Find a Particular Solution of 9y''+5y'+2y=sin^2(x)

[Tom McCurdy]

What did I do Wrong??

Homework Statement

Find a particular solution yp of the differential equation
$$9y''+5y'+2y=sin^2(x)$$

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

Homework Equations

$$sin^2(x) = \frac{1-cos(2x)}{2}$$

The Attempt at a Solution

Possible derivatives

• A
• B cos(2x)
• C sin(2x)

$$y_p = A + Bcos(2x) + Csin(2x)$$

$$y_p' = -2Bsin(2x) + 2Ccos(2x)$$

$$y_p'' = -4Bcos(2x) + -4Csin(2x)$$

Sub back into original modified with cos substitution on the right

$$9y''+5y'+2y = \frac{1-cos(2x)}{2}$$

Becomes

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

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

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

Where
B = 17/1056
and
C = 5/1056

 

[arildno]

Spare us the decimals. Please!

 

[gammamcc]

Did you check your answer by plugging it into the diff. eq.? (I'm sure you will see that A is wrong at least.)

 

[Tom McCurdy]

I realized I type it wrong when i had A defined, but i had it correct or so I thought in the equation A=1/4

 

[AKG]

-36B + 10C + 2B = -1/2
-36C + -10B + 2C = 0

Where
B = 17/1056
and
C = 5/1056

You solved for A correctly, but B and C are wrong.

 

[Tom McCurdy]

-36B + 10C + 2B = -1/2
-36C + -10B + 2C = 0

-34B + 10 C = -1/2
-10B + -34C = 0

34B + 115.6C = 0
125.6C=-1/2

C = -5/1256

Therefore B equal

-10B + -34C = 0

B = 17/1256


I am down to one final attempt at this problem can someone confirm this correct?


$$y_p = (1/4)+(17/1256)*cos(2*x)+(-5/1256)*sin(2*x)$$

 

[Dick]

That's what I get.

 

[Tom McCurdy]

Algerbra FTW

... eh wow I feel stupid

it worked!