\[ a+x y'(x)+x y(x)^2=0 \] ✓ Mathematica : cpu = 0.0841315 (sec), leaf count = 157
\[\left \{\left \{y(x)\to -\frac {c_1 J_1\left (2 i \sqrt {-a} \sqrt {x}\right )+i \sqrt {-a} \sqrt {x} \left (-2 J_0\left (2 i \sqrt {-a} \sqrt {x}\right )+c_1 J_0\left (2 i \sqrt {-a} \sqrt {x}\right )-c_1 J_2\left (2 i \sqrt {-a} \sqrt {x}\right )\right )}{2 x \left (J_1\left (2 i \sqrt {-a} \sqrt {x}\right )-c_1 J_1\left (2 i \sqrt {-a} \sqrt {x}\right )\right )}\right \}\right \}\] ✓ Maple : cpu = 0.097 (sec), leaf count = 59
\[y \relax (x ) = \frac {\sqrt {a}\, \left (\BesselJ \left (0, 2 \sqrt {a}\, \sqrt {x}\right ) c_{1}+\BesselY \left (0, 2 \sqrt {a}\, \sqrt {x}\right )\right )}{\sqrt {x}\, \left (c_{1} \BesselJ \left (1, 2 \sqrt {a}\, \sqrt {x}\right )+\BesselY \left (1, 2 \sqrt {a}\, \sqrt {x}\right )\right )}\]
Hand solution
\begin {align*} xy^{\prime }+xy^{2}+a & =0\\ y^{\prime } & =-\frac {a}{x}-y^{2} \end {align*}
This is Riccati first order non-linear. Let \(y=-\frac {u^{\prime }}{uR}=\frac {u^{\prime }}{u}\), hence \(y^{\prime }=\frac {u^{\prime \prime }}{u}-\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}}\). Equating this to RHS of the above gives\begin {align*} \frac {u^{\prime \prime }}{u}-\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}} & =-\frac {a}{x}-\left (\frac {u^{\prime }}{u}\right ) ^{2}\\ \frac {u^{\prime \prime }}{u} & =-\frac {a}{x}\\ u^{\prime \prime }+\frac {a}{x}u & =0 \end {align*}
This is linear second order, an Emden Fowler ODE, with removal singularity. Solved using power series method. The solution is\[ u=C_{1}\sqrt {x}\operatorname {BesselJ}\left (1,2\sqrt {ax}\right ) +C_{2}\sqrt {x}\operatorname {BesselY}\left (1,2\sqrt {ax}\right ) \] But \[ \frac {d}{dx}\operatorname {BesselJ}\left (1,2\sqrt {ax}\right ) =\frac {\sqrt {a}}{\sqrt {x}}\left (\operatorname {BesselJ}\left (0,2\sqrt {ax}\right ) -\frac {1}{2}\frac {1}{\sqrt {ax}}\operatorname {BesselJ}\left (1,2\sqrt {ax}\right ) \right ) \] And\[ \frac {d}{dx}\operatorname {BesselY}\left (1,2\sqrt {ax}\right ) =\frac {\sqrt {a}}{\sqrt {x}}\left (\operatorname {BesselY}\left (0,2\sqrt {ax}\right ) -\frac {1}{2}\frac {1}{\sqrt {ax}}\operatorname {BesselY}\left (1,2\sqrt {ax}\right ) \right ) \] Therefore, \begin {align*} u^{\prime } & =C_{1}\left (\frac {1}{2\sqrt {x}}\operatorname {BesselJ}\left ( 1,2\sqrt {a}\sqrt {x}\right ) +\sqrt {x}\frac {\sqrt {a}}{\sqrt {x}}\left ( \operatorname {BesselJ}\left (0,2\sqrt {a}\sqrt {x}\right ) -\frac {1}{2}\frac {1}{\sqrt {ax}}\operatorname {BesselJ}\left (1,2\sqrt {a}\sqrt {x}\right ) \right ) \right ) \\ & +C_{2}\left (\frac {1}{2\sqrt {x}}\operatorname {BesselY}\left (1,2\sqrt {a}\sqrt {x}\right ) +\sqrt {x}\frac {\sqrt {a}}{\sqrt {x}}\left ( \operatorname {BesselY}\left (0,2\sqrt {ax}\right ) -\frac {1}{2}\frac {1}{\sqrt {ax}}\operatorname {BesselY}\left (1,2\sqrt {a}\sqrt {x}\right ) \right ) \right ) \end {align*}
Which is simplified to\[ u^{\prime }=C_{1}\sqrt {a}\operatorname {BesselJ}\left (0,2\sqrt {a}\sqrt {x}\right ) +C_{2}\sqrt {a}\operatorname {BesselY}\left (0,2\sqrt {a}\sqrt {x}\right ) \] Therefore, from \(y=\frac {u^{\prime }}{u}\), the solution is\[ y=\frac {C_{1}\sqrt {a}\operatorname {BesselJ}\left (0,2\sqrt {a}\sqrt {x}\right ) +C_{2}\sqrt {a}\operatorname {BesselY}\left (0,2\sqrt {a}\sqrt {x}\right ) }{C_{1}\sqrt {x}\operatorname {BesselJ}\left (1,2\sqrt {a}\sqrt {x}\right ) +C_{2}\sqrt {x}\operatorname {BesselY}\left (1,2\sqrt {a}\sqrt {x}\right ) }\] Let \(C=\frac {C_{1}}{C_{2}}\), hence\[ y=\frac {C\sqrt {a}\operatorname {BesselJ}\left (0,2\sqrt {a}\sqrt {x}\right ) +\ \sqrt {a}\operatorname {BesselY}\left (0,2\sqrt {a}\sqrt {x}\right ) }{C\sqrt {x}\operatorname {BesselJ}\left (1,2\sqrt {a}\sqrt {x}\right ) +\ \sqrt {x}\operatorname {BesselY}\left (1,2\sqrt {a}\sqrt {x}\right ) }\] Verification