find the values ofm we merely need to solve the quadratic equation.

However there are three cases to consider, depending upon the sign of the discriminant2 4b ac =

Case I

If 2 4 0b ac = > then there are two real distinct roots say 1 2andm m m m= = and in this case the generalsolution of the differential equation is of the form

1 2

1 2

m x m xy C e C e= + where 1 2andC C are the constants of integration and can only befound if there is extra information called initial values or boundary values.

Case II

If 2 4 0b ac = = then the two roots are equal say 1m m= and in this case to get two linearly independentsolutions it can be shown that the general solution of the differential equation is of theform

( ) 11 2m xy C C x e= + where 1 2andC C are again the constants of integration and can only

be found if there is extra information called initial values or boundary values.


24/37


Case III

If 2 4 0b ac = < then since a,b and c are real, the two roots are complex conjugatesof each other and can be written in the form say 1 2andm i m i = + =

where

24and2 2

b ac b

a a

= =

In this case the general solution of the differential equation is similar to case I( ) ( )

( )

1 2

1 2

1 2

1 2

but this can be simplified

using index laws

taking out the common factor of

now using the definition of

i x i x

x i x x i x

x i x x i x x

x i x i x

y C e C e

y C e C e

y C e e C e e e

y e C e C e

+

+

= +

= +

= +

= +

( ) ( )( ) ( ) ( )( )( )

( ) ( ) ( ) ( )( )( ) ( )( )

1 2

1 2 1 2

1 2

the complex exponential

cos sin cos sin simplifying

cos sincos sin

the constants and are real since and are complex conjugates.

x

x

x

y e C x i x C x i x

y e C C x i C C xy e A x B x

A B C C

= + +

= + + = +

The following worked examples will illustrate all these possible cases and some special

situations which can apply.

Summary

To solve2

20

d y dya b cy

dx dx+ + = Let mxy Ae=

2Then solve the auxilary equation 0am bm c+ + =

Case I

If 2 4 0b ac = > 1 2

1 2

m x m xy C e C e= +

Case II

If 2 4 0b ac = =

( ) 11 2m xy C C x e= +

Case IIIIf 2 4 0b ac = <

1 2andm i m i = + =

( ) ( )( )1 2cos sinxy e C x C x = +

where 1 2andC C are the constants of integration and can only be found if there is extra

information called initial values or boundary values.


25/37


Example Expressy in terms ofx given ( ) ( )2

25 36 0 0 5 0 8

d y dyy y y

dx dx+ = = =

Solution

( ) ( )

( ) ( )

2

2

2

5 36 0 0 5 0 8

Let substituting gives the auxilary equation

5 36 0

9 4 0 this corresponds to case I with two distinct roots

4, 9 in this case the general solution can

mx

d y dyy y y

dx dxy Ae

m m

m m

m

+ = = =

=

+ =

+ =

=

( ) ( )

( ) ( )

( )

4 9

1 2 1 2

1 2

4 9

1 2

1 2

1 2 1

1 2

be written as

to find and

using 0 5 1

differentiating 4 9

using 0 8 4 9 2

9 x 1 45 9 9 adding gives 13 53

53 12and the particular so

13 13

x x

x x

y C e C e C C

y C C

dyC e C e

dxy C C

C C C

C C

= +

= = +

=

= =

= + =

= =

( )4 9

lution is

153 12

13

x xy e e= +


25 0 0 5 0 8

d y dyy y

dx dx+ = = =

Solution

( ) ( )

( )

2

2

2

5 0 0 5 0 8


5 0

5 0 this corresponds to case I with two distinct roots

0, 5 in this case the general solution can be written

mx

d y dyy y

dx dx

y Ae

m m

m m

m

+ = = =

=

+ =

+ =

=

( )

( )

( )

5

1 2 1 2

5

1 2 2

2 2 1

5

as

to find and

using 0 5 differentiating 5

8 33using 0 8 5 so that and

5 5

1the particular solution is 33 8

5

x

x

x

y C C e C C

dyy C C C edx

y C C C

y e

= +

= = + =

= = = =

=


26/37



236 0 0 5 0 8

d yy y y

dx = = =

Solution Method I

( ) ( )

