The first is an "Euler type" or "equi-potential" equation. The change of variable t= ln(x) changes it into an equation with constant coefficients:
\frac{du}{dx}= \frac{du}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{du}{dt}
\frac{d^2u}{dx^2}= \frac{d }{dx}(\frac{1}{x}\frac{du}{dt})
= -\frac{1}{x^2}\frac{du}{dt}+ \frac{1}{x}\frac{d }{dt}\frac{du}{dt}
and changing that last d/dt to d/ds introduces another 1/x
= -\frac{1}{x^2}\frac{du}{dt}+ \frac{1}{x^2}\frac{d^2u}{dt^2}
Thus
x^2\frac{d^2u}{dx^2}= x^2(\frac{1}{x^2}\frac{d^2u}{dt^2}- \frac{1}{x^2}\frac{du}{dt}
= \frac{d^2u}{dt^2}- \frac{du}{dt}
So the first equation is just
\frac{d^2u}{dt^2}- \frac{du}{dt}= ku
a linear equation with constant coefficients which has characteristic equation
r^2- r- k= 0
That has roots
\frac{1\pm\sqrt{1+4k}}{2}[/itex] <br />
so the general solution of the equation in terms of t is<br />
u(t)= e^t\left(C_1e^{\sqrt{1+4k}t}+ C_2e^{-\sqrt{1+4k}t}\right)<br />
<br />
In terms of x, <br />
u(x)= e^{ln x}\left(C_1e^{\sqrt{1+4k}(ln x)}+ C_2e^{-\sqrt{1+4k}(ln x)}\right)<br />
= x\left(C_1x^{\sqrt{1+4k}}+ C_2x^{-\sqrt{1+4k}}\right)<br />
<br />
There is no general method for the general equation.
