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

$\displaystyle x^2$ $\displaystyle *$$\displaystyle \frac{d^2u}{dx^2} +\displaystyle x *\displaystyle \frac{du}{dx} -\displaystyle 4u =\displaystyle 3x 2. For the general solution of the 'incomplete' DE... \displaystyle x^{2}\cdot \frac{d^{2} u}{du^{2}} + x\cdot \frac{du}{dx} - 4\cdot x =0 (1) ... try functions like \displaystyle u(x)= x^{\alpha}.For the particular solution of the 'complete' DE... \displaystyle x^{2}\cdot \frac{d^{2} u}{du^{2}} + x\cdot \frac{du}{dx} - 4\cdot x = 3\cdot x (2) ... try a function like \displaystyle u(x)= \beta\cdot x... Kind regards \displaystyle \chi \displaystyle \sigma 3. Originally Posted by ulysses123 I re posted because i stuffed up the latex code. I am stuck on this ODE. Boundary conditions are: U(0)=U(1)=0 \displaystyle x^2 \displaystyle *$$\displaystyle \frac{d^2u}{dx^2}$ +$\displaystyle x$ *$\displaystyle \frac{du}{dx}$ -$\displaystyle 4u$ =$\displaystyle 3x$
This is an "Euler-type" or "equipotential" equation. As chisigma said, you can try a solution of the form $\displaystyle y= x^\alpha$ 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.