Prog 25 Second-Order Differential Equations

7/27/2019 Prog 25 Second-Order Differential Equations

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Worked examples and exercises are in the textSTROU

PROGRAMME 25

SECOND-ORDER

DIFFERENTIAL

EQUATIONS


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Programme 25: Second-order differential equations

Introduction

Homogeneous equations

The auxiliary equation

Summary

Inhomogeneous equations


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Introduction



Summary



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Introduction

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

are solutions to the quadratic equation:

with the properties:

2 2

1 24 4and2 2

b b ac b b acm ma a

+ = =

2 0am bm c+ + =

1 2 1 2and

b cm m m m

a a+ = =


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Introduction

The differential equation:

can be re-written to read:

that is:

2

2

0d y dy

a b cydx 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

+ + =


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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

+ = = =


7/26 Worked examples and exercises are in the textSTROU


Introduction


has solution:

This means that:

That is:

2 0dz

m z

dx

=

2

1

m x

dyz m y

dx

Ce

=

=

2 : being the integration constantm xz Ce C=

2

1

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

= +

= + =

2

1

m xdy m y Ce

dx

=




Introduction



Summary



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Is asecond-order, constant coefficient, linear, homogeneous differential

equation. Its solution is found from the solutions to the auxiliary equation:

These are:

2

2 0

d y dy

a b cydx dx+ + =

2

0am bm c+ + =

2 2

1 2

4 4and

2 2

b b ac b b acm m

a a

+ = =


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Introduction



Summary



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Real and different roots

Real and equal roots

Complex roots


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Real and different roots

If the auxiliary equation:

with solution:

where:

then the solution to:

2 0am bm c+ + =

2 2

1 2

4 4and

2 2

b b ac b b acm m

a a

+ = =

1 2 1 2and are real andm m m m

1 2

2

20 is

m x m xd y dya b cy y Ae Be

dx dx+ + = = +


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Real and equal roots


with solution:

where:

then the solution to:

2 0am bm c+ + =

2 2

1 2

4 4and

2 2

b b ac b b acm m

a a

+ = =

1 2 1 2and are real andm m m m=

1

2

20 is ( )

m xd y dya b cy y A Bx e

dx dx+ + = = +


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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 4and

2 2

b b ac b b acm m

a a

+ = =

1 2and arem m complex

1 2andm j m j = + =


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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

+

= +

= +


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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

+ = + + = +


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Introduction



Summary



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Summary

Differential equations of the form:

Auxiliary equation:

Roots real and different: Solution

Roots real and the same: Solution

Roots complex (j): Solution

2

2

0 where , and are contantsd y dy

a b cy a b cdx dx

+ + =

2

1 20 with roots andam bm c m m+ + =

1 2m x m xy Ae Be= +

1( )m x

y A Bx e= +

( )cos sinxy e C x D x = +


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Introduction



Summary



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The second-order, constant coefficient, linear, inhomogeneous differential

equation is an equation of the type:

The solution is in two partsy1 +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 theparticular integral.

2

2( )d y dya b cy f x

dx dx+ + =


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Complementary function

Example, to solve:

(a) Complementary function

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

Complementary functiony1

=Ae2x +Be3x where:

22

25 6

d y dyy x

dx dx + =

2

1 112

5 6 0d y dy

ydx dx

+ =


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Particular integral

(b)Particular integral

Assume a form fory2 asy2 = Cx2 +Dx +Ethen 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 = + +


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Complete solution

(c) The complete solution to:

consists of:

complementary function + particular integral

That is:

2

2 3

1 25 19

6 18 108

x x x xy y y Ae Be= + = + + + +

22

2

5 6d y dy

y xdx dx

+ =


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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 cossinh or cosh sinh cosh

kx kx

f x y

k C

kx Cx D

kx Cx Dx E

k x k x C x D xk x k x C x D x

e Ce

++ +

++

K

K

K

K


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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


Prog 25 Second-Order Differential Equations

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