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Who can help me? Variation of parameters \[y''-2y'+y=e^{2x}\] \[r^2-2r+1=0\] \[(r-1)^2=0\] \[r=1\] \[y_c(x)=c_1e^x+c_2xe^x\] \[y_p(x)=u_1e^x+u_2xe^x\] \[y_p'(x)=u_1'e^x+u_2'xe^x+u_1e^x+u_2e^x+u_2xe^x\] \[u_1'e^x+u_2'xe^x=0\] \[y_p'=u_1e^x+u_2e^x+u_2xe^x\] \[y_p''(x)=u_1e^x+u_1e^x+u_2'e^x+u_2e^x+u_2'xe^x+u_2e^x+u_2xe^x\]

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@hartnn I'm almost done with the question

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Other answers:

glad to see you my friend =D
the next step...everything I did so far seems right
ohhh...I plug them back into the original formula!!!
and then compare the co-efficients.
and then I....? Yeah that's where I'm stuck :P
oh ok
what am I looking for when I compare the coeffients....sorry don't mean to ask dumb questions...but what is my eventual goal? Do I have to manipulate the coeffients if they're diffferent?
sry :S
oh sorry I forgot to write \[y_p'=u_1e^x+u_2e^x+u_2xe^x\]
Hey, is that the Stewart book?
I solved it using undetermined coefficients and I got \[y=y_c+y_p\] \[y=c_1e^x+c_2xe^x+e^{2x}\] and then I was asked to solve the same problem using variation of parameters
you have y_p, y'_p and y''_p insert them into t your original DE and simplify

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