Solve the Differential Equation: A(x)u'' + A'(x)u' + u/A(x) = 0

[buffordboy23]

Homework Statement

Solve the differential equation:

$$A\left(x\right)\frac{d^{2}u}{dx^{2}} + A'\left(x\right)\frac{du}{dx} + \frac{1}{A\left(x\right)}u = 0$$

where

$$u\left(x\right) = exp\left(c\int^{x}A\left(x'\right)^{q}dx'\right)$$

for some contants c and q.

The Attempt at a Solution

I tried substitution to obtain the constants c and q and also tried solving for A(x). I did not post my work since I don't even know if the approach is correct. I never saw a problem like this. The textbook does not offer any assistance, nor could I find anything on the internet. I do know that there should be two linear independent functions since it is a second-order equation. Any hints on solving the problem or suggestion of topics to research to help solve the problem would be appreciated.

 

[Pere Callahan]

Maybe it helps to express u'' and u' in terms of u...

 

[buffordboy23]

I did. That is what I meant by saying I tried substitution.

$$\frac{du}{dx} = -cA\left(x\right)^{q}u\left(x\right)$$

$$\frac{d^{2}u}{dx^{2}} = -cqA\left(x\right)^{q-1}A'\left(x\right)u\left(x\right) + \left[cA\left(x\right)^{q}\right]^{2}u\left(x\right)$$

 

[Pere Callahan]

You can cancel the u's then... maybe it helps using A^qA'=(A^q+1)' (I neglected constant factors)...however there is still the A^2q+1...

 

[buffordboy23]

First, I forgot a sign in u(x) of the original problem statement. It should be

$$u\left(x\right) = exp\left(-c\int^{x}A\left(x\right)^{q}dx\right)$$

My first and second derivatives of u w.r.t x are correct though. So, I substituted these into the equation, and the u(x) terms easily cancel out. After some manipulating, I get

$$c^{2}A^{2\left(q+1\right)}-c\left(q+1\right)A^{q+1}A'+1=0$$

where A is A(x) and A' is A'(x) from earlier. I then chose q = -1, which leads to c = i. So,

$$u_{1}\left(x\right) = exp\left(-i\int^{x}\frac{dx'}{A\left(x'\right)}\right)$$

If you choose $$u_{2}\left(x\right) = u_{1}\left(x\right)h\left(x\right)$$ and then manipulate the equation through substitution, I found that

$$u_{2}\left(x\right) = Bexp\left(-i\int^{x}\frac{dx'}{A\left(x'\right)}\right) \left[\int^{x}exp\left(2i\int^{x'}\frac{dx''}{A\left(x''\right)}\right)\frac{dx'}{A\left(x'\right)}\right]$$

where B is some constant. So, u(x) is just a linear combination of these two solutions for some coefficient function A(x).