anonymous
  • anonymous
I am having a problem in arriving at the correct answer to a second order inhomogeneous differential equation (below):
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ d ^{2}y }{ dx ^{2} }-6\frac{ dy }{ dx }+8y=8e^{4x}\]
anonymous
  • anonymous
I know that the complimentary function is \[Ae^{4x}+Be^{2x}\] however i seem to arrive at an incorrect answer for the particular integral.
anonymous
  • anonymous
use the undetermined coefficients method

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anonymous
  • anonymous
For the particular integral i get \[16Ce^{4x}-24Ce^{4x}+8Ce^{4x}=8e^{4x}\] Based on derivatives of the general function \[Ce^{4x}\] I therefore get \[e^{4x}(16C-24C+8C)=8e^{4x}\] Therefore \[0=8\] Which means there is no solution. The answer in my book for the particular integral however is \[4e^{4x}\]. Where did i go wrong?
anonymous
  • anonymous
What is the undetermined coefficients method?
anonymous
  • anonymous
Oh, i believe that is what i am doing
anonymous
  • anonymous
your should assume the particular\[y = Ce^{kx}\]then you solve for C and k
anonymous
  • anonymous
thats what i tried, but i get C = 0
anonymous
  • anonymous
I believe 'k' is actually 4 in this instance.
anonymous
  • anonymous
Can you see where i went wrong in working through the answer? Or do you think it is an error in my book?
anonymous
  • anonymous
wait I am working on it too, if Ce^(kx) won't work, try Cxe^(kx)
anonymous
  • anonymous
Oh... i'll try that only i thought you only did that when there was already a \[e^{kx}\] on the LHS.
anonymous
  • anonymous
try\[y = Cxe^{4x}\]it works
anonymous
  • anonymous
Just trying that now...
anonymous
  • anonymous
Awesome, as you said, that works. I didn't realize that you could use that method when there is no solution of a coefficient. I thought it was only when an element of the general solution appeared in the LHS. Good to know, i've learned a new trick =)
anonymous
  • anonymous
Oh and thanks heaps... =)
anonymous
  • anonymous
lol, you are welcome

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