\[ a x^m+y'(x)+y(x)^2=0 \] ✓ Mathematica : cpu = 0.085408 (sec), leaf count = 254
\[\left \{\left \{y(x)\to -\frac {i \sqrt {-a} x^{\frac {m+2}{2}} \left (-2 J_{\frac {1}{m+2}-1}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )+c_1 J_{\frac {m+1}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m}{2}+1}}{m+2}\right )-c_1 J_{-\frac {m+3}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )\right )-c_1 J_{-\frac {1}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )}{2 x \left (J_{\frac {1}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )+c_1 J_{-\frac {1}{m+2}}\left (\frac {2 i \sqrt {-a} x^{\frac {m+2}{2}}}{m+2}\right )\right )}\right \}\right \}\] ✓ Maple : cpu = 0.087 (sec), leaf count = 187
\[ \left \{ y \left ( x \right ) ={\frac {1}{x} \left ( -{{\sl J}_{{\frac {3+m}{m+2}}}\left (2\,{\frac {\sqrt {a}{x}^{m/2+1}}{m+2}}\right )}\sqrt {a}{x}^{{\frac {m}{2}}+1}{\it \_C1}-{{\sl Y}_{{\frac {3+m}{m+2}}}\left (2\,{\frac {\sqrt {a}{x}^{m/2+1}}{m+2}}\right )}\sqrt {a}{x}^{{\frac {m}{2}}+1}+{\it \_C1}\,{{\sl J}_{ \left ( m+2 \right ) ^{-1}}\left (2\,{\frac {\sqrt {a}{x}^{m/2+1}}{m+2}}\right )}+{{\sl Y}_{ \left ( m+2 \right ) ^{-1}}\left (2\,{\frac {\sqrt {a}{x}^{m/2+1}}{m+2}}\right )} \right ) \left ( {\it \_C1}\,{{\sl J}_{ \left ( m+2 \right ) ^{-1}}\left (2\,{\frac {\sqrt {a}{x}^{m/2+1}}{m+2}}\right )}+{{\sl Y}_{ \left ( m+2 \right ) ^{-1}}\left (2\,{\frac {\sqrt {a}{x}^{m/2+1}}{m+2}}\right )} \right ) ^{-1}} \right \} \]
\begin {align} y^{\prime }\left ( x\right ) +y^{2}\left ( x\right ) +ax^{m} & =0\nonumber \\ y^{\prime }\left ( x\right ) & =-ax^{m}-y^{2}\left ( x\right ) \tag {1} \end {align}
This is Riccati first order non-linear ODE of the form \begin {equation} y^{\prime }\left ( x\right ) =P\left ( x\right ) +Q\left ( x\right ) y+R\left ( x\right ) y^{2}\left ( x\right ) \tag {2} \end {equation} where in this case \(Q\left ( x\right ) =0,R\left ( x\right ) =-1,P\left ( x\right ) =-ax^{m}\). We can solve this in two ways. If we know one particular solution \(y_{p}\left ( x\right ) \) for (1) then we use the substitution \(y=y_{p}+\frac {1}{u}\) and convert (1) to new associated linear ODE of the form \(u^{\prime }+\left ( Q\left ( x\right ) +2R\left ( x\right ) \right ) y_{p}+R\left ( x\right ) =0\). If we do not know a particular solution, then we use the standard substitution \(y=\frac {-u^{\prime }}{uR\left ( x\right ) }=\frac {u^{\prime }}{u}\) since \(R\left ( x\right ) =-1\) 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 \(u^{\prime \prime }+q\left ( x\right ) u=0.\)
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\left ( x\right ) =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 \(u^{\prime }\left ( x\right ) \) 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 }\left ( x\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}}\\ +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
\[ y=\frac {1}{x}\frac {C_{1}\left [ BesselJ\left ( \frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselJ\left ( \frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) \right ] +BesselY\left ( \frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) -\sqrt {a}x^{\frac {m+1}{2}}BesselY\left ( \frac {m+3}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }{C_{1}BesselJ\left ( \frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) +BesselY\left ( \frac {1}{m+2},\frac {2\sqrt {a}x^{\frac {m+1}{2}}}{m+2}\right ) }\]