Laplace solution of a first-order equation with exponential memory ($x^{\prime }+e^{-2t}\ast x=1,x(0)=0$)

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 $x^{\prime }=(1+t)e^{-t}$ and $(e^{-2t}\ast x)(t)=1-(1+t)e^{-t}$. Introducing the memory variable $w=e^{-2t}\ast x$ gives the equivalent local system $x^{\prime }+w=1$, $w^{\prime }=x-2w$, $w(0)=0$, explaining why the transform denominator has a repeated pole at $p=-1$.