1. ## ODE

I re posted because i stuffed up the latex code.

I am stuck on this ODE.

Boundary conditions are:
U(0)=U(1)=0

$x^2$ $*$ $\frac{d^2u}{dx^2}$ + $x$ * $\frac{du}{dx}$ - $4u$ = $3x$

2. For the general solution of the 'incomplete' DE...

$x^{2}\cdot \frac{d^{2} u}{du^{2}} + x\cdot \frac{du}{dx} - 4\cdot x =0$ (1)

... try functions like $u(x)= x^{\alpha}$.For the particular solution of the 'complete' DE...

$x^{2}\cdot \frac{d^{2} u}{du^{2}} + x\cdot \frac{du}{dx} - 4\cdot x = 3\cdot x$ (2)

... try a function like $u(x)= \beta\cdot x$...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by ulysses123
I re posted because i stuffed up the latex code.

I am stuck on this ODE.

Boundary conditions are:
U(0)=U(1)=0

$x^2$ $*$ $\frac{d^2u}{dx^2}$ + $x$ * $\frac{du}{dx}$ - $4u$ = $3x$
This is an "Euler-type" or "equipotential" equation. As chisigma said, you can try a solution of the form $y= x^\alpha$ to get the "characteristic equation".

It is also true that the substitution t= ln(x) converts an "Euler-type" equation into an equation with constant coefficients having the same characteristic equation.