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Solve differential equation

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation:

    [tex] 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 [/tex]


    [tex] u\left(x\right) = exp\left(c\int^{x}A\left(x'\right)^{q}dx'\right) [/tex]

    for some contants c and q.

    3. 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.
  2. jcsd
  3. Nov 18, 2008 #2
    Maybe it helps to express u'' and u' in terms of u....
  4. Nov 18, 2008 #3
    I did. That is what I meant by saying I tried substitution.

    [tex] \frac{du}{dx} = -cA\left(x\right)^{q}u\left(x\right) [/tex]

    [tex] \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) [/tex]
    Last edited: Nov 18, 2008
  5. Nov 18, 2008 #4
    You can cancel the u's then... maybe it helps using AqA'=(Aq+1)' (I neglected constant factors)...however there is still the A2q+1...
  6. Nov 19, 2008 #5
    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

    [tex] c^{2}A^{2\left(q+1\right)}-c\left(q+1\right)A^{q+1}A'+1=0 [/tex]

    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 [tex] u_{2}\left(x\right) = u_{1}\left(x\right)h\left(x\right) [/tex] and then manipulate the equation through substitution, I found that

    [tex] 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] [/tex]

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