My recent complex analysis exam had the following problem as the last question, which I had a hard time solving.
The problem
Use the Laplace transform to solve the following differential equation for $u(t)$.
$$\dfrac{du(t)}{dt}+\dfrac{1}{2}\int_{0}^{t} e^{-t'}u(t-t')\,dt'=0$$
with the initial condition $u(0)=1.$
My attempt
When applying the Laplace transform, the first term becomes $s\hat{u}(s)-u(0)=s\hat{u}(s)-1$
For the second term I used the formula for a Laplace transform of a convolution integral
$$ L\bigg\{ \int_{0}^{t} g(\tau)f(t-\tau)\,d\tau\bigg\} = \hat{f}(s)\hat{g}(s) $$
This approach meant that the second term would be $\dfrac{1}{2}e^{-s}\hat{u}(s)$
After isolationg for $\hat{u}(s)$ I had
$$ \hat{u}(s)= \dfrac{1}{s+\dfrac{1}{2}e^{-s}} $$
I then tried to apply the inverse Laplace transform
$$u(s)=\dfrac{1}{2\pi i}\int_{\lambda - i\infty}^{\lambda + i\infty} \dfrac{e^{st}}{s+\dfrac{1}{2}e^{-s}}\,ds$$
When trying to find a singular point in the integrand, I found the solution $s=\mathrm{LambertW}\left(-\dfrac{1}{2}\right)$
I am not entirely familiar with the LambertW-function, and my attempt ended here.
My question
Did I make any mistakes leading up to the inverse Laplace?
Is my approach even correct?
How would you go about solving?
Is this considered an Integro-differential-equation?
Thanks for your time. :)
We can do \begin{align*} \dfrac{du(t)}{dt}+\dfrac{1}{2}\int_{0}^{t} e^{-t'}u(t-t')\,dt'&=0\\ sU(s)-u(0^{-})+\frac12\,\mathcal{L}[e^{-t}]\,\mathcal{L}[u(t)]&=0\\ sU(s)-1+\frac{U(s)}{2(s+1)}&=0\\ U(s)&=\frac{2s+2}{2s^2+2s+1}\\ &=\frac{s+1}{s^2+s+1/2}\\ &=\frac{s+1/2+1/2}{(s+1/2)^2+1/4}\\ &=\frac{s+1/2}{(s+1/2)^2+1/4}+\frac{1/2}{(s+1/2)^2+1/4}\\ u(t)&=e^{-t/2}\cos(t/2)\operatorname{UnitStep}(t)+e^{-t/2}\sin(t/2)\operatorname{UnitStep}(t)\\ &=[\cos(t/2)+\sin(t/2)]\,e^{-t/2}\operatorname{UnitStep}(t). \end{align*}