I have done some attempts to solve the following ODE: $$y'''(x)+ax\,y(x)=0\,,$$ with $a >0$.
I put $y(x)= e^{rx}$, with $r$ is an arbitrary real number, then $y'''=r^3 e^{rx}$, by substitution in the titled equation I got: $r^3 e^x +a x e^{rx}=0$ implies $ e^{rx}(r^3+a x)=0$ implies $x=\frac{-a}{r^3}$ this yield to $y(x)= \exp(\frac{-a x}{r^3})$, But I didn't got the same result using wolfram alpha as shown here , The result were given by Generalized hypergeometric function, Any help?