Variation of Parameters Problem
y"+36y=csc^3(6t)

Question

Variation of Parameters Problem
y"+36y=csc^3(6t)

anonymous · Answer

I've been using y1=cos6t and y2=sin6t, but I'm not sure if that's right. it's not working out, so I think probably it isn't.

turingtest · Answer

that is the correct complimentary solution, so your mistake must lie elsewhere

turingtest · Answer

what do you have for the wronskian?

amistre64 · Answer

the wronskian, if you aint heard of it, is the shorter version of the system of equations part

anonymous · Answer

I'm doing it without wronskians :S

anonymous · Answer

been working on this ALL DAY... just keep running into roadblocks.

anonymous · Answer

damn now i wish i came here to study differental equuations...

turingtest · Answer

it is the determinant|dw:1335557269955:dw|and I don't see how you're going to do variation of parameters without the wronskian

turingtest · Answer

$$W=\left|\begin{matrix}y_1&y_2\y_1'&y_2'\end{matrix}\right|$$can you take that determinant?

anonymous · Answer

i'm not sure. it's the diagonals, right?

turingtest · Answer

yeah, sort of|dw:1335557569768:dw|$$W=\left|\begin{matrix}y_1&y_2\y_1'&y_2'\end{matrix}\right|=y_1y_2'-y_1'y_2$$

anonymous · Answer

so it's cos6t*cos6t-(-sin6t)*sin6t
sin^2(6t)+cos^2(6t) = ...1?

turingtest · Answer

gotta use the chain rule on those derivatives

anonymous · Answer

oh snap. it'll be 6, then

turingtest · Answer

yep
now do you think you can proceed from here?

anonymous · Answer

possibly. i'll see what i can do. thanks again