= Solution
For the Laplace-type contour ansatz, differentiating under the integral and putting $P(t)=t^2-t-2$ gives the differential-equation residual
$$
\int_\gamma e^{zt}\{zP(t)f(t)+(2-t)f(t)\}\,dt.
$$
<Integration by parts> turns this into $[e^{zt}Pf]_{\partial\gamma}+\int_\gamma e^{zt}[(2-t)f-(Pf)']dt$. Thus it vanishes whenever the endpoint contribution vanishes and $(Pf)'=(2-t)f$. The latter condition is $Pf'=3(1-t)f$, solved by
$$
f(t)=\frac1{(t-2)(t+1)^2}
$$
up to a constant multiplier. Closed contours avoiding the poles make the endpoint contribution zero.
Choose separate positively oriented small circles around 2 and $-1$. The <residue theorem> gives, after division by $2\pi i$,
$$
\operatorname{Res}_{t=2}(e^{zt}f)=\frac{e^{2z}}9,\qquad
\operatorname{Res}_{t=-1}(e^{zt}f)=-\frac{(3z+1)e^{-z}}9.
$$
Therefore
$$
\boxed{w(z)=Ae^{2z}+B(3z+1)e^{-z}.}
$$
The <Wronskian> of the two basis functions is $-9ze^z$, nonzero for $z\ne0$. They span the two-dimensional solution space on every regular domain, and both extend entire across the apparent singular point at zero. Their initial values satisfy $w(0)=A+B$ and $w'(0)=2(A+B)$. Thus \b[every solution has $w'(0)=2w(0)$], and the requested combination $w(0)=0$, $w'(0)\ne0$ is impossible. No additional logarithmic solution is lost: the two independent functions already span the solution space away from zero.
Back to article page
