General solution of ode using fourier transform

[naspek]

ok well I'm pretty much home and dry in this problem

the aim of this problem is to get the general solution for the ode below..

2u'' - xu' + u = 0 = g(x)
i started to solve it by rearranging the equation..

2u'' + u = xu'

apply Fourier transform..
2F(u'') + u^ = g^

(-2k^2)u^ + u^ = g^

u^ [1- 2(k^2)] = g^
u^ = {1/ [1- 2(k^2)]}g^

the problem is, i can't find any of the them in the transform table..

 

[vela]

ok well I'm pretty much home and dry in this problem

the aim of this problem is to get the general solution for the ode below..

2u'' - xu' + u = 0 = g(x)

So g(x)=0?

i started to solve it by rearranging the equation..

2u'' + u = xu'

apply Fourier transform..
2F(u'') + u^ = g^

Where did the xu' term go, and where did g come from?

(-2k^2)u^ + u^ = g^

u^ [1- 2(k^2)] = g^
u^ = {1/ [1- 2(k^2)]}g^

the problem is, i can't find any of the them in the transform table..

 

[naspek]

rearranging my equations...

2u'' + u = xu' = g(x)
where xu' = g(x)

F{ 2u'' + u = g(x) }
-2k^2(u^) + (u^) = (g^)
(1 - k^2 )(u^) = (g^)
u^ = [1 / (1 - k^2 )] (g^)

correct?

 

[vela]

Sort of, but it's not what you want to do. What exactly is g(x) supposed to be? According to your first equation (in your original post), g(x)=0. Or is it supposed to be the source term/forcing function, i.e. the term that results in the particular solution?

Try looking up a property of the Fourier transform relating $xf(x)$ to $$\frac{d}{dk}\hat{f}(k)$$.

 

[naspek]

xu' --->> [- (1/ i2pi)][d/dx]u

correct?
should i substitute it to my previous equation?

 

[vela]

Try it out.

 

[naspek]

Try it out.

F(u'' -xu' + u = 0)
F(-xu') = -[(-1 / i2pi)(d/dx)u^])
F(-xu') = [(1 / i2pi)(d/dx)u^])

F(u'' -xu' + u = 0)
[1 -2k^2 + (1 / i2pi)(d/dx)]u^ = 0

correct?

 

[vela]

F(u'' -xu' + u = 0)
F(-xu') = -[(-1 / i2pi)(d/dx)u^])
F(-xu') = [(1 / i2pi)(d/dx)u^])

F(u'' -xu' + u = 0)
[1 -2k^2 + (1 / i2pi)(d/dx)]u^ = 0

correct?

No, you seem to be missing a factor of k. First, take care of the effect of the x:

$$F[xu']=\frac{1}{-i}\frac{d}{dk}F[u']$$

And then take the Fourier transform of u':

$$F[xu']=\frac{1}{-i}\frac{d}{dk}(-ikF<u>)=\frac{d}{dk}[k\hat{u}(k)]</u>$$

(If you're using x and k, I think there are no factors of 2π.)