12/19/2015 1 Non- homogeneous Differential Equation Chapter 4.

04/21/23 1Non- homogeneous Differential Equation Chapter 4

A differential equation of the type ay’’+by’+cy=0, a,b,c real numbers,is a homogeneous linear second order differential equation.

Second Order Differential EquationsDefinition

2e and e .mx mx mxy y me y m

2 20 e e e 0 0.mx mx mxay by cy am bm c am bm c

To solve the equation ay’’+by’+cy=0 substitute y = emx and try to determine m so that this substitution is a solution to the differential equation.

Compute as follows:

Homogeneous linear second order differential equations can always be solved by certain substitutions.

This follows since emx≠0 for all x.

Definition The equation am2 + bm+ c = 0 is the Characteristic Equation of the differential equation ay’’ + by’ + cy = 0.

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Solving Homogeneous 2nd Order Linear Equations: Case I

Equation

Case I

1 21 2e em x m xy C C

CE has two different real solutions m1 and m2.

ay’’+by’+cy=0 CE am2+bm+c=0

In this case the functions y = em1x and y = em2x are both solutions to the original equation.

General Solution

Example 0y y CE 2 1 0m 1 or m 1.m

1 2e ex xy C C General Solution

The fact that all these functions are solutions can be verified by a direct calculation.

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Solving Homogeneous 2nd Order Linear Equations: Case II

Equation

Case II

1 2e emx mxy C C x

CE has real double root m.

ay’’+by’+cy=0 CE

In this case the functions y = emx and y = xemx are both solutions to the original equation.

General Solution

Example 2 0y y y CE 2 2 1 0m m 1 (double root).m

1 2e ex xy C C xGeneral Solution

am2+bm+c=0

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Solving Homogeneous 2nd Order Linear Equations: Case III

Equation

Case III

1 2e sin e cosx xy C x C x

ay’’+by’+cy=0 CE

General Solution

Example 2 5 0y y y

CE 2 2 5 0m m 1 2m i

1 2e sin 2 cos 2xy C x C xGeneral Solution

CE has two complex solutions .m i

In this case the functions are both solutions to the original equation.

e sin and e cosx xy x y x

am2+bm+c=0

1 2e sin cosxy C x C x

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Real and Unequal Roots

If roots of characteristic polynomial P(m) are real and unequal, then there are n distinct solutions of the differential equation:

If these functions are linearly independent, then general solution of differential equation is

The Wronskian can be used to determine linear independence of solutions.

21 , , , nm x m xm xe e e

211 2( ) nm x m xm x

ny x c e c e c e

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Non- homogeneous Differential Equation Chapter 4

Example 1: Distinct Real Roots (1 of 3)

Solve the differential equation

Assuming exponential soln leads to characteristic equation:

Thus the general solution is

4 3 2( ) 2 13 14 24 0

1 2 3 4 0

mxy x e m m m m

m m m m

2 3 41 2 3 3( ) x x x xy x c e c e c e c e

(4) 2 13 14 24 0y y y y y

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Complex Roots

If the characteristic polynomial P(r) has complex roots, then they must occur in conjugate pairs,

Note that not all the roots need be complex. i

In this case the functions are both solutions to the original equation.

e sin and e cosx xy x y x

1 2e sin cosxy C x C x

General Solution

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Example 2: Complex Roots

Consider the equation

Then

Now

Thus the general solution is

3 2( ) 1 0 1 1 0mty t e m m m m

0 yy

2 1 1 4 1 3 1 31 0

2 2 2 2

im m m i

/21 2 3( ) cos 3 / 2 sin 3 / 2x xy t c e e c x c x

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Example 3: Complex Roots (1 of 2)

Consider the initial value problem

Then

The roots are 1, -1, i, -i. Thus the general solution is

Using the initial conditions, we obtain

The graph of solution is given on right.

4 2 2( ) 1 0 1 1 0mxy x e r r r

2)0(,2/5)0(,4)0(,2/7)0(,0)4( yyyyyy

1 2 3 4( ) cos sinx xy x c e c e c x c x

1( ) 0 3 cos sin

2x xy x e e x x

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Repeated Roots

Suppose a root m of characteristic polynomial P(r) is a repeated root with multiplicity n. Then linearly independent solutions corresponding to this repeated root have the form

2 1, , , ,m x m x m x s mxe xe x e x e

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Example 4: Repeated Roots

Consider the equation

Then

The roots are 2i, 2i, -2i, -2i. Thus the general solution is

4 2 2( ) 8 16 0 4 4 0mxy x e m m m m

0168)4( yyy

1 2 3 4( ) cos 2 sin 2 cos 2 sin 2y x c x c x c x x c x x

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04/21/23 13Non- homogeneous Differential Equation Chapter 4

The general solution of the non homogeneous differential equation

There are two parts of the solution:1. solution of the homogeneous part of DE

2. particular solution

( )ay by cy f x

cy

py

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General solution

c py y y

Complementary Function, solution of Homgeneous part

Particular Solution


The method can be applied for the non – homogeneous differential equations , if the

f(x) is of the form:

1. A constant C

2.A polynomial function

3.

4.

5.A finite sum, product of two or more functions of type (1- 4)

( )ay by cy f x

mxe

sin ,cos , sin , cos ,...x xx x e x e x

12/19/2015 1 Non- homogeneous Differential Equation Chapter 4.

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