Question

Question: Variation of parameters

Answer 1

I'm using the variations of parameters method
\[y''-2y'+y=e^{2x}\]
\[y_c=c_1e^x+c_2xe^x\]
\[y_p=u_1e^x+u_2xe^x\]
\[y_p'=u_1e^x+u_2e^x+u_2xe^x\]
\[y_p''=u_1'e^x+u_1e^x+u_2'e^x+u_2e^x+u_2'xe^x+u_2e^x+u_2xe^x\]
now substitute into \[y''-2y'+y=e^{2x}\]
After I substituted I simplified and got the following:

Answer 2

\[u_1'e^x+u_2'e^x+u_2'xe^x=e^{2x}\]
\[u_1'e^x+u_2'xe^x=0\]
\[u_2'e^x=e^{2x}\]
\[u_2'=e^x\]

Answer 3

and through the same process I got
\[u_1'=-xe^x\]

Answer 4

and now I'm stuck

Answer 5

@zepdrix
@TuringTest
@hartnn

Answer 6

oh hi @lalaly =)

Answer 7

\[y_p=u_1 e^x +u_2xe^x\]from u1' find u1 (integratiion by parts)
let u=-x dv=e^xdx
du=-dx v=e^x
\[u_1=-xe^x+e^x\]\[u_1=(1-x)e^x\]
now same way find u2
\[u_2=e^x\]

from what uve written\[y_p=u_1e^x+u_2xe^x\]substitute u1 and u2\[y_p=(1-x)e^x(e^x)+xe^x(e^x)=(1-x)e^{2x}+xe^{2x}\]\[y_p=e^{2x}-xe^{2x}+xe^{2x}=e^{2x}\]

Answer 8

therefore\[y=y_h+y_p = c_1e^x+c_2xe^x+e^{2x}\]

Answer 9

give me a min. I'm looking through it....

Answer 10

take ur time:)

Answer 11

AWESOME!!!!!!!!!!

Answer 12

thanks @lalaly

Answer 13

your welcome:D