Seeking the solution of an integro-differential equation

[Judas503]

I'm seeking the solution to the following integro-differential equation:
$$ \frac{\partial c(x,t)}{\partial t}=-xc(x,t)+2\int_{x}^{\infty} c(y,t)dy$$

I know that to solve this, the Laplace transform must be taken term by term. Let
$$ \mathcal{L}\{c(x,t)\}=\int_{0}^{\infty}e^{-st}c(x,s) $$
Then, $$ \mathcal{L}\{ \frac{\partial c(x,t)}{\partial t} \}=s\mathcal{L}\{c(x,t)\}-c(x,0)$$
and, $$ \mathcal{L}\{xc(x,t)\}=x\mathcal{L}\{c(x,t)\} $$

The problem arises with the Laplace transform of the integral.
\begin{align*}
\mathcal{L}\{ \int_{x}^{\infty}c(y,t)dt \} &= \mathcal{L}\{ \int_{0}^{\infty}c(y,t)dt \}-\mathcal{L}\{ \int_{0}^{x}c(y,t)dt \}
\\&=\mathcal{L}\{ \int_{0}^{\infty}c(y,t)dt \}-\frac{\mathcal{L}\{ c(x,t) \}}{s}
\end{align*}
My question is what do I do with the first integral of the last equation?

 

[fzero]

I'm seeking the solution to the following integro-differential equation:
$$ \frac{\partial c(x,t)}{\partial t}=-xc(x,t)+2\int_{x}^{\infty} c(y,t)dy$$

I know that to solve this, the Laplace transform must be taken term by term. Let
$$ \mathcal{L}\{c(x,t)\}=\int_{0}^{\infty}e^{-st}c(x,s) $$

This should be
$$ \mathcal{L}\{c(x,t)\}=\int_{0}^{\infty}e^{-st}c(x,t) dt.$$

Then, $$ \mathcal{L}\{ \frac{\partial c(x,t)}{\partial t} \}=s\mathcal{L}\{c(x,t)\}-c(x,0)$$
and, $$ \mathcal{L}\{xc(x,t)\}=x\mathcal{L}\{c(x,t)\} $$

The problem arises with the Laplace transform of the integral.
\begin{align*}
\mathcal{L}\{ \int_{x}^{\infty}c(y,t)dt \} &= \mathcal{L}\{ \int_{0}^{\infty}c(y,t)dt \}-\mathcal{L}\{ \int_{0}^{x}c(y,t)dt \}
\\&=\mathcal{L}\{ \int_{0}^{\infty}c(y,t)dt \}-\frac{\mathcal{L}\{ c(x,t) \}}{s}
\end{align*}
My question is what do I do with the first integral of the last equation?

You've made a mistake in swapping the original integration over the position domain with one over the time domain. If we let $\mathcal{L}\{c(x,t)\} = C(x,s)$, then
$$\mathcal{L}\{ \int_{x}^{\infty}c(y,t)dy \} = \int_{x}^{\infty}C(y,s)dy.$$
You'll still end up with an integral equation, but without the time derivative. You should be able to differentiate the equation with respect to $x$ to get a first-order differential equation for $C(x,s)$.