<s>[INST] $x^2y''+(2x^2+x)y'+(2x^2+x)y=0$ A Bessel equation $$x^2y''+(2x^2+x)y'+(2x^2+x)y=0$$
The solution is $$e^{-x}J_o(x)+e^{-x}Y_o(x)$$
How does one approach a problem like this?
 [/INST] One method is to recognize that $x^2 y''$ is frequently seen in many differential equations whereas the other terms are not. With this then one can consider a function of the type $y(x) = f(x) g(x)$.
\begin{align}
y(x) &= f g \\
y' &= f g' + f' g \\
y'' &= f g'' + 2 f' g' + f''
\end{align}
Now,
\begin{align}
0 &= x^2y''+(2x^2+x)y'+(2x^2+x)y \\
&= x^2 f g'' + 2 x^2 f' g' + x^2 f'' g + (2 x^2 + x) f g' + (2 x^2 + x) f' g + (2x^2 + x) f g \\
&= x^2 f g'' + (2 x^2 f' + 2 x^2 f + x f) g' +(x^2 f'' +(2 x^2 + x)f' + (2x^2 + x)f )g \\
&= x^2 g'' + \left( 2 x^2 \frac{f'}{f} + 2x^2 + x \right) g' + \left( x^2 \frac{f''}{f} + (2 x^2 + x) \frac{f'}{f} + 2 x^2 + x \right) g. 
\end{align}
Let, from the coefficient of $g'$, 
\begin{align}
\frac{f'}{f} = -1
\end{align}
which yields $f = e^{-x}$, $f' = - f$, $f'' = f$ and
\begin{align}
0 &= x^2 g'' + x g' + x^2 g
\end{align}
Compare this to the known Bessel function equation which is 
\begin{align}
x^2 w'' + x w' + (x^2 + \nu^2) w = 0 \hspace{10mm} w = A J_{\nu}(x) + B Y_{\nu}(x)
\end{align}
to obtain $g = A J_{0}(x) + B Y_{0}(x)$. Placing the components in the proper order it has then been found that
\begin{align}
y(x) = e^{-x} \, \left( A J_{0}(x) + B Y_{0}(x) \right).
\end{align}
</s>