How to solve this inhomogeneous partial differential equation

Question

Any help to solve this second-order inhomogeneous partial differential equation

$ \frac{1}{f(y)}~ \frac{d^2 f(y)}{dy^2} - b ~\frac{1}{k(t)}~ \frac{d^2 k(t)}{dt^2} - b B ~\frac{1}{k(t)}~ \frac{d k(t)}{dt } =0 $

Where b and B are constants.

Here is my trail:

I'm using the separation of variables method in Griffith’s book of electrodynamics, so I have assumed

$ \frac{1}{f(y)}~ \frac{d^2 f(y)}{dy^2} = c_1, ~~~~~ (1) $

$ b ~\frac{1}{k(t)}~ \frac{d^2 k(t)}{dt^2} + b B ~\frac{1}{k(t)}~ \frac{d k(t)}{dt } = c_2 ~~~~~ (2) $

With: $ ~~~~ c_1- c_2 = 0 $

Then (1) and (2) become

$ \frac{d^2 f(y)}{dy^2} = n^2 f(y) $

$ \frac{d^2 k(t)}{dt^2} + B \frac{d k(t)}{dt } = n^2 \frac{k(t)}{b}~~~~~~~ <-- ~ \text{is this correct?} $

The solution of (1) is:

$ f(y) = A e^{ny} + D e^{-ny} $

While (2) I have solved as in:

And found that:

$ k= E e^{\frac{1}{2} (- B + \sqrt{B^2+ 4 \frac{n^2}{b} } ) t} + G e^{-\frac{1}{2} (B + \sqrt{B^2+ 4 \frac{n^2}{b} } ) t} $

From the initial conditions: $f(0)= 0$ and $k(t) =0$, one can find that : $E+ G= 0$ and $A+D =0$, but anyhow, I will choose $G=D =0.$

Please let me know if these results are right or not.

score 1 · Answer 1 · answered May 14 '21 at 10:10

Your calculations seem correct. Just a few remarks :

there is no reason to think that $c_1 \geq 0$, so if you write $c_1 = n^2$, you might end up with $n$ being imaginary
since you are dealing with second order differential equations, initial conditions need to specify $f(0),\text df/\text dx(0),k(0)$, and $\text dk/\text dt(0)$

1 Answers1