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- anonymous

I am having a problem in arriving at the correct answer to a second order inhomogeneous differential equation (below):

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- anonymous

- jamiebookeater

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- anonymous

\[\frac{ d ^{2}y }{ dx ^{2} }-6\frac{ dy }{ dx }+8y=8e^{4x}\]

- 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

use the undetermined coefficients method

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- 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

What is the undetermined coefficients method?

- anonymous

Oh, i believe that is what i am doing

- anonymous

your should assume the particular\[y = Ce^{kx}\]then you solve for C and k

- anonymous

thats what i tried, but i get C = 0

- anonymous

I believe 'k' is actually 4 in this instance.

- 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

wait I am working on it too, if Ce^(kx) won't work, try Cxe^(kx)

- anonymous

Oh... i'll try that only i thought you only did that when there was already a \[e^{kx}\] on the LHS.

- anonymous

try\[y = Cxe^{4x}\]it works

- anonymous

Just trying that now...

- 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

Oh and thanks heaps... =)

- anonymous

lol, you are welcome

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