A differential equation : x^2 * u'' = k*u

1. Sep 7, 2008

obomov2

Other than guessing what is the formal way of solving following types of DE :

x^2 * u'' = k*u

or more generally :

x^n * u'' + x^m * u' = k*u

u is a function of x.
thanks.

2. Sep 8, 2008

HallsofIvy

Staff Emeritus
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
$$\frac{1\pm\sqrt{1+4k}}{2}[/itex] so the general solution of the equation in terms of t is [tex]u(t)= e^t\left(C_1e^{\sqrt{1+4k}t}+ C_2e^{-\sqrt{1+4k}t}\right)$$

In terms of x,
$$u(x)= e^{ln x}\left(C_1e^{\sqrt{1+4k}(ln x)}+ C_2e^{-\sqrt{1+4k}(ln x)}\right)$$
$$= x\left(C_1x^{\sqrt{1+4k}}+ C_2x^{-\sqrt{1+4k}}\right)$$

There is no general method for the general equation.

3. Sep 8, 2008

obomov2

That was nice. Thanks.