( )( )

2

2

2

36 0 0 5 0 8


36 0

6 6 0 this corresponds to case I with two distinct roots

6, 6 in this case the general solution can be written

mx

d yy y y

dxy Ae

m

m m

m

= = =

=

=

+ =

=

( ) ( )

( ) ( )

( )

( )

6 6

1 2 1 2

1 2

6 6

1 2

1 2

1 2 1

1 2

6 6

as

to find and

using 0 5 1

differentiating 6 6

using 0 8 6 6 2

6 x 1 30 6 6 adding gives 12 38

19 11and

6 6

119 11

6

x x

x x

x x

y C e C e C C

y C C

dyC e C e

dxy C C

C C C

C C

y e e

= +

= = +

=

= =

= + =

= =

= +

Solution Method II

( ) ( )

( )( )

( ) ( )

2

2

2

1 2 1

36 0 0 5 0 8


36 0

6 6 0

6, 6 alternatively this can be written in terms of hyperbolic functions

sinh 6 cosh 6 to find

mx

d y y y ydx

y Ae

m

m m

m

y C x C x C

= = =

=

=

+ =

=

= +

( )

( ) ( )

( )

( ) ( )

2

2

1

1 1

and

using 0 5

differentiating 6 cosh 6 30 sinh 6

4using 0 8 6

3

4sinh 6 5cosh 6

3

C

y C

dy C x xdx

y C C

y x x

= =

= +

= = =

= +


27/37


Example Expressy in terms ofx given

( ) ( )2

236 0 0 5 0 8

d yy y y

dx+ = = =

Solution

( ) ( )

( )( )

2

2

2

36 0 0 5 0 8


36 0

6 6 0 this corresponds to case III with 0

6 in this case the general solution can be written in terms

mx

d y y y ydx

y Ae

m

m i m i

m i

+ = = =

=

+ =

+ = =

=

( ) ( )

( )

( ) ( )

( )

( ) ( )

1 2 1 2

2

1 2

1 1

of trigonometric functions

sin 6 cos 6 to find and

using 0 5

differentiating 6 cos 6 6 sin 6

4using 0 8 6

3

4sin 6 5cos 6

3

y C x C x C C

y C

dy C x C xdx

y C C

y x x

= +

= =

=

= = =

= +

Example Expressy in terms ofx given

( ) ( )2

212 36 0 0 5 0 8

d y dyy y y

dx dx+ + = = =

Solution

( ) ( )

( )

2

2

2

2

12 36 0 0 5 0 8


12 36 0

6 0 this corresponds to case II with

6 as a repeated factor, in this case the general solution ca

mx

d y dyy y y

dx dx

y Ae

m m

m

m

+ + = = =

=

+ + =

+ =

=

( )( )

( )

( )

( )

6

1 2 1 2

1

6 6

2 1 2

2 1 2

6

n be written as

to find and

using 0 5

differentiating using the product rule 6

using 0 8 6 so that 38

5 38

x

x x

x

y C C x e C C

y C

dyC e C C x e

dx

y C C C

y x e

= +

= =

= +

= = =

= +


28/37


Example


( ) ( )2

24 13 0 0 5 0 8

d y dyy y y

dx dx+ + = = =

Solution

( ) ( )

( )

2

2

2

2

2 2

4 13 0 0 5 0 8


4 13 0 completing the square

4 4 13 4

2 9 9

2 3

2 3 this corresponds to case III and the genera

mx

d y dyy y y

dx dx

y Ae

m m

m m

m i

m i

m i

+ + = = =

=

+ + =

+ + = +

+ = =

+ = =

( ) ( )( )

( )

( ) ( )( ) ( ) ( )( )

( )

2

1 2 1 2

1

2 2

1 2 1 2

1 2

l solution can be written as

cos 3 sin 3 to find and

using 0 5

differentiating using the product rule

2 cos 3 sin 3 3 sin 3 3 cos 3

using 0 8 2 3 so t

x

x x

y e C x C x C C

y C

dye C x C x e C x C x

dx

y C C

= +

= =

= + + +

= = +

( ) ( )( )

