anonymous
  • anonymous
Who can help me? Variation of parameters \[y''-2y'+y=e^{2x}\] \[r^2-2r+1=0\] \[(r-1)^2=0\] \[r=1\] \[y_c(x)=c_1e^x+c_2xe^x\] \[y_p(x)=u_1e^x+u_2xe^x\] \[y_p'(x)=u_1'e^x+u_2'xe^x+u_1e^x+u_2e^x+u_2xe^x\] \[u_1'e^x+u_2'xe^x=0\] \[y_p'=u_1e^x+u_2e^x+u_2xe^x\] \[y_p''(x)=u_1e^x+u_1e^x+u_2'e^x+u_2e^x+u_2'xe^x+u_2e^x+u_2xe^x\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@UnkleRhaukus
anonymous
  • anonymous
@radar
anonymous
  • anonymous
@hartnn I'm almost done with the question

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anonymous
  • anonymous
glad to see you my friend =D
anonymous
  • anonymous
the next step...everything I did so far seems right
anonymous
  • anonymous
ohhh...I plug them back into the original formula!!!
hartnn
  • hartnn
and then compare the co-efficients.
anonymous
  • anonymous
and then I....? Yeah that's where I'm stuck :P
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
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?
anonymous
  • anonymous
sry :S
anonymous
  • anonymous
anonymous
  • anonymous
oh sorry I forgot to write \[y_p'=u_1e^x+u_2e^x+u_2xe^x\]
anonymous
  • anonymous
Hey, is that the Stewart book?
anonymous
  • anonymous
yes
anonymous
  • anonymous
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
UnkleRhaukus
  • UnkleRhaukus
you have y_p, y'_p and y''_p insert them into t your original DE and simplify

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