How do I go about solving this problem? Let f(t) = |t| for -1<t<1; find periodic soln to x"+(pi)^2 x = f(t)

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corolla

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If anyone can recommend or suggest anything, that would be great. Any website or video I can go to and learn to solve this would be awesome too.
 
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I haven't tried to solve this yet, but my guts tell me to look at solutions for [imath]-1\leq t\leq 0[/imath] and [imath]0 \leq t \leq 1[/imath], then see if they can be spliced into a single periodic function.
 
For the particular integral, begin by calculating the Fourier series for f(t), it will be a series of cosine terms since f(t) is an even function.
Assume a matching series for x(t), same cosine terms but unknown coefficients.
Substitute into the de and equate coefficients across the equation. That should get you the unknown coefficients in the x(t) series.
For the general solution of the equation add that to the solution of the homogeneous equation.
 
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blamocur said:
I haven't tried to solve this yet, but my guts tell me to look at solutions for [imath]-1\leq t\leq 0[/imath] and [imath]0 \leq t \leq 1[/imath], then see if they can be spliced into a single periodic function.
Click to expand...
This approach worked for me, here is one solution from the family:
058.png
 
BobP said:
For the particular integral, begin by calculating the Fourier series for f(t), it will be a series of cosine terms since f(t) is an even function.
Assume a matching series for x(t), same cosine terms but unknown coefficients.
Substitute into the de and equate coefficients across the equation. That should get you the unknown coefficients in the x(t) series.
For the general solution of the equation add that to the solution of the homogeneous equation.
Click to expand...
I might be missing something, but I wouldn't expect this to work: Fourier series for [imath]f(t)[/imath] will have infinite number of terms, so we'd end up with an infinite number of unknowns. The solution I found is, IMHO, simpler than that.
 
blamocur said:
I might be missing something, but I wouldn't expect this to work: Fourier series for [imath]f(t)[/imath] will have infinite number of terms, so we'd end up with an infinite number of unknowns. The solution I found is, IMHO, simpler than that.
Click to expand...
The question asks for the solution as a series, (assumed infinite in length).
The series for x(t) will have an infinite number of terms but so does the Fourier series on which it is based.
The relationship between the two sets of coefficients is fairly simple making it easy to write down the general term of the x(t) series.
 
BobP said:
The question asks for the solution as a series, (assumed infinite in length).
The series for x(t) will have an infinite number of terms but so does the Fourier series on which it is based.
The relationship between the two sets of coefficients is fairly simple making it easy to write down the general term of the x(t) series.
Click to expand...
You are right, I missed that part.
 
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