2

2

hat 6 the particular solution is

5cos 3 6sin 3x

C

y e x x

=

= +


29/37


1.15 Second order constant coefficients non-homogeneous

In this section we solve second order differential equations with constant coefficients of

the form ( )2

2

d y dya b cy f x

dx dx+ + = where a, b and c are real constants and ( )f x is now

a non zero given function ofx. Since we have already solved the corresponding second

order homogeneous differential equation2

20

d y dya b cy

dx dx+ + = when the right hand side

was identically zero, it follows that the solution of the homogeneous case is still a

solution of the non-homogenous case. Let the solution of2

20

d y dya b cy

dx dx+ + = be

hy y= the homogeneous part of the solution is called the complementary function and

the solution of ( )2

2

d y dya b cy f x

dx dx+ + = is h py y y= + where py y= is called the

particular solution or integral of the non-homogenous equation.To determine the form of py we use the method of undetermined coefficients, by

guessing the form of py up to arbitrary multiplicative constants.

These arbitrary constants are then evaluated by substituting the proposed solution into the

given differential equation and equating coefficients. Again there are a couple of general

cases to consider, depending on the form for ( )f x which can be a polynomial,

exponential or trigonometric function.

Case I

( ) ( )

1

1 1 0

an th degree polynomial in .

Assume a solution of the form

......

n

n n

p n n

f x p x n x

y A x A x A x A

=

= + + + +

Case II

( )


x

x

p

f x Ce

y Ae

=

=

Case III

( ) ( ) ( )

( ) ( )

sin cos


sin cosp

f x C x D x

y A x B x

= +

= +


30/37


The aim is to equate coefficients and find the values of As andBs given the Cs and

Ds. All of the above will work fine provided that the form of ( )f x is not already part

of the solution of the homogeneous case hy y= . If any form of the assumed solution for

py y= is already part of hy y= then the assumed solution must be modified and

multiplied by a factor ofx, ( possibly even 2x ) such that the assumed solution with

the product ofxs now has no terms in common with hy y=

Notice that to determine the constants of integration, first we find the general solution of

( )2

2

d y dya b cy f x

dx dx+ + = as h py y y= + then proceed to use the boundary values to

determine the constants 1 2andC C .

The following examples will illustrate the above cases.

Example

Expressy in terms ofx given ( ) ( )2

236 72 0 5 0 8

d yy y y

dx+ = = =

Solution

( ) ( )

( ) ( )

( )

2

2

2

2

1 2

36 72 0 5 0 8

first we solve the homogeneous equation 36 0

from previously sin 6 cos 6

Since 72 a constant, we now try

since 0 substituting gives

36 72

h

p

d yy y y

dx

d yy

dx

y C x C x

f x

dyy A

dx

A

+ = = =

+ =

= +

=

= =

=

( ) ( )

( )

( ) ( )

( )

( ) ( )

1 2

1 2

2 2

1 2

1 1

so that 2 the general solution is

2 sin 6 cos 6

Now to find and

using 0 5 2 3

differentiating 6 cos 6 6 sin 6

4using 0 8 6 so that

3

42 sin 6 3cos 6

3

A

y C x C x

C C

y C C

dyC x C x

dx

y C C

y x x

=

= + +

= = + =

=

= = =

= + +


31/37


Example

a) Find the general solution of2

3

25 36 21 x

d y dyy e

dx dx

+ =

b) Find the general solution of2

2

25 36 576

d y dyy x

dx dx

+ =

c) Hence find the general solution of2

3 2

25 36 84 288x

d y dyy e x

dx dx

+ =

Solution

2

2

4 9

1 2

In both parts a) and b) first we solve the

homogeneous equation 5 36 0

this has been solved previously to givethe general solution

x x

h

d y dyy

dx dx

y C e C e

+ =

= +

( )

( )

23 3

2

23 3 3

2

3 3

a) 5 36 21 Since 21 an exponential function

let then 3 and 9

substituting gives 9 15 36 21 so that

142 21 and the general solu

2

x x

x x x

p

x x

d y dyy e f x e

dx dx

dy d yy Ae Ae Ae

dx dx

Ae e

A A

+ = =

= = =

=

= = 4 9 31 21

tion is2

x x xy C e C e e = +

( )

