# Thread: Differential equation in a distribution space

1. ## Differential equation in a distribution space

Hi everybody!

I need your help to solve the following exercise:

find the general solution of the equation
x^2*u'(x) = pv(1/x)
where u'(x) is the derivative of u with respect to x and pv(1/x) stands for 'principal value of 1/x'.

The answer should be u(x) = c1 + c2*theta(x) + c3*delta(x) - 1/2*fp(1/x^2), where theta(x) is the Heaviside's function and fp(1/x^2) is the 'finite part of 1/x^2'.

For the sake of clarity, I solved similar equations in distributions spaces making use of Green's functions, Fourier transforms and convolutions, etc., but I can't figure out how to approach this one.

2. I hope someone can help me...

3. Most people here seem to do Pure Mathematics, so you will be hard pressed to find help. Unfortunately, we don't cover Green's Functions until tomorrow, so I can't help yet.

4. I think it's too late for givin a usefull answer to mambutu, however i write for next viewer:
First observe that: if we indicate with $\displaystyle f^{(m)}$ the $\displaystyle m$-order derivative of $\displaystyle f$
if $\displaystyle m<n$
then $\displaystyle x^n \delta^{(m)}=0$
In fact $\displaystyle (x^n \delta^{(m)}, \phi)=(\delta^{(m)},x^n \phi)=\pm(\delta,(x^n\phi)^{(m)}) = \pm(\delta, Ax^n+...+Bx^{n-m})=0$
So if u want to solve
$\displaystyle x^2 u^{(1)}=0$
we remember that $\displaystyle \theta^{(1)}=\delta$
then
$\displaystyle u=c_1 + c_2\theta(x) + c_3\delta(x)$
this is the general solution.
The particular solution is $\displaystyle - 1/2 fp(\frac 1 {x^2})$
The full solution is the sum of both....

i hope it's clear.