Worked examples and exercises are in the text STROUD PROGRAMME 25 SECOND-ORDER DIFFERENTIAL EQUATIONS

Worked examples and exercises are in the textSTROUD

PROGRAMME 25

SECOND-ORDER DIFFERENTIAL

EQUATIONS


Programme 25: Second-order differential equations

Introduction

Homogeneous equations

The auxiliary equation

Summary

Inhomogeneous equations



Introduction



Summary




Introduction

For any three numbers a, b and c, the two numbers:

are solutions to the quadratic equation:

with the properties:

2 2

1 2

4 4 and

2 2

b b ac b b acm m

a a

2 0am bm c

1 2 1 2 and b c

m m m ma a



Introduction

The differential equation:

can be re-written to read:

that is:

2

20

d y dya b cy

dx dx

2

20 provided 0

d y b dy cy a

dx a dx a

2

20

d y b dy ca y

dx a dx a



Introduction

The differential equation can again be re-written as:

where:

2 2

1 2 1 22 2

1 2 1

2

0

d y b dy c d y dyy m m m m y

dx a dx a dx dxd dy dy

m y m m ydx dx dx

dzm z

dx

2 2

1 2 1

4 4, and

2 2

b b ac b b ac dym m z m y

a a dx



Introduction


has solution:

This means that:

That is:

2 0dz

m zdx

2

1

m x

dyz m y

dx

Ce

2 : being the integration constantm xz Ce C

21

m xdym y Ce

dx



Introduction


has solution:

where: and are constantsA B

1 2

1

1 2

1 2

: if

( ) : if

m x m x

m x

y Ae Be m m

A Bx e m m

21

m xdym y Ce

dx



Introduction



Summary






Is a second-order, constant coefficient, linear, homogeneous differential equation. Its solution is found from the solutions to the auxiliary equation:

These are:

2

20

d y dya b cy

dx dx

2 0am bm c

2 2

1 2

4 4 and

2 2

b b ac b b acm m

a a



Introduction



Summary





Real and different roots

Real and equal roots

Complex roots




Real and different roots

If the auxiliary equation:

with solution:

where:

then the solution to:

2 0am bm c

2 2

1 2

4 4 and

2 2

b b ac b b acm m

a a

1 2 1 2 and are real and m m m m

1 2

2

20 is m x m xd y dy

a b cy y Ae Bedx dx




Real and equal roots


with solution:

where:

then the solution to:

2 0am bm c

2 2

1 2

4 4 and

2 2

b b ac b b acm m

a a

1 2 1 2 and are real and m m m m

1

2

20 is ( ) m xd y dy

a b cy y A Bx edx dx




Complex roots


with solution:

where:

Then the solutions to the auxiliary equation are complex conjugates. That is:

2 0am bm c

2 2

1 2

4 4 and

2 2

b b ac b b acm m

a a

1 2 and are m m complex

1 2 and m j m j




Complex roots

Complex roots to the auxiliary equation:

means that the solution of the differential equation:

is of the form:

2 0am bm c

2

20

d y dya b cy

dx dx

( ) ( )j x j x

x j x j x

y Ae Be

e Ae Be




Complex roots

Since:

then:

The solution to the differential equation whose auxiliary equation has complex roots can be written as::

cos sin and cos sinj x j xe x j x e x j x

cos sinxy e C x D x

( )cos ( )sin

cos sin

j x j xAe Be A B x j A B x

C x D x



Introduction



Summary




Summary

Differential equations of the form:

Auxiliary equation:

Roots real and different: Solution

Roots real and the same: Solution

Roots complex ( j): Solution

2

20 where , and are contants

d y dya b cy a b c

dx dx

21 20 with roots and am bm c m m

1 2m x m xy Ae Be

1( ) m xy A Bx e

cos sinxy e C x D x



Introduction



Summary





The second-order, constant coefficient, linear, inhomogeneous differentialequation is an equation of the type:

The solution is in two parts y1 + y2:

(a) part 1, y1 is the solution to the homogeneous equation and is called the complementary function which is the solution to the homogeneous equation

(b) part 2, y2 is called the particular integral.

2

2( )

d y dya b cy f x

dx dx




Complementary function

Example, to solve:

(a) Complementary function

Auxiliary equation: m2 – 5m + 6 = 0 solution m = 2, 3

Complementary function y1 = Ae2x + Be3x where:

22

25 6

d y dyy x

dx dx

21 1

125 6 0

d y dyy

dx dx




Particular integral

(b) Particular integral

Assume a form for y2 as y2 = Cx2 + Dx + E then substitution in:

gives:

yielding:

so that:

222 2

225 6

d y dyy x

dx dx

2 26 (6 10 ) (2 5 6 ) 0 0Cx D C x C D E x x

1/ 6 : 5 /18 : 19 /108C D E

2

2

5 19

6 18 108

x xy




Complete solution

(c) The complete solution to:

consists of:

complementary function + particular integralThat is:

22 3

1 2

5 19

6 18 108x x x x

y y y Ae Be

22

25 6

d y dyy x

dx dx




Particular integrals

The general form assumed for the particular integral depends upon the form of the right-hand side of the inhomogeneous equation. The following table can be used as a guide:

2 2

( ) Assume

sin or cos sin cos

sinh or cosh sinh coshkx kx

f x y

k C

kx Cx D

kx Cx Dx E

k x k x C x D x

k x k x C x D x

e Ce


Learning outcomes

Use the auxiliary equation to solve certain second-order homogeneous equations

Use the complementary function and the particular integral to solve certain second-order inhomogeneous equations


Documents

Worked examples and exercises are in the text STROUD PROGRAMME 25 SECOND-ORDER DIFFERENTIAL EQUATIONS