Nonhomogeneous Differential Equation

[Zem]

I got a particular solution y_p(x) that is different from what the book has.

y'' + 9y = 2cos3x + 3sin3x

Characteristic equation: r^2 + 9 = 0
(r+3i)(r-3i) = 0
y_c = c_1cos3x + c_2sin3x
y_p = Acos3x + Bsin3x (not linearly independent, so I'll try another y_p)
y_p = Axcos3x + Bxsin3x
y'_p = Acos3x - 3Axsin3x + Bsin3x + 3Bxcos3x
y''_p = -3Asin3x - 3Asin3x - 9Axcos3x + 3Bcos3x + 3Bcos3x - 9Bxsin3x
x = 0
y(x) = Acos3x + Bsin3x - 6Asin3x + 6Bcos3x + Acos3x + Bsin3x - 6Asin3x + 6Bcos3x = 2cos3x + 3sin3x

Acos3x + 6Bcos3x = 2cos3x
A + 6B = 1
Bsin3x - 6Asin3x = 3sin3x
-12A + 2B = 3

My ti89 says A = -8/37 , B = 15/74
y_p = (15/37sin3x - 8/37xcos3x)
But the book's answer is y_p = 1/6(2xsin3x - 3xcos3x)

What have I missed?

 

[TD]

Your derivatives look right but I don't think you substituted them well in the equation, where is the factor 9 of y_p? You need to do y''_p+9y_p = ...

 

[Zem]

And I should not have included y'_p (oops).
Here is my new solution.
y(x) = 6Bcos3x - 6Asin3x + 9Acos3x + 9Bsin3x = 2cos3x + 3sin3x
9A + 6B = 2
-2A + 3B = 1
[ 9 6 2 ]
[ -2 3 1 ]~

[ 1 (2/3) (2/9) ](2)
[ -2 3 1 ]~

[ 1 (2/3) (2/9) ] (13)
[ 0 (13/3) (13/9) ](-2) ~

[1 0 0 ]
[ 0 3 1 ]
A = 0, B= 1/3
The book says A = 1/2, B = 1/3. So I'm still a little off here, but can't see how. Thanks!

 

[Zem]

Actually, I did not include y_p is that x = 0, so y_p = A(0)cos3x + B(0)sin3x. I included my original try for y_p in my above answer. But you helped me find the solution. Thanks so much!

6Bcos3x = 2cos3x
3B = 1
-6Asin3x = 3sin3x
-2A = 1
y_p(x) = 1/6(2xsin3x - 3xcos3x)

 

[TD]

Nice