\[ a y(x)^2-b x^{\nu }+y'(x)=0 \] ✓ Mathematica : cpu = 0.123375 (sec), leaf count = 277
DSolve[-(b*x^nu) + a*y[x]^2 + Derivative[1][y][x] == 0,y[x],x]
\[\left \{\left \{y(x)\to -\frac {\sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}} \left (c_1 J_{\frac {\nu +1}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu }{2}+1}}{\nu +2}\right )-2 J_{\frac {1}{\nu +2}-1}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )-c_1 J_{-\frac {\nu +3}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )\right )-c_1 J_{-\frac {1}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )}{2 a x \left (J_{\frac {1}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )+c_1 J_{-\frac {1}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )\right )}\right \}\right \}\] ✓ Maple : cpu = 0.067 (sec), leaf count = 214
dsolve(diff(y(x),x)+a*y(x)^2-b*x^nu = 0,y(x))
\[y \left (x \right ) = \frac {-\BesselJ \left (\frac {3+\nu }{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right ) \sqrt {-a b}\, x^{\frac {\nu }{2}+1} c_{1}-\BesselY \left (\frac {3+\nu }{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right ) \sqrt {-a b}\, x^{\frac {\nu }{2}+1}+c_{1} \BesselJ \left (\frac {1}{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right )+\BesselY \left (\frac {1}{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right )}{x a \left (c_{1} \BesselJ \left (\frac {1}{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right )+\BesselY \left (\frac {1}{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right )\right )}\]
Hand solution
\begin {align} y^{\prime }+ay^{2}-bx^{v} & =0\nonumber \\ y^{\prime } & =bx^{v}-ay^{2}\tag {1}\\ & =P\left ( x\right ) +Q\left ( x\right ) y+R\left ( x\right ) y^{2}\nonumber \end {align}
This is Riccati first order non-linear ODE with \(P\left ( x\right ) =bx^{v},Q\left ( x\right ) =0,R\left ( x\right ) =-a\). Using the standard substitution
\[ y=-\frac {u^{\prime }}{uR\left ( x\right ) }=\frac {u^{\prime }}{au}\]
Hence
\[ y^{\prime }=\frac {u^{\prime \prime }}{au}-\frac {\left ( u^{\prime }\right ) ^{2}}{au^{2}}\]
Therefore (1) becomes
\begin {align*} \frac {u^{\prime \prime }}{au}-\frac {\left ( u^{\prime }\right ) ^{2}}{au^{2}} & =bx^{v}-ay^{2}\\ & =bx^{v}-a\left ( \frac {u^{\prime }}{au}\right ) ^{2}\\ & =bx^{v}-\frac {\left ( u^{\prime }\right ) ^{2}}{au^{2}} \end {align*}
Hence
\begin {align*} \frac {u^{\prime \prime }}{au} & =bx^{v}\\ u^{\prime \prime }-abx^{v}u & =0 \end {align*}
This is an Emden-Fowler equation, of the general form \(u^{\prime \prime }=Ax^{n}u^{m}\), where here \(m=1\) and \(n=v\) and \(A=ab\).
For any \(n\), the solution uses Bessel functions and modified Bessel functions of first and second kind. From Handbook of exact solutions for ODE, page 237, equation 2.1.2.7 we see the solution is given as
\[ u=\left \{ \begin {array} [c]{ccc}C_{1}\sqrt {x}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +C_{2}\sqrt {x}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) & & ab<0\\ C_{1}\sqrt {x}I_{\frac {1}{2q}}\left ( \frac {\sqrt {ab}}{q}x^{q}\right ) +C_{2}\sqrt {x}K_{\frac {1}{2q}}\left ( \frac {\sqrt {ab}}{q}x^{q}\right ) & & ab>0 \end {array} \right . \]
Where \(q=\frac {n+1}{2}\). \(J\) is Bessel function of first kind and \(Y\) is Bessel function of second kind. \(I\) is modified Besself function of first kind and \(K\) is modified Besself function of second kind. To find \(y\) we now use \(y=\frac {u^{\prime }}{au}\). Derivative of Bessel functions is given by
\begin {align*} J_{m}^{\prime }\left ( x\right ) & =\frac {1}{2}\left ( J_{m-1}\left ( x\right ) -J_{m+1}\left ( x\right ) \right ) \\ Y_{m}^{\prime }\left ( x\right ) & =\frac {1}{2}\left ( Y_{m-1}\left ( x\right ) -Y_{m+1}\left ( x\right ) \right ) \\ I_{m}^{\prime }\left ( x\right ) & =\frac {1}{2}\left ( I_{m-1}\left ( x\right ) +I_{m+1}\left ( x\right ) \right ) \\ K_{m}^{\prime }\left ( x\right ) & =-\frac {1}{2}\left ( K_{m-1}\left ( x\right ) +K_{m+1}\left ( x\right ) \right ) \end {align*}
Using these, then
\[ u^{\prime }=\left \{ \begin {array} [c]{ccc}C_{1}\left [ \frac {1}{2\sqrt {x}}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +\sqrt {x}J_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ] +C_{2}\left [ \frac {1}{2\sqrt {x}}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +\sqrt {x}Y_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ] & & ab<0\\ C_{1}\left [ \frac {1}{2\sqrt {x}}I_{\frac {1}{2q}}\left ( \frac {\sqrt {ab}}{q}x^{q}\right ) +\sqrt {x}I_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {ab}}{q}x^{q}\right ) \right ] +C_{2}\left [ \frac {1}{2\sqrt {x}}K_{\frac {1}{2q}}\left ( \frac {\sqrt {ab}}{q}x^{q}\right ) +\sqrt {x}K_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {ab}}{q}x^{q}\right ) \right ] & & ab>0 \end {array} \right . \]
Hence for \(ab<0\)
\begin {align*} y & =\frac {u^{\prime }}{au}\\ & =\frac {C_{1}\left [ \frac {1}{2\sqrt {x}}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +\sqrt {x}J_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ] +C_{2}\left [ \frac {1}{2\sqrt {x}}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +\sqrt {x}Y_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ] }{aC_{1}\sqrt {x}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +aC_{2}\sqrt {x}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) }\\ & =\frac {\sqrt {x}C_{1}\left [ \frac {1}{2x}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +J_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ] +\sqrt {x}C_{2}\left [ \frac {1}{2x}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +Y_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ] }{aC_{1}\sqrt {x}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +aC_{2}\sqrt {x}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) }\\ & =\frac {C_{1}\left [ \frac {1}{2x}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +J_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ] +C_{2}\left [ \frac {1}{2x}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +Y_{\frac {1}{2q}}^{\prime }\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ] }{aC_{1}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +aC_{2}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) } \end {align*}
Using derivatives the above becomes
\begin {align*} y & =\frac {C_{1}\left [ \frac {1}{2x}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +\frac {1}{2}\left ( J_{\frac {1}{2q}-1}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) -J_{\frac {1}{2q}+1}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ) \right ] }{aC_{1}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +aC_{2}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) }\\ & +\frac {C_{2}\left [ \frac {1}{2x}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +\frac {1}{2}\left ( Y_{\frac {1}{2q}-1}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) -Y_{\frac {1}{2q}+1}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) \right ) \right ] }{aC_{1}J_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) +aC_{2}Y_{\frac {1}{2q}}\left ( \frac {\sqrt {-ab}}{q}x^{q}\right ) } \end {align*}
Similar result can be found for \(ab>0\)