What are the steps for finding u1 and u2 in the variation of parameters method?

[Cafka]

Hello,
I'm trying to understand this concept. Jere's the problem I'm doing.
I have to find the general solution for:
y'' + 36y = -4xsin(6x)

So you then solve for your characteristic equation and get lamda = +/- 6
so y1 = e^-6x and y2 = e^6x
You get your matrix for w, w1, and w2.
w = 12
w1 = 4xe^6xsin(6x)
w2 = -4xe^-6xsin(6x)

I have the problem at getting u1 and u2.
u1 = 1/3 the integral of xe^6xsin(6x) dx
u2 = -1/3 the integral of xe^-6xsin(6x) dx

How do you do that integration?

Thanks for your help,
Tom

 

[estalniath]

Your characteristic equation should be of the form y=Asin6x+Bcos6x since the the squareroot of -36 is 6i and -6i =)

 

[dextercioby]

Yeah,and Lagrange's method kicks ass.So apply it.

Daniel.

 

[rick1138]

I was wondering if anyone knew of any good references on solving systems of equations involving trigonometric equations. Any information would be appreciated.

 

[rick1138]

Found the answer - I had forgotten Cramer's rule.