Question

y'' +4y' +4y = e^(3t) using variation of parameters

I think I did it wrong because I got W(y1,y2) = 0

Answer 1

Variation of parameters says if you have a homogenous solution:
\[y_H(t)=y_1(t)+y_2(t)\]
Then the solution is:
\[y(t)=-y_1 \int\limits y_2 g(t)/W(t)dt +y_2 \int\limits y_1 g(t)/W(t)dt + y_H(t)\]
So:
\[y_H(t) = c_1 e^{-2t} + c_2 t e^{-2t}\]
Because of the redundancy in the roots right?

Answer 2

I get W(t)=e^(-4t)

Answer 3

Does all that seem good?

Answer 4

W(t)= y1 x y'2 - y2 x y'1

Answer 5

i used this and i get:
y1 x y'2 - y2 x y'1
e^(-2t) x -2e^(-2t) -e^(-2t) x -2e^(-2t) = 0?
that is where i get 0

Answer 6

I get the particular solution cutting itself down to:\[\frac {e^{3x}}{25}\]

Anybody else got this?