For
x>0, substituting
y(x)=∫γextf(t)dt into
xy(3)+2y=0 and integrating by parts gives
[extt3f(t)]∂γ+∫γext{2f−(t3f)′}dt=0.
The amplitude equation
(t3f)′=2f has solution
f(t)=Ct−3e−1/t2. On
γ=(−∞,0) both endpoint terms vanish, and hence
y(x)=C∫−∞0ext−t−2t−3dt=C1∫0∞ue−u2−x/udu.