( ) ( )

( ) ( )

22 2

2

22

2

2 2

2 2

b) 5 36 576 Since 576 a quadratic polynomial

let then 2 and 2

substituting gives 2 5 2 36 576

36 10 36 5 2 36 576 equating c

p

d y dy

y x f x xdx dx

dy d yy Ax Bx C Ax B A

dx dx

A Ax B Ax Bx C x

Ax x A B B A C x

+ = =

= + + = + =

+ + + + =

+ + + =

( )

2

4 9

1 2

oefficients of 's gives

5 4036 576 so that 16 coefficients of ' 10 36 0 and

18 9

1 1 200 122also 5 2 36 0 so that 5 2 32

36 36 9 811

the general solution is x x

x

A A x s A B B A

B A C C B A

y C e C e

= = = = =

+ = = + = =

= + 2

22 4016

81 9x x

23 2

2

4 9 3 2

1 2

c) 5 36 84 288

1the particular integral is 4 and the general solution is thus

2

61 202 8

81 9

x

pa pb

x x x

d y dyy e x

dx dx

y y

y C e C e e x x

+ =

= + + + +


32/37


Example


( ) ( ) ( )2

25 58sin 2 0 5 0 8

d y dyx y y

dx dx

+ = = =

Solution

( ) ( )

( ) ( )

( )

2

2

5

1 2

First solve the homogeneous solution 5 0

this has been solved previously the general solution is

Since 58sin 2 a trigonometric function

let sin 2 cos 2

2 cos 2

x

h

p

d y dy

dx dx

y C C e

f x x

y A x B x

dyA x

dx

+ =

= +

=

= +

= ( )

( ) ( )

( ) ( ) ( )( ) ( )

( ) ( )

( ) ( )

( )

( )

2

2

2 sin 2

4 sin 2 4 cos 2 substituting gives

sin 2 4 10 cos 2 4 10 58sin 2

4 10 58 1 coefficients of sin 2

4 10 0 2 coefficients of cos 2

2 x 1 8 20 116

5x 2 50 20 0 58 116 so that

B x

d yA x B x

dx

x A B x B A x

A B x

B A x

A B

A B A

=

+ + =

=

+ =

=

+ = =( ) ( )

( )

( ) ( )

( )

5

1 2 1 2

1 2 1 2

5

2

2

2 and 5

the general solution is 2sin 2 5cos 2 now to find and


differentiating gives 5 4cos 2 10sin 2

using 0 8 5 4

x

x

A B

y C C e x x C C

y C C C C

dyC e x x

dx

y C

= = = +

= = + = +

= +

= =

( ) ( ) ( )

2 1

5

12 62so that and

5 5

162 12 2sin 2 5cos 2

5

x

C C

y e x x

= =

=


33/37


Example


( ) ( ) ( )2

236 4cos 6 0 5 0 8

d yy x y y

dx+ = = =

Solution

( ) ( ) ( )

( ) ( )

2

2

2

2

1 2

36 4cos 6 0 5 0 8

first we solve the homogeneous equation 36 0

from previously the complementary function is sin 6 cos 6

since the form for the particular integral for

h

d yy x y y

dx

d yy

dx

y C x C x

+ = = =

+ =

= +

( ) ( )

( ) ( )( )

( ) ( )( ) ( ) ( )( )

( ) ( )( ) ( )2

2

4 cos 6

is already part of the complementary function we must

try sin 6 cos 6 differentiating using the product rule

6 cos 6 6 sin 6 sin 6 cos 6

36 sin 6 36 cos 6 6 cos 6

p

f x x

y x A x B x

dyx A x B x A x B x

dx

d yx A x B x A x

dx

=

= +

= + +

= + ( )( ) ( ) ( )( )

( ) ( )( ) ( ) ( )( )

( )

( ) ( )( ) ( ) ( )( ) ( ) ( )( )

( ) ( ) ( )

