# Integro-Differential equation from my Complex analysis exam

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. :)

• The Laplace transform of $e^{-t}$ is $\frac{1}{s+1}$, not $e^{-s}$, so what you would actually get is $\hat{u}(s) = \frac{2s+2}{2s^2+2s+1}$ if I'm not mistaken. In this case, you should be able to rewrite the denominator in a more convenient form; the final answer should involve $\sin,\cos,\exp$ with appropriate arguments. Commented Jun 24, 2021 at 2:50
• Honestly, while you can certainly use the LT on an integro-differential equation, I would have just differentiated this equation once to get to a pure DE. Commented Jun 24, 2021 at 14:32
• Oops, my bad. Need to rework. I would have some strong words for anyone who would, in the context of LT's, where the unit step function is ubiquitous, write a DE in terms of $u(t)!$ Commented Jun 24, 2021 at 15:18

## 1 Answer

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*}

• Ah, it seems I misunderstood the Laplace of a convolution integral. Thank you very much for the answer, have a good weekend:) Commented Jun 26, 2021 at 23:04