\[ \left (4 x^2+2\right ) y(x)+y''(x)+4 x y'(x)=0 \] ✓ Mathematica : cpu = 0.0100467 (sec), leaf count = 27
\[\left \{\left \{y(x)\to c_1 e^{-x^2}+c_2 e^{-x^2} x\right \}\right \}\] ✓ Maple : cpu = 0.027 (sec), leaf count = 16
\[y \relax (x ) = {\mathrm e}^{-x^{2}} \left (x c_{2}+c_{1}\right )\]
Hand solution
\begin {equation} y^{\prime \prime }+4xy^{\prime }+\left (4x^{2}+2\right ) y=0\tag {1} \end {equation}
Second order with varying coefficient. Using power series, let \(y=\sum _{n=0}^{\infty }c_{n}x^{n}\), hence\begin {align*} y^{\prime } & =\sum _{n=0}^{\infty }nc_{n}x^{n-1}=\sum _{n=1}^{\infty }nc_{n}x^{n-1}=\sum _{n=0}^{\infty }\left (n+1\right ) c_{n+1}x^{n}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }n\left (n+1\right ) c_{n+1}x^{n-1}=\sum _{n=1}^{\infty }n\left (n+1\right ) c_{n+1}x^{n-1}=\sum _{n=0}^{\infty }\left (n+1\right ) \left (n+2\right ) c_{n+2}x^{n} \end {align*}
Substituting back in the original ODE gives\begin {align*} \sum _{n=0}^{\infty }\left (n+1\right ) \left (n+2\right ) c_{n+2}x^{n}+4x\sum _{n=0}^{\infty }\left (n+1\right ) c_{n+1}x^{n}+\left (4x^{2}+2\right ) \sum _{n=0}^{\infty }c_{n}x^{n} & =0\\ \sum _{n=0}^{\infty }\left (n+1\right ) \left (n+2\right ) c_{n+2}x^{n}+\sum _{n=0}^{\infty }4\left (n+1\right ) c_{n+1}x^{n+1}+\sum _{n=0}^{\infty }4c_{n}x^{n+2}+\sum _{n=0}^{\infty }2c_{n}x^{n} & =0\\ \sum _{n=0}^{\infty }\left (n+1\right ) \left (n+2\right ) c_{n+2}x^{n}+\sum _{n=1}^{\infty }4nc_{n}x^{n}+\sum _{n=2}^{\infty }4c_{n-2}x^{n}+\sum _{n=0}^{\infty }2c_{n}x^{n} & =0 \end {align*}
For \(n=0\)\begin {align*} \left (n+1\right ) \left (n+2\right ) c_{n+2}+2c_{n} & =0\\ \relax (1) \relax (2) c_{2}+2c_{0} & =0\\ c_{2} & =-c_{0} \end {align*}
For \(n=1\)\begin {align*} \left (n+1\right ) \left (n+2\right ) c_{n+2}+4nc_{n}+2c_{n} & =0\\ \relax (2) \relax (3) c_{3}+4c_{1}+2c_{1} & =0\\ c_{3} & =-c_{1} \end {align*}
For \(n\geq 2\)\begin {align*} \left (n+1\right ) \left (n+2\right ) c_{n+2}+4nc_{n}+4c_{n-2}+2c_{n} & =0\\ c_{n+2} & =\frac {\left (-4n-2\right ) c_{n}-4c_{n-2}}{\left (n+1\right ) \left (n+2\right ) } \end {align*}
Hence for \(n=2\)\[ c_{4}=\frac {\left (-8-2\right ) c_{2}-4c_{0}}{\relax (3) \left ( 4\right ) }=\frac {\left (-8-2\right ) \left (-c_{0}\right ) -4c_{0}}{\left ( 3\right ) \relax (4) }=c_{0}\frac {6}{\relax (3) \left ( 4\right ) }\] For \(n=3\)\[ c_{5}=\frac {\left (-12-2\right ) c_{3}-4c_{1}}{\relax (4) \left ( 5\right ) }=\frac {\left (-12-2\right ) \left (-c_{1}\right ) -4c_{1}}{\relax (4) \relax (5) }=c_{1}\frac {10}{\relax (4) \relax (5) }\] For \(n=4\)\[ c_{6}=\frac {\left (-16-2\right ) c_{4}-4c_{2}}{\relax (5) \left ( 6\right ) }=\frac {\left (-16-2\right ) \left (c_{0}\frac {6}{\left ( 3\right ) \relax (4) }\right ) -4\left (-c_{0}\right ) }{\left ( 5\right ) \relax (6) }=c_{0}\frac {-5}{\relax (5) \left ( 6\right ) }\] For \(n=5\)\[ c_{7}=\frac {\left (-20-2\right ) c_{5}-4c_{3}}{\relax (6) \left ( 7\right ) }=\frac {\left (-20-2\right ) \left (c_{1}\frac {10}{\left ( 4\right ) \relax (5) }\right ) -4\left (-c_{1}\right ) }{\left ( 6\right ) \relax (7) }=c_{1}\frac {-7}{\relax (6) \left ( 7\right ) }\] For \(n=6\)\[ c_{8}=\frac {\left (-24-2\right ) c_{6}-4c_{4}}{\relax (7) \left ( 8\right ) }=\frac {\left (-24-2\right ) \left (c_{0}\frac {-5}{\left ( 5\right ) \relax (6) }\right ) -4\left (c_{0}\frac {6}{\left ( 3\right ) \relax (4) }\right ) }{\relax (7) \relax (8) }=c_{0}\frac {7}{3}\frac {1}{\relax (7) \relax (8) }\] And so on. Hence\begin {align*} y & =\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =c_{0}+c_{1}x+c_{2}x^{2}+\cdots \\ & =c_{0}+c_{1}x-c_{0}x^{2}-c_{1}x^{3}+c_{0}\frac {6}{\relax (3) \left ( 4\right ) }x^{4}+c_{1}\frac {10}{\relax (4) \relax (5) }x^{5}-c_{0}\frac {5}{\relax (5) \relax (7) }x^{6}-c_{1}\frac {7}{\relax (6) \relax (7) }x^{7}+c_{0}\frac {12}{7}\frac {1}{\relax (7) \relax (8) }x^{8}+\cdots \\ & =c_{0}\left (1-x^{2}+\frac {6}{\relax (3) \relax (4) }x^{4}-\frac {5}{\relax (5) \relax (6) }x^{6}+\frac {7}{3}\frac {1}{\relax (7) \relax (8) }x^{8}+\cdots \right ) +c_{1}\left ( x-x^{3}+\frac {10}{\relax (4) \relax (5) }x^{5}-\frac {7}{\left ( 6\right ) \relax (7) }x^{7}+\cdots \right ) \\ & =c_{0}\left (1-x^{2}+\frac {1}{2}x^{4}-\frac {1}{6}x^{6}+\frac {1}{24}x^{8}+\cdots \right ) +c_{1}\left (x-x^{3}+\frac {1}{2}x^{5}-\frac {1}{6}x^{7}+\cdots \right ) \end {align*}
But Taylor series for \(e^{-x^{2}}=1-x^{2}+\frac {1}{2}x^{4}-\frac {x^{6}}{4}+\cdots \), therefore the above becomes\[ y=c_{0}e^{-x^{2}}+c_{1}xe^{-x^{2}}\]
Verification