More Fourier Series (and differential equations) Watch

trm90
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#1
Report Thread starter 7 years ago
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I have the equation:

y'' - y = f(x)

where f(x) = e^{-\alpha x}

I'm asked to solve this equation using FTs.

Let Y(k) be the FT of y(x). Then, we have:

-k^2 Y(k) - Y(k) = \dfrac{1}{\alpha + ik}

as the ft of d^n y/dx^n is (ik)^n Y(k) and the ft of f(x) is \dfrac{1}{\alpha + ik}.


I rearranged this to:

Y(k) = - \dfrac{1}{\alpha +ik} \dfrac{1}{k^2 + 1}

The next step is to expand both the above fractions with partial fractions so I can get recognisable functions which can be fourier transformed one more time to find y(x). But I've never been good at partial fractions at all, and I'm not quite sure how to split these.

The answer is in the spoiler below...
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y = exp(- \alpha x) / (\alpha^2 - 1) + exp(-x)/[2(1 - \alpha)] for x \geq 0

 y = exp(x)/[2(1 + \alpha)] for x < 0.


Lots and lots of thanks for any help!
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suneilr
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Report 7 years ago
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(Original post by trm90)
Y(k) = - \dfrac{1}{\alpha +ik} \dfrac{1}{k^2 + 1}
Which bit of the partial fractions are you stuck on?

If you expand the denominator you should find that it factorises nicely which then makes it quite easy to get the partial fractions.
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DFranklin
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There's some stuff missing from what you've posted.

The general solution of y''-y = e^{-\alpha x} is A e^x + B e^{-x} + \frac{e^{-\alpha x}}{\alpha^2-1}. (when alpha isn't 1 or -1). So why don't you have any arbitrary constants?

Also, the Fourier Transform of e^{-\alpha x} is undefined.
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trm90
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Report Thread starter 7 years ago
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D'oh, looks like I actually just solved it; thought I couldn't do the partial fractions properly. Thanks anyway for the posts guys!

P.S. @DFranklin I (once again) forgot to specify that it was actually f(x) = H(x)exp(-ax), sorry :p:
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