- #1

- 1,020

- 1

**What did I do Wrong??**

## Homework Statement

Find a particular solution yp of the differential equation

[tex] 9y''+5y'+2y=sin^2(x) [/tex]

using the Method of Undetermined Coefficients. Primes denote derivatives with respect to x.

## Homework Equations

[tex] sin^2(x) = \frac{1-cos(2x)}{2} [/tex]

## The Attempt at a Solution

Possible derivatives

- A
- B cos(2x)
- C sin(2x)

[tex] y_p = A + Bcos(2x) + Csin(2x) [/tex]

[tex] y_p' = -2Bsin(2x) + 2Ccos(2x) [/tex]

[tex] y_p'' = -4Bcos(2x) + -4Csin(2x) [/tex]

Sub back into original modified with cos substitution on the right

[tex] 9y''+5y'+2y = \frac{1-cos(2x)}{2} [/tex]

Becomes

[tex] 9[-4Bcos(2x) + -4Csin(2x)] + 5[-2Bsin(2x) + 2Ccos(2x)] + 2[A + Bcos(2x) + Csin(2x)] = \frac{1-cos(2x)}{2} [/tex]

From that I get

2A = 1/2 => A=1/4

-36B + 10C + 2B = -1/2

-36C + -10B + 2C = 0

Solving for B and C

Leads to

[tex]y_p = (1/4)+(17/1056)*cos(2*x)+(5/1056)*sin(2*x) [/tex]

Where

B = 17/1056

and

C = 5/1056

Last edited: