Nonhomogeneous Linear Differential

EquationsAP CALCULUS BC

Nonhomogeneous Differential EquationsRecall 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.

( )ay by cy G x

Complementary EquationThe related homogeneous equation is called the complementary equation, and it is part of the general solution to the nonhomogeneous equation.

0ay by cy

General SolutionThe 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 yc

is the general solution to the complementary equation.

Proof for Grant

yc(x) = y(x) – yp(x)( ) ( ) ( )p p pa y y b y y c y y

p p pay ay by by cy cy

( ) ( )p p pay by cy ay by cy ( ) ( ) 0g x g x

Method of Undetermined Coefficients

There are two methods for solving nonhomogeneous equations:Method of Undetermined CoefficientsVariation of Parameters

First, 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, yp 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.

Example 1 Solve the equation yʹʹ + yʹ – 2y = x2.

The auxiliary equation is r2 + r – 2 = 0

Factor (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

yp(x) = Ax2 + Bx + C

So ypʹ(x) = 2Ax + B and ypʹʹ(x) = 2A

Example 1 (continued) Substitute these into the differential equation (2A) + (2Ax + B) – 2(Ax2 + Bx + C) = x2

= – 2Ax2 + (2A – 2B)x + (2A + B – 2C) = 1x2 + 0x + 0

–2A = 1

2A – 2B = 0

1

2A

12 2 0

2B

1

2B

2 2 0A B C 1 1

2 2 02 2

C

3

4C

Example 1 (FINAL) Therefore, our particular solution is

So our general solution is:

y = yc + yp = c1ex + c2e–2x

21 1 3

2 2 4py x x

21 1 3

2 2 4x x

Example 2 Solve yʹʹ + 4y = e3x

When G(x) is of the form ekx, we use yp = Aekx because the derivatives of ekx are constant multiples of ekx and work out nicely.

Complementary equation is r2 + 4 = 0

Therefore, yc = c1 cos 2x + c2 sin 2x

Solve for yp:

y = Ae3x yʹ = 3Ae3x yʹʹ = 9Ae3x

Substitute

2r i

3 3 39 4x x xAe Ae e

Example 2 (continued)

13A = 1

A = 1/13

General solution is

3 3 39 4x x xAe Ae e 3 313 x xAe e

31

13x

py e

31 2

1cos2 sin 2

13xy c x c x e

Example 3 Solve the equation yʹʹ + yʹ – 2y = sin x.

When G(x) is of the form C sin kx or C cos kx, we use yp = A cos kx + B sin kx

Complementary equation is r2 + r – 2 = 0 r = –2, 1

yp = A cos x + B sin x

ypʹ = –A sin x + B cos x

ypʹʹ = –A cos x – B sin x

( 2)( 1) 0r r 2

1 2x x

cy c e c e

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 x

Therefore, –3A + B = 0 and –A – 3B = 1

Solve as a system

General solution is

1 3

10 10A B

1 3cos sin

10 10py x x

21 2

1 3cos sin

10 10x xy c e c e x x

Review and More Rules for Method of Undetermined Coefficients

Form is ayʹʹ + byʹ + cy = G(x)

1. If G(x) is a polynomial, use yp = Axn + Bxn–1 + … + C.

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

3. If G(x) = C sin kx or C cos kx, use yp = A cos kx + B sin kx

4. If G(x) is a product of functions, multiply them for yp.

Example: G(x) = x cos 3x yp = (Ax + B) cos 3x + (Cx + D) sin 3x

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)ex and yp2 = C cos 2x + D sin 2x

Then add y(x) = yc + yp1 + yp2

6. If yp is a solution to the complementary equation (yc), multiply yp by x or x2, so yc and yp are linearly independent.

7. The particular solutions we try to find using yp (the ones with the A, B, C, etc. in them) are called “trial solutions”.