Nonhomogeneous Linear Differential Equations - PowerPoint Presentation

Slide1

Nonhomogeneous Linear Differential Equations

AP Calculus BCSlide2

Nonhomogeneous Differential Equations

Recall that second order linear differential equations with constant coefficients have the form:

Now we will solve equations where

G(x) ≠ 0, which are non-homogeneous differential equations.Slide3

Complementary Equation

The related homogeneous equation is called the

complementary equation,

and it is part of the general solution to the nonhomogeneous equation.Slide4

General Solution

The general solution to a

nonhomogenous

differential equation is y(x) = yp(x) + yc(x)

where yp

is a particular solution to the

nonhomogenous

equation, and

y

c

is the general solution to the complementary equation.Proof for Grantyc(x) = y(x) – yp(x)Slide5

Method of Undetermined Coefficients

There are two methods for solving nonhomogeneous equations:

Method of Undetermined Coefficients

Variation of ParametersFirst, we will learn about the Method of Undetermined Coefficients to solve the equation ayʹʹ + byʹ + cy = G(x) when G(

x) is a polynomial.Since

G

(

x

) is a polynomial,

y

p

is also a polynomial of the same degree as G. Therefore, we substitute yp = a polynomial (of the same degree as G) into the equation and determine the coefficients.Slide6

Example 1

Solve the equation

y

ʹʹ + yʹ – 2y = x2.The auxiliary equation is r2 + r – 2 = 0Factor  (

r – 1)(r

+ 2) = 0

r

= 1,

r

= –2

Solution of complementary equation is

yc = c1ex + c2e–2x

Since G(

x) =

x2, we want a particular solution where

y

p(x) =

Ax

2 + Bx

+

C

So

y

p

ʹ

(

x

) =

2

Ax

+

B

and

y

p

ʹʹ

(

x

) = 2

ASlide7

Example 1 (continued)

Substitute these into the differential equation



(2A) + (2Ax + B) – 2(Ax2 + Bx + C) = x2

= – 2Ax2

+

(

2

A

– 2

B

)x + (2A + B – 2C) = 1x2 + 0x + 0–2A = 1 

2A

– 2

B = 0  Slide8

Example 1 (FINAL)

Therefore, our particular solution is

So our general solution is:

y = yc + yp = c1

ex

+

c

2

e

–2

xSlide9

Example 2

Solve

y

ʹʹ + 4y = e3xWhen G(x) is of the form ekx, we use y

p = Ae

kx

because the derivatives of

e

kx

are constant multiples of

e

kx and work out nicely.Complementary equation is r2 + 4 = 0 Therefore, yc = c1 cos 2x +

c2

sin 2x

Solve for y

p

:y =

Ae

3x

y

ʹ = 3

Ae

3

x

y

ʹʹ =

9

Ae

3

x

Substitute  Slide10

Example 2 (continued)

13

A = 1A = 1/13General solution is Slide11

Example 3

Solve the equation

y

ʹʹ + yʹ – 2y = sin x.When G(x) is of the form C sin kx or

C cos

kx

,

we use

y

p

= A cos kx + B sin kxComplementary equation is r2 + r – 2

= 0 r

= –2, 1 

yp =

A cos

x

+ B

sin x

y

p

ʹ = –

A

sin

x

+

B

cos

x

y

p

ʹʹ =

–

A

cos

x

–

B

sin

xSlide12

Example 3 (continued)

Substitute back into original equation:

(–

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

) – 2(A cos

x

+

B

sin

x

) = sin x(–3A + B) cos x + (–A – 3B) sin x = sin xTherefore, –3A + B

= 0 and –A – 3

B = 1

Solve as a system 

General solution is Slide13

Review and More Rules for Method of Undetermined Coefficients

Form is

ay

ʹʹ + byʹ + cy = G(x)1. If G(x) is a polynomial, use yp =

Axn +

Bx

n

–1

+ … +

C.

2.

If G(x) = Cekx, use yp = Aekx .3. If G(

x) =

C sin

kx or

C cos

kx,

use y

p =

A

cos

kx

+

B

sin

kx

4.

If

G

(

x

) is a product of functions, multiply them for

y

p

.

Example:

G

(

x

) =

x

cos 3

x



y

p

= (

Ax

+

B

) cos 3

x

+

(

Cx

+

D

) sin

3

x

Slide14

5. If

G

(

x) is a sum of functions, find separate particular solutions and add them together at the end. Example: G(x) = xex + cos 2x

Use yp1

= (

Ax

+

B

)

e

x and yp2 = C cos 2x + D sin 2x Then add  y(x

) = y

c

+ yp

1

+ y

p2

6. If

y

p

is a solution to the complementary equation (

y

c

), multiply

y

p

by

x

or

x

2

, so

y

c

and

y

p

are linearly independent

.

7. The particular solutions we try to find using

y

p

(the ones with the A, B, C, etc. in them) are called “trial solutions”.