Question

Need some help with Variation of Parameters please:
y''+2y'+2y=e^(-x) cscx

Answer 1

@SithsAndGiggles the problem shows as:
\[y''+2y'+2y=e^{-x}cscx\]

Answer 2

Characteristic solutions:
\[r^2+2r+2=0\\
r=\frac{-2\pm2i}{2}=-1\pm i\]
yielding
\[y_c=C_1e^{-x}\cos x+C_2e^{-x}\sin x\]
If I'm not mistaken...

This makes
\[y_1=e^{-x}\cos x\\
y_2=e^{-x}\sin x\]

Answer 3

I am a bit confused on what Yc would be..
\[Yc=e^x(c _{1} \cos x+c _{2} sinx) \]?

Answer 4

My bad, you're right

Answer 5

Im working on the Wronskian..

Answer 6

Okay, so now you can use the formulas for VoP:
\[u_1=-\int\frac{e^{-x}\sin xe^{-x}\csc x}{W(y_1,y_2)}~dx=-\int\frac{e^{-2x}}{W(y_1,y_2)}~dx\\
u_2=\int\frac{e^{-x}\cos xe^{-x}\csc x}{W(y_1,y_2)}~dx=\int\frac{e^{-2x}\cot x}{W(y_1,y_2)}~dx\]
where
\[\begin{align*}W(y_1,y_2)&=\begin{vmatrix}e^{-x}\cos x&e^{-x}\sin x\\
-e^{-x}\cos x-e^{-x}\sin x&-e^{-x}\sin x+e^{-x}\cos x\end{vmatrix}\\\\
&=e^{-2x}\bigg[-\cos x\sin x+\cos^2 x+\sin x\cos x+\sin^2 x\bigg]\\
&=e^{-2x}
\end{align*}\]
Hmm, convenient.

Answer 7

Would the Wronskian not equal \[2e^{-2x}\]?

Answer 8

Ok i see, my bad you're right..

Answer 9

http://www.wolframalpha.com/input/?i=Det%5B%7B%7BExp%5B-x%5D*Cos%5Bx%5D%2CExp%5B-x%5D*Sin%5Bx%5D%7D%2C%7BD%5BExp%5B-x%5D*Cos%5Bx%5D%2Cx%5D%2CD%5BExp%5B-x%5D*Sin%5Bx%5D%2Cx%5D%7D%7D%5D

Answer 10

Awesome! Thank you so much!

Answer 11

You're welcome!