How to solve this integro-differential equation? I came across this integro-differential equation to solve
$$\frac{du(x;t)}{dt}=-\lambda\int_0^xu(\xi;t)\;d\xi\tag{1}$$
under the initial condition $u(x;0)=f(x)$ where $x$ is a parameter, $\lambda$ is a constant, and $0<t<\infty$.
My first thought is that I can just directly integrate the equation to obtain
$$u(x;t)=-\lambda\int_0^t\int_0^xu(\xi;\tau)\;d\xi\,d\tau\;.\tag{2}$$
This equation is highly implicit and an explicit expression is desired. So then I thought of using Laplace transforms instead. Let $U(\cdot)$ be the Laplace transform of $u(\cdot)$, and $s$ be the complex co-variable of the real variable $t$, then
$$s\,U(x;s)-f(x)=-\lambda\,\int_0^xU(\xi;s)\,d\xi\tag{3}$$
which can be converted to the differential equation
$$s\,U'(x;s)+\lambda\,U(x;s)=f'(x)\tag{4}$$
where the derivative is now with respect to $x$. Equation $(4)$ is easily solvable. Though I am uncertain if I did the following Laplace transform correctly,
$$\mathscr{L}\left\{\int_0^xu(\xi;t)\,d\xi\right\}=\int_0^xU(\xi;s)\,d\xi\;.\tag{5}$$
I figured since the integration is over the parameter instead of the transforming variable I could bring it into the integral under the heuristic that the Laplace transform of a sum is the sum of the Laplace transforms, but I am unsure of this.
Any insight on any of this or alternative methods to solve Equation $(1)$ is welcome.
 A: Incomplete attempt at a solution but thought I would offer it anyways as food for thought.
Differentiate both sides of the integral equation with respect to $x$.
$$
\frac{\partial ^2 u}{\partial x\partial t}=-\lambda u(x,t)
$$
$$
u_{\alpha}(x,t)=A(\alpha)\cos(\alpha x+\frac{\lambda}{\alpha} t)+B(\alpha)\sin(\alpha x+\frac{\lambda}{\alpha}t),\;\;\;\alpha>0
$$
Now take advantage of the linearity of the equation, usually we sum but here we have a continuous parameter $\alpha$.
$$
u(x,t)=\int_{0}^{\infty}u_{\alpha}(x,t)\mathbb{d}\alpha
$$
$$
f(x)=u(x,0)=\int_{0}^{\infty}u_{\alpha}(x,0)\mathbb{d}\alpha
$$
$$
f(x)=\int_{0}^{\infty}A(\alpha)\cos(\alpha x)\mathbb{d}\alpha+\int_{0}^{\infty}B(\alpha)\sin(\alpha x)\mathbb{d}\alpha
$$
I would like to now solve for $A(\alpha),B(\alpha)$ in terms of $f(x)$. Perhaps if one converted the above trigonometric expressions to complex exponentials a certain Laplace/Fourier transform structure would emerge that would allow one to use the established formalism to invert (solve for $A,B$ in terms of $f$).
A: The solution detailed below is :
With $\quad F(s)=$ Laplace transform of $f(x)$.
$$\Phi(s,t)=e^{-\frac{\lambda \:t}{s}}F(s)$$
$$\boxed{u(x,t)= \text{ Inverse Laplace Transform of } \Phi(s,t)}$$
The result cannot be expressed more explicitly until the function $f(x)$ be explicitly given.

