<s>[INST] How to solve $'''+ ''+  = 0$? $$'''+ ''+  = 0$$
What method can I use to further solve the equation?
 [/INST] We can write the first as
\begin{align}
(xy'')^{'} + \left(\dfrac{x^2}{2}\right)' = 0
\end{align}
so it becomes a problem of integration, since of course the integrating process can be applied linearly. As such
\begin{align}
xy'' + \dfrac{x^2}{2} = C_1 &\implies y'' = \dfrac{C_1}{x} - \dfrac{x}{2}\\
&\implies \int y''(x) dx = \int \left(\dfrac{C_1}{x} - \dfrac{x}{2}\right) dx\\
&\implies y' = C_1 \ln x- \dfrac{x^2}{4}+C_2\\
&\implies \int y'(x) dx = \int \left( C_1 \ln x- \dfrac{x^2}{4}+C_2 \right)dx\\
&\implies y(x) = C_1 \int \ln x dx - \dfrac{1}{4} \int x^2 dx + C_2 \int 1 dx\\
&\implies y(x) = C_1 (x \ln x - x ) - \dfrac{x^3}{12} + C_2 x + C_3.
\end{align}
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