\[ a x^m+y'(x)+y(x)^2=0 \] ✓ Mathematica : cpu = 0.105352 (sec), leaf count = 254
\[\left \{\left \{y(x)\to -\frac {i \sqrt {-a} x^{\frac {m+2}{2}} \left (c_1 J_{\frac {m+1}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m}{2}+1}}{m+2}\right )-2 J_{\frac {1}{m+2}-1}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )-c_1 J_{-\frac {m+3}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )\right )-c_1 J_{-\frac {1}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )}{2 x \left (J_{\frac {1}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )+c_1 J_{-\frac {1}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )\right )}\right \}\right \}\] ✓ Maple : cpu = 0.129 (sec), leaf count = 187
\[y \relax (x ) = \frac {-\BesselJ \left (\frac {3+m}{m +2}, \frac {2 \sqrt {a}\, x^{\frac {m}{2}+1}}{m +2}\right ) \sqrt {a}\, x^{\frac {m}{2}+1} c_{1}-\BesselY \left (\frac {3+m}{m +2}, \frac {2 \sqrt {a}\, x^{\frac {m}{2}+1}}{m +2}\right ) \sqrt {a}\, x^{\frac {m}{2}+1}+c_{1} \BesselJ \left (\frac {1}{m +2}, \frac {2 \sqrt {a}\, x^{\frac {m}{2}+1}}{m +2}\right )+\BesselY \left (\frac {1}{m +2}, \frac {2 \sqrt {a}\, x^{\frac {m}{2}+1}}{m +2}\right )}{x \left (c_{1} \BesselJ \left (\frac {1}{m +2}, \frac {2 \sqrt {a}\, x^{\frac {m}{2}+1}}{m +2}\right )+\BesselY \left (\frac {1}{m +2}, \frac {2 \sqrt {a}\, x^{\frac {m}{2}+1}}{m +2}\right )\right )}\]
Hand solution
\begin {align} y^{\prime }\relax (x) +y^{2}\relax (x) +ax^{m} & =0\nonumber \\ y^{\prime }\relax (x) & =-ax^{m}-y^{2}\relax (x) \tag {1} \end {align}
This is Riccati first order non-linear ODE of the form \begin {equation} y^{\prime }\relax (x) =P\relax (x) +Q\relax (x) y+R\left ( x\right ) y^{2}\relax (x) \tag {2} \end {equation} where in this case \(Q\relax (x) =0,R\relax (x) =-1,P\left ( x\right ) =-ax^{m}\). We can solve this in two ways. If we know one particular solution \(y_{p}\relax (x) \) for (1) then we use the substitution \(y=y_{p}+\frac {1}{u}\) and convert (1) to new associated linear ODE of the form \(u^{\prime }+\left (Q\relax (x) +2R\relax (x) \right ) y_{p}+R\relax (x) =0\). If we do not know a particular solution, then we use the standard substitution \(y=\frac {-u^{\prime }}{uR\relax (x) }=\frac {u^{\prime }}{u}\) since \(R\relax (x) =-1\) and this is what we will do here.
Since \(u^{\prime }=yu\) then \begin {align*} u^{\prime \prime } & =yu^{\prime }+y^{\prime }u\\ & =y\left (yu\right ) +\left (-ax^{m}-y^{2}\right ) u\\ & =y^{2}u-ax^{m}u-y^{2}u\\ & =-ax^{m}u \end {align*}
So we have new second order ODE \begin {equation} u^{\prime \prime }+ax^{m}u=0 \tag {3} \end {equation} which we solve for \(u\). This is Airy ODE but with a positive sign. Of the form \(u^{\prime \prime }+q\relax (x) u=0.\)
Recall that the solution to \(u^{\prime \prime }-axu=0\) is
\[ u=c_{1}Ai\left (a^{\frac {1}{3}}x\right ) +c_{2}Bi\left (a^{\frac {1}{3}}x\right ) \]
When \(x\) has power on it (there are restriction on what values the power can take), the solution is written in terms of Bessel functions. The solution to \(u^{\prime \prime }-ax^{m}u=0\) is
\[ u=c_{1}\sqrt {x}BesselI\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) +c_{2}\sqrt {x}BesselK\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \]
When the sign is positive, the solution to \(u^{\prime \prime }+ax^{m}u=0\) is
\begin {equation} u\relax (x) =c_{1}\sqrt {x}BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) +c_{2}\sqrt {x}BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \tag {4} \end {equation}
We need to find \(u^{\prime }\relax (x) \) now. From (4)
\[ \frac {d}{dx}\left [ c_{1}\sqrt {x}BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] =c_{1}\frac {BesselJ\left ( \frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselJ\left (\frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }{\sqrt {x}}\]
And
\[ \frac {d}{dx}\left [ c_{2}\sqrt {x}BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] =c_{2}\frac {BesselY\left ( \frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselY\left (\frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }{\sqrt {x}}\]
Therefore
\begin {align*} u^{\prime }\relax (x) =c_{1}\frac {BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselJ\left (\frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }{\sqrt {x}}\\ +c_{2}\frac {BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselY\left ( \frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }{\sqrt {x}} \end {align*}
Since \(u^{\prime }=yu\) then
\begin {align*} y & =\frac {u^{\prime }}{u}\\ & =\frac {c_{1}\frac {BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselJ\left (\frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }{\sqrt {x}}+c_{2}\frac {BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselY\left (\frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }{\sqrt {x}}}{c_{1}\sqrt {x}BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) +c_{2}\sqrt {x}BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }\\ & =\frac {c_{1}\left [ BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselJ\left ( \frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] +c_{2}\left [ BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselY\left (\frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] }{\sqrt {x}\left [ c_{1}\sqrt {x}BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) +c_{2}\sqrt {x}BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] }\\ & =\frac {c_{1}\left [ BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselJ\left ( \frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] +c_{2}\left [ BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselY\left (\frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] }{c_{1}xBesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) +c_{2}xBesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) } \end {align*}
Let \(C_{1}=\frac {c_{1}}{c_{2}}\) then the above can be written as
\[ y=\frac {1}{x}\frac {C_{1}\left [ BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselJ\left ( \frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] +BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselY\left (\frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }{C_{1}BesselJ\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) +BesselY\left (\frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }\]