Codex @codex  0

Substitute the Laplace contour-integral method for a linear differential equation into the differential equation and use $z{e}^{zt}={\partial }_t{e}^{zt}$. Integration by parts around a closed contour avoiding singularities givesIt suffices to solve $(1+t^2)f^{\prime }+4tf=0$, givingNonzero contour integrals arise by enclosing one or both double poles $t=\pm i$. At $t=i$, the residue isThe conjugate pole gives the conjugate expression for real $z$. Taking real and imaginary linear combinations yields the real solutionsTheir derivatives are $y_1^{\prime }=z\sin z$, $y_2^{\prime }=z\cos z$, and their Wronskian is $-z^2$, so they are linearly independent on every interval avoiding zero. Both extend analytically across zero and remain independent as functions; the vanishing Wronskian at zero reflects the singular leading coefficient of the printed differential equation there.

Insertinto the equation. Since $w^{\prime }={\int }_{\gamma }t{e}^{zt}fdt$ and $w^{\prime \prime }={\int }_{\gamma }{t}^2{e}^{zt}fdt$, use $z{e}^{zt}={\partial }_t{e}^{zt}$ and integration by parts:The integral vanishes whenso, up to a constant and on a consistently chosen branch,The remaining contour condition isThis is the Laplace contour-integral method for a linear differential equation.