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Need help solving a differential equation

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  1. Feb 28, 2016 #1
    Are there any known analytical method to solve the equation
    $$
    A\frac{d^2f(x)}{dx^2}+B\frac{df(x)}{dx}+Ce^{igx}f(x) = 0\hspace{1cm}?
    $$
    All quantities appearing in that equation are complex except for ##g## and ##x##.
     
  2. jcsd
  3. Feb 28, 2016 #2
    The solution is [itex]f_{(x)}=e^{a+bx}[/itex]. You need to find the complex constants a and b.
     
  4. Feb 28, 2016 #3
    Thanks, but how do you come up with that solution? Have you ever found the above equation somewhere else and the solution is given like that?
     
  5. Feb 28, 2016 #4
    It is a trivial solution taught at any basic course of differential equations. If you have a differential equation written as [itex]\sum_{n=0} ^m c_n \frac{\mathrm{d} ^n f_{(x)}}{\mathrm{d} x^n}=0[/itex], then is is known that the solution is an exponential function.
     
  6. Feb 28, 2016 #5
    Please take a look again at the original equation, it is different from the one you suggest due to the presence of ##e^{igx}## in the third term.
     
  7. Feb 28, 2016 #6
    There are many ways to solve it without previous knowledge of the solution. One is to make a Laplace or Fourier transform. It turns the problem into a very simple equation without any derivatives. You solve it, and reverse the transform.
     
  8. Feb 28, 2016 #7
    The exponential is easy to deal because exponents are added on multiplication.
     
  9. Feb 28, 2016 #8
    I know what you mean there, but do those methods work for homogenous equation? For the moment just remove ##e^{igx}## from the third term, and then perform Fourier transform on each term. I will then get
    $$
    Ak^2F(k) + iBkF(k) + CF(k) = 0
    $$
    where ##F(k)## is the Fourier transform of ##f(x)##. How will you solve for a non-trivial solution of ##F(k)## in the above equation since the right hand side is zero? Even when ##e^{igx}## is retained, the equation in Fourier domain becomes
    $$
    Ak^2F(k) + iBkF(k) + CF(k-g) = 0
    $$
    I don't see how I can get an explicit expression for ##F(k)## in this case.
     
  10. Feb 28, 2016 #9
    Sorry, you are right. Is not trivial. I was wrong.

    Is better to use Laplace transform, and even then you get a nasty work.

    Your solution is

    $$
    f(x) = \left ( c_1 \Gamma_{\left(\frac{B}{A g i}+1\right)} J_{\left({\frac{B}{A g i}} , \frac{2 \sqrt{A C e^{g i x}}}{A g i}\right)}+c_2 \Gamma_{\left(1-\frac{B}{A g i}\right)} J_{\left({-\frac{B}{A g i}} , \frac{2 \sqrt{A C e^{g i x}}}{A g i}\right)} \right) e^{-\frac{B x}{2 A}}
    $$

    .. where [itex]\Gamma[/itex] is the Gamma function, [itex]J_{(\alpha,x)}[/itex] is the Bessel function; [itex]c_1[/itex] and [itex]c_2[/itex] are constants to adjust to your initial conditions.
     
    Last edited: Feb 28, 2016
  11. Feb 28, 2016 #10
    Thank you very much, I really appreciate your effort to help me. Just, can you please point out the steps you took to reach that solution?
     
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