2

2

2

2

6 sin 6 6 cos 6 6 sin 6

36 sin 6 36 cos 6 12 cos 6 12 sin 6

substituting into 36 4cos 6 gives

36 sin 6 36 cos 6 12 cos 6 12 sin 6 36 sin 6 cos 6

12 cos 6 12 sin 6 4cos 6 equatin

B x A x B x

d yx A x B x A x B x

dx

d yy x

dxx A x B x A x B x x A x B x

A x B x x

+

= +

+ =

+ + +

= =

( ) ( ) ( )

( )

( )

1 2

1 2

2

1 2

1 1

1g coefficients and 0

3

the general solution is sin 6 cos 6 sin 63

Now to find and

using 0 5

1differentiating 6 cos6 6 sin 6 sin 6 2 cos 6

3

using 0 8 6 so that

A B

xy C x C x x

C C

y C

dyC x C x x x x

dx

y C C

= =

= + +

= =

= + +

= =

( ) ( ) ( )

4

3

14 sin 6 5cos 6

3y x x x

=

= + +


34/37


Example


( ) ( )2

6

212 36 6 0 5 0 8x

d y dyy e y y

dx dx

+ + = = =

Solution

( )

2

2

6

1 2

First consider the homogeneous case when 12 36 0

In this case the solution for the complementary function has previously found to be

since this already contains thex

h

d y dyy

dx dx

y C C x e e

+ + =

= +

( )

( ) ( ) ( )

6 6

2 6

6 2 6 2 6

26 2 6 6 2

2

and

for the particular integral we must now try

differentiating using the product rule

2 6 2 6

2 12 6 2 6 2 24 36

substit

x x

x

p

x x x

x x x

xe

y Ax e

dy xAe Ax e A x x edx

d yA x e A x x e Ae x x

dx

=

= =

= = +

( ) ( )

( )

( ) ( )

26

2

6 2 2 6 2 6 6 6

2 6

1 2

61 2 1 2

uting into 12 36 6 gives

2 24 36 12 2 6 36 2 6 so that 3

the general solution is 3

using 0 5 differentiating 6 6

x

x x x x x

x

x

d y dyy e

dx dx

Ae x x A x x e Ax e Ae e A

y C C x x e

dyy C C x e C C x

dx

+ + =

+ + + = = =

= + +

= = = + + +( )( )

( )

2 6

2 1 2

2 6

3


5 38 3

x

x

x e

y C C C

y x x e

= = =

= + +

Applications to electrical circuits

In anRLCseries circuit consisting of a resistorR ohms an inductanceL Henries and acapacitanceL Farads, the rise of current i Amperes at time tseconds satisfies

( )

( )

2

02

0

sin

where sin Volts is an external voltage source.

d i di iL R V t dt dt c

V t

+ + =


35/37


Example

In a particular circuit 2 010 0.5 10 12.5 and 5R L H C F V V = = = = = Find the

current assuming that initially the current is zero and its initial charging rate is also zero.

Solution

( )

( )

( )

2

2

2

2

2 2

2 2

0.5 10 100 12.5sin 5

20 200 25sin 5 let first finding the complementary function

20 200 0 completing the square 20 100 200 100

10 100 100 so that 10 10

mt

d i dii t

dt dt

d i dii t i Ae

dt dt

m m m m

m i m i

m

+ + =

+ + = =

+ + = + + = +

+ = = + =

( ) ( )( )( ) ( )

( ) ( )

10

1 2

2

2

10 10 this corresponds to case III

cos 10 sin 10 as the complementary function.

For the particular integral try sin 5 cos 5 differentiating gives

5 cos 5 5 sin 5 and

t

h

p

i

i e C t C t

i A t B t

di d iA t B t

dt dt

=

= += +

= ( ) ( )

( ) ( ) ( ) ( ) ( )

( )

( )

25 sin 5 25 cos 5 substituting gives

25 100 200 sin 5 25 100 200 cos 5 25sin 5 so that

25 100 200 25 coefficients of sin 5 and

25 100 200 0 coefficients of cos 5 the solution of these

A t B t

A B A t B A B t t

A B A t

B A B t

=

+ + + + =

+ =

+ + =

( ) ( ) ( )( ) ( ) ( )( )10 1 2

1 1

simultaneous equations is

7 4and so that the general solution is

65 651

cos 10 sin 10 7sin 5 4cos 565

4 4Now initially when 0 0 so that 0 this gives , now differenti

65 65

t

A B

i i t e C t C t t t

t i C C

= =

= = + +

= = = =

( ) ( )( ) ( ) ( )( )

