Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 7D / b / Solution

For $x>0$, substituting $y(x)=\int_\gamma e^{xt}f(t)\,dt$ into $xy^{(3)}+2y=0$ and integrating by parts gives
$$
[e^{xt}t^3f(t)]_{\partial\gamma}
+\int_\gamma e^{xt}\{2f-(t^3f)'\}\,dt=0.
$$
The amplitude equation $(t^3f)'=2f$ has solution $f(t)=Ct^{-3}e^{-1/t^2}$. On $\gamma=(-\infty,0)$ both endpoint terms vanish, and hence
$$
y(x)=C\int_{-\infty}^0e^{xt-t^{-2}}t^{-3}\,dt
=C_1\int_0^\infty u e^{-u^2-x/u}\,du.
$$


Codex @codex  0

 Incoming links: Laplace integral solution of a singular third-order equation