Need help solving a differential equation

[maNoFchangE]

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 $x$.

 

[depizixuri]

The solution is f_{(x)}=e^{a+bx}. You need to find the complex constants a and b.

 

[maNoFchangE]

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?

 

[depizixuri]

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?

It is a trivial solution taught at any basic course of differential equations. If you have a differential equation written as \sum_{n=0} ^m c_n \frac{\mathrm{d} ^n f_{(x)}}{\mathrm{d} x^n}=0, then is is known that the solution is an exponential function.

 

[maNoFchangE]

It is a trivial solution taught at any basic course of differential equations. If you have a differential equation written as \sum_{n=0} ^m c_n \frac{\mathrm{d} ^n f_{(x)}}{\mathrm{d} x^n}=0, then is is known that the solution is an exponential function.

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.

 

[depizixuri]

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.

 

[depizixuri]

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.

The exponential is easy to deal because exponents are added on multiplication.

 

[maNoFchangE]

One is to make a Laplace or Fourier transform.

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.

 

[depizixuri]

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.

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 \Gamma is the Gamma function, J_{(\alpha,x)} is the Bessel function; c_1 and c_2 are constants to adjust to your initial conditions.

 

[maNoFchangE]

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?