( ) ( )( )

10 10

1 2 1 2

1 2 2

ating

10 cos 10 sin 10 10 sin 10 10 cos 10

135cos 5 20sin 5

65

Now initially when 0 0 the initial charging rate is also zero, this gives

35 10 10 10 so that t

65 130

t tdi e C t C t e C t C t dt

t t

di

t dt

C C C

= + + +

+ +

= =

= + + =

( ) ( ) ( )( ) ( ) ( )( )10

he particular solution is

18cos 10 sin 10 7sin 5 4cos 5

130 65

tei i t t t t t

= = + +


36/37


Applications to vibrating springs

Consider a particle of mass m attached horizontally to a spring, if it is acted upon by a

restoring force from the spring given by Hookes law as kx and a force due to airresistance being proportional to the velocity cx and a time dependant external force of

( )f t then by Newtons second law of motion we have that ( )mx kx cx f t = + or

( )

( )

2

2that is

d x dxm c kx f t

dt dt

mx cx kx f t

+ + =

+ + =

wherex is the displacement at a time t. If initially the displacement is

( ) ( )0 00 and the initial velocity is 0x x x v= = then we can find the particular solution of

the differential equation. There are a number of cases to consider

if ( ) 0 and 0f t c= = the motion is free and undamped,

if ( ) 0 and 0f t c= the motion is free and damped.

for damped motion the are three cases to consider as to the value of 2 4c mk =

if 2 4 0c mk = > the roots of the auxiliary equation are real and distinct, the motion isclassified as overdamped or strongly damped.

if 2 4 0c mk = = the roots of the auxiliary equation are equal then the motion isclassified as critically damped.

if 2 4 0c mk = < the roots of the auxiliary equation are complex conjugates then themotion is classified as oscillatory damped or weakly damped.

if ( ) 0f t the motion is forced and termed transient if as t the complementary

function component diminishes ( and dies away ) leaving only the forced component of

the particular integral component, called the steady state solution.


37/37


Example

A particle of mass 10 kilograms is attached to a spring having a stiffness of 140 N/m, if

( ) ( )90 and 5sinc f t t = = classify the motion and find solution if initially the mass is

started in the equilibrium position with a initial velocity of 1 m/s

Solution

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

2

2

2

becomes with 10 90 140 and 5sin

9 14 0.5sin 0 0 and 0 1

Letting gives the auxilary equation as

9 14 0

2 7 0 two distinct real roots, so the motio

mt

d x dxm c kx f t m c k f t t

dt dt

x x x t x x

x Ae

m m

m m

+ + = = = = =

+ + = = =

=

+ + =

+ + =

( ) ( )

2 7

1 2

n is overdamped

2 and 7as the homogenenous part of the solution.

To find the particular integral using the method of undetermined coefficients,

let sin cos it follows t

t t

h

p

m mx C e C e

x A t B t

= = = +

= +

( )( )

( )( )

( ) ( )

( ) ( )

( ) 2 71 2

hat

9 14 0.5 coefficient of sin and that

9 14 0 coefficient of cos

13 9 0.5 1 Now 13x 1 169 117 6.5

9 13 0 2 9x 2 81 117 0

13 9and that

500 500

the general solution is

1t t

A B A t

B A B t

A B A B

A B A B

A B

x x t C e C e

+ =

+ + =

= =

+ = + =

= =

= = + + ( ) ( )( )

( )

( ) ( )( )

( )

( )

1 2

2 7

1 2

1 2

1 2

2

13sin 9cos500

9Applying 0 0 gives 0

500

12 7 13cos 9sin

500

13and 0 1 2 7 solving gives500

90 99and the particular solution is

500 500

190 99

500

t t

t

t t

x C C

dxC e C e t t

dt

x C C

C C

x x t e

= = +

= + +

= = +

= =

= = +

( ) ( )( )7 13sin 9coste t t +

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Notes Differential Equations