# Solving ODE using Fourier Transform

by absolute76
Tags: fourier, solving, transform
P: 21
 Quote by gabbagabbahey Actually, I made an error in my previous post: $$i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]\neq-k\frac{d\tilde{u}}{dk}[/itex] You will need to use the product rule to carry out the derivative. error??? error in which part?? yes i got the answer of 2k³-4k²-k/(-2k²+1)² using....(u'v-uv')/v² am i right?? HW Helper P: 5,003  Quote by absolute76 error??? error in which part?? [tex]\frac{d}{dk}\left(k\tilde{u}\right)\neq k\frac{d\tilde{u}}{dk}$$
P: 21
 Quote by gabbagabbahey Actually, I made an error in my previous post: $$i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]\neq-k\frac{d\tilde{u}}{dk}[/itex] You will need to use the product rule to carry out the derivative. [tex] i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]$$

owk i think i got it, is it correct....i(iu~(k)-k²u~(k))

that will give me -u~(k)+ik²u~(k)---->u~(k)[-1+ik²]??

owk how do i separate u~(k) so that it don't cancel each other(left and right side)??
HW Helper
P: 5,003
 Quote by absolute76 $$i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]$$ owk i think i got it, is it correct....i(iu~(k)-k²u~(k))
No.

$$i\frac{d}{dk}\left[(ik)\tilde{u}\right]=-\frac{d}{dk}\left[k\tilde{u}\right]=-\left(\tilde{u}\frac{dk}{dk}+k\frac{d\tilde{u}}{dk}\right)=-\tilde{u}-k\frac{d\tilde{u}}{dk}$$

This is a basic application of the product rule.
P: 21
 Quote by gabbagabbahey No. $$i\frac{d}{dk}\left[(ik)\tilde{u}\right]=-\frac{d}{dk}\left[k\tilde{u}\right]=-\left(\tilde{u}\frac{dk}{dk}+k\frac{d\tilde{u}}{dk}\right)=-\tilde{u}-k\frac{d\tilde{u}}{dk}$$ This is a basic application of the product rule.
Ok my mistake again,

so now the equation will be u~(k)[-2k²+2]=-k du~/dk

so u~(k)= [-k/(-2k²+2) du~/dk

Is it correct if i diffrentiate towards k on the right side???
HW Helper
P: 5,003
 Quote by absolute76 so now the equation will be u~(k)[-2k²+2]=-k du~/dk so u~(k)= [-k/(-2k²+2) du~/dk
That's correct. So now you have a 1st order ODE to solve instead of the 2nd order one you started with. This one is separable:

$$\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^2-1}{k}dk$$
P: 21
 Quote by gabbagabbahey That's correct. So now you have a 1st order ODE to solve instead of the 2nd order one you started with. This one is separable: $$\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^2-1}{k}dk$$
Owk now is this correct?

u~(k)=e^[(k²)-(2 ln k)]

....u~(k)=e^(k²)/e^(2 ln k)

then i transform it using the table?? right?
HW Helper
P: 5,003
 Quote by absolute76 Owk now is this correct? u~(k)=e^[(k²)-(2 ln k)] ....u~(k)=e^(k²)/e^(2 ln k) then i transform it using the table?? right?
Right. Although, I doubt your table will give you the inverse FT of this function, so you will probably want to use the convolution theorem with $\tilde{g}(k)=\frac{1}{k^2}$
P: 21
 Quote by gabbagabbahey Right. Although, I doubt your table will give you the inverse FT of this function, so you will probably want to use the convolution theorem with $\tilde{g}(k)=\frac{1}{k^2}$
ok, so if i want to find the general solution,

should i just let it be in terms of convolution plus the homogenous equation right?

that should be my final general solution.
 HW Helper P: 5,003 Right, although in this case, I think your second solution comes from solving $u''(x)=0$ and $xu'-u=0[/tex]. which isn't really "the homogeneous equation". P: 21  Quote by gabbagabbahey Right, although in this case, I think your second solution comes from solving [itex]u''(x)=0$ and [itex]xu'-u=0[/tex]. which isn't really "the homogeneous equation".
owh ok..i think i get it...
anyway thanks a lot for your help!
i really appreciate it..

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