\[ a x^2 y(x)+x y''(x)+2 y'(x)=0 \] ✓ Mathematica : cpu = 0.0053405 (sec), leaf count = 42
\[\left \{\left \{y(x)\to \frac {c_1 \text {Ai}\left (-\frac {a x}{(-a)^{2/3}}\right )}{x}+\frac {c_2 \text {Bi}\left (-\frac {a x}{(-a)^{2/3}}\right )}{x}\right \}\right \}\] ✓ Maple : cpu = 0.034 (sec), leaf count = 33
\[y \relax (x ) = \frac {c_{2} \BesselY \left (\frac {1}{3}, \frac {2 \sqrt {a}\, x^{\frac {3}{2}}}{3}\right )+c_{1} \BesselJ \left (\frac {1}{3}, \frac {2 \sqrt {a}\, x^{\frac {3}{2}}}{3}\right )}{\sqrt {x}}\]
Hand solution
\begin {equation} xy^{\prime \prime }+2y^{\prime }+ax^{2}y=0\tag {1} \end {equation}
Since there is a term \(2y\), we can use \(y=\frac {u\relax (x) }{x}\), hence\begin {align*} y^{\prime } & =\frac {u^{\prime }}{x}-\frac {u}{x^{2}}\\ y^{\prime \prime } & =\frac {u^{\prime \prime }}{x}-\frac {u^{\prime }}{x^{2}}-\frac {u^{\prime }}{x^{2}}+2\frac {u}{x^{3}} \end {align*}
And (1) becomes\begin {align} x\left (\frac {u^{\prime \prime }}{x}-\frac {u^{\prime }}{x^{2}}-\frac {u^{\prime }}{x^{2}}+2\frac {u}{x^{3}}\right ) +2\left (\frac {u^{\prime }}{x}-\frac {u}{x^{2}}\right ) +ax^{2}\left (\frac {u}{x}\right ) & =0\nonumber \\ u^{\prime \prime }-2\frac {u^{\prime }}{x}+2\frac {u}{x^{2}}+\frac {2u^{\prime }}{x}-\frac {2u}{x^{2}}+axu & =0\nonumber \\ u^{\prime \prime }+axu & =0\tag {2} \end {align}
This is Emdon-Fowler. (form is \(u^{\prime \prime }+x^{n}u=0\)) with \(n=1\). Assume that \[ u=\sum _{n=0}^{\infty }c_{n}x^{n}\] Hence\begin {align*} u^{\prime } & =\sum _{n=0}nc_{n}x^{n-1}=\sum _{n=1}nc_{n}x^{n-1}=\sum _{n=0}\left (n+1\right ) c_{n+1}x^{n}\\ u^{\prime \prime } & =\sum _{n=0}n\left (n+1\right ) c_{n+1}x^{n-1}=\sum _{n=1}n\left (n+1\right ) c_{n+1}x^{n-1}=\sum _{n=0}\left (n+1\right ) \left (n+2\right ) c_{n+2}x^{n} \end {align*}
Substituting back in (2) gives\begin {align*} \sum _{n=0}\left (n+1\right ) \left (n+2\right ) c_{n+2}x^{n}+\sum _{n=0}ac_{n}x^{n+1} & =0\\ \sum _{n=0}\left (n+1\right ) \left (n+2\right ) c_{n+2}x^{n}+\sum _{n=1}ac_{n-1}x^{n} & =0 \end {align*}
For \(n=0\)\[ \relax (1) \relax (2) c_{2}=0 \] Hence \(c_{2}=0\). For \(n\geq 1\)\begin {align} \left (n+1\right ) \left (n+2\right ) c_{n+2}+ac_{n-1} & =0\nonumber \\ c_{n+2} & =\frac {-ac_{n-1}}{\left (n+1\right ) \left (n+2\right ) }\tag {3} \end {align}
