I have derived two equations of the following type
$$
\dfrac{\partial A}{\partial x}=a\dfrac{\partial B}{\partial t}-b\dfrac{\partial^3 B}{\partial x^2 \partial t}$$
and
$$
\dfrac{\partial B}{\partial x}=\int \_0^l e^{-\lambda|x-x'|}\dfrac{\partial A(x')}{\partial t} dx'$$
Where $A$ and $B$ are functions of $x$ and $t$, $x$ and $x'$ are any point between $0$ and $l$ and $a, b, \lambda$ are constants.  

Is it possible to transform these two equations into a single partial differential equation for $B$?



Coupled partial differential and integro-differential equation
I am going to elaborate on Nemo's comment. You first need to note that $\frac{d^2}{dx^2}e^{|x|} = 2\delta\_0(x)\cdot e^{|x|} + e^{|x|}$ in the sense of distributions, where $\delta\_0(x)$ represents the dirac delta function at $0$. This follows from a simple computation and noting that $\frac{d}{dx}\mathrm{sign}(x) = 2\delta\_0(x)$. 


Using the above fact, we have that 
$$\frac{\partial^3 B}{\partial x^3} = \frac{\partial^2}{\partial x^2}\int\_0^l e^{-\lambda|x-x'|}\frac{\partial A(x')}{\partial t} dx'$$
If we commute the integral with the derivative, we get that 
$$ 
\begin{align}
\frac{\partial^3 B}{\partial x^3} &= \int\_0^l \left(\frac{\partial^2}{\partial x^2}e^{-\lambda|x-x'|}\right)\frac{\partial A(x')}{\partial t} dx' \\
&= \int\_0^l \left(-2\lambda \delta\_0(|x-x'|)e^{-\lambda|x-x'|} + \lambda^2 e^{-\lambda|x-x'|}\right) \frac{\partial A(x')}{\partial t} dx' \\
&= -2\lambda \frac{\partial A}{\partial t} + \lambda^2 \frac{\partial B}{\partial x} 
\end{align}
$$


Solving this, we get that 
$$ \frac{\partial A}{\partial t} = \frac{1}{2\lambda}\left(\lambda^2 \frac{\partial B}{\partial x} - \frac{\partial^3 B}{\partial x^3} \right) $$
Differenting the above equation with respect to $x$ and the first equation with respect to $t$ and setting the two equal to each other, we get 
$$ 
\frac{1}{2\lambda}\left(\lambda^2 \frac{\partial^2 B}{\partial x^2} - \frac{\partial^4 B}{\partial x^4} \right) = a \frac{\partial^2 B}{\partial t^2} - b \frac{\partial^4 B}{\partial x^2 \partial t^2}
$$


I hope this solves your problem.


