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. :)

  • $\begingroup$ 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. $\endgroup$
    – lc2r43
    Commented Jun 24, 2021 at 2:50
  • $\begingroup$ 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. $\endgroup$ Commented Jun 24, 2021 at 14:32
  • $\begingroup$ 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 $u(t)!$ $\endgroup$ Commented Jun 24, 2021 at 15:18

1 Answer 1


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*}

  • $\begingroup$ Ah, it seems I misunderstood the Laplace of a convolution integral. Thank you very much for the answer, have a good weekend:) $\endgroup$
    – Lbark05
    Commented Jun 26, 2021 at 23:04

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