For \(n=1\,\), from (3)\[ c_{3}=\frac {-ac_{0}}{\relax (2) \relax (3) }\] For \(n=2\), from (3)\[ c_{4}=\frac {-ac_{1}}{\relax (3) \relax (4) }\] For \(n=3\), from (3)\[ c_{5}=\frac {-ac_{2}}{\relax (4) \relax (5) }=0 \] For \(n=4\), from (3)\[ c_{6}=\frac {-ac_{3}}{\relax (5) \relax (6) }=\frac {-a}{\left ( 5\right ) \relax (6) }\left (\frac {-ac_{0}}{\relax (2) \left ( 3\right ) }\right ) =\frac {a^{2}c_{0}}{\relax (2) \relax (3) \relax (5) \relax (6) }\] For \(n=5\), from (3)\[ c_{7}=\frac {-ac_{4}}{\relax (6) \relax (7) }=\frac {-a}{\left ( 6\right ) \relax (7) }\left (\frac {-ac_{1}}{\relax (3) \left ( 4\right ) }\right ) =\frac {a^{2}c_{1}}{\relax (3) \relax (4) \relax (6) \relax (7) }\] For \(n=6\), from (3)\[ c_{8}=\frac {-ac_{5}}{\relax (7) \relax (8) }=0 \] For \(n=7\), from (3)\[ c_{9}=\frac {-ac_{6}}{\relax (8) \relax (9) }=\frac {-a}{\left ( 8\right ) \relax (9) }\left (\frac {a^{2}c_{0}}{\relax (2) \relax (3) \relax (5) \relax (6) }\right ) =\frac {-a^{3}c_{0}}{\relax (2) \relax (3) \relax (5) \left ( 6\right ) \relax (8) \relax (9) }\] For \(n=8\), from (3)\[ c_{10}=\frac {-ac_{7}}{\relax (9) \left (10\right ) }=\frac {-a}{\left ( 9\right ) \left (10\right ) }\left (\frac {a^{2}c_{1}}{\relax (3) \relax (4) \relax (6) \relax (7) }\right ) =\frac {-a^{3}c_{1}}{\relax (3) \relax (4) \relax (6) \left ( 7\right ) \relax (9) \left (10\right ) }\] And so on. Hence,\begin {align*} u & =\sum _{n=0}^{\infty }c_{n}x^{n}=c_{0}+c_{1}x^{1}+c_{2}x^{2}+c_{3}x^{3}+\cdots \\ & =c_{0}+c_{1}x+0+c_{3}x^{3}+c_{4}x^{4}+0+c_{6}x^{6}+c_{7}x^{7}+0+c_{9}x^{9}+\cdots \\ & =c_{0}+c_{1}x-\frac {ac_{0}}{\relax (2) \relax (3) }x^{3}-\frac {ac_{1}}{\relax (3) \relax (4) }x^{4}+\frac {a^{2}c_{0}}{\relax (2) \relax (3) \relax (5) \left ( 6\right ) }x^{6}+\frac {a^{2}c_{1}}{\relax (3) \relax (4) \left ( 6\right ) \relax (7) }x^{7}-\frac {a^{3}c_{0}}{\relax (2) \left ( 3\right ) \relax (5) \relax (6) \relax (8) \left ( 9\right ) }x^{9}-\frac {a^{3}c_{1}}{\relax (3) \relax (4) \left ( 6\right ) \relax (7) \relax (9) \left (10\right ) }x^{10}+\cdots \\ & =c_{0}\left (1-\frac {a}{6}x^{3}+\frac {a^{2}}{180}x^{6}-\frac {a^{3}}{12\,960}x^{9}+\cdots \right ) +xc_{1}\left (1-\frac {a}{12}x^{3}+\frac {a^{2}}{504}x^{6}-\frac {a^{3}}{45\,360}x^{9}+\cdots \right ) \\ & =c_{0}\left (1-\frac {1}{6}\left (a^{\frac {1}{3}}x\right ) ^{3}+\frac {1}{180}\left (a^{\frac {1}{3}}x\right ) ^{6}-\frac {1}{12\,960}\left ( a^{\frac {1}{3}}x\right ) ^{9}+\cdots \right ) +xc_{1}\left (1-\frac {1}{12}\left (a^{\frac {1}{3}}x\right ) ^{3}+\frac {1}{504}\left (a^{\frac {1}{3}}x\right ) ^{6}-\frac {1}{45\,360}\left (a^{\frac {1}{3}}x\right ) ^{9}+\cdots \right ) \end {align*}
Comparing the above to the series expansion of Airy functions, we see that\[ u=c_{0}\operatorname {AiryAI}\left (-a^{\frac {1}{3}}x\right ) +c_{1}\operatorname {AiryBI}\left (-a^{\frac {1}{3}}x\right ) \]
And since \(y=\frac {u\relax (x) }{x}\) then\[ y=\frac {1}{x}\left (c_{0}\operatorname {AiryAI}\left (-a^{\frac {1}{3}}x\right ) +c_{1}\operatorname {AiryBI}\left (-a^{\frac {1}{3}}x\right ) \right ) \]
Verification