= Laplace solution of a first-order equation with exponential memory
{c}
{title2=$x'+e^{-2t}*x=1,\quad x(0)=0$}
The <derivative> and <convolution theorem for Laplace transforms> give $[p+1/(p+2)]\bar x=1/p$. Partial fractions yield $\bar x=2/p-2/(p+1)-1/(p+1)^2$, so $x(t)=2-(2+t)e^{-t}$ tends to two. A direct check gives $x'=(1+t)e^{-t}$ and $(e^{-2t}*x)(t)=1-(1+t)e^{-t}$. Introducing the memory variable $w=e^{-2t}*x$ gives the equivalent local system $x'+w=1$, $w'=x-2w$, $w(0)=0$, explaining why the transform denominator has a repeated pole at $p=-1$.
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