#### 2.24   ODE No. 24

$a y(x)^2-b x^{\nu }+y'(x)=0$ Mathematica : cpu = 0.0970147 (sec), leaf count = 277

$\left \{\left \{y(x)\to -\frac {\sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}} \left (c_1 J_{\frac {\nu +1}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu }{2}+1}}{\nu +2}\right )-c_1 J_{-\frac {\nu +3}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )-2 J_{\frac {1}{\nu +2}-1}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )\right )-c_1 J_{-\frac {1}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )}{2 a x \left (c_1 J_{-\frac {1}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )+J_{\frac {1}{\nu +2}}\left (\frac {2 \sqrt {-a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )\right )}\right \}\right \}$ Maple : cpu = 0.05 (sec), leaf count = 214

$\left \{ y \left ( x \right ) ={\frac {1}{ax} \left ( -{{\sl J}_{{\frac {3+\nu }{\nu +2}}}\left (2\,{\frac {\sqrt {-ab}{x}^{\nu /2+1}}{\nu +2}}\right )}\sqrt {-ab}{x}^{{\frac {\nu }{2}}+1}{\it \_C1}-{{\sl Y}_{{\frac {3+\nu }{\nu +2}}}\left (2\,{\frac {\sqrt {-ab}{x}^{\nu /2+1}}{\nu +2}}\right )}\sqrt {-ab}{x}^{{\frac {\nu }{2}}+1}+{\it \_C1}\,{{\sl J}_{ \left ( \nu +2 \right ) ^{-1}}\left (2\,{\frac {\sqrt {-ab}{x}^{\nu /2+1}}{\nu +2}}\right )}+{{\sl Y}_{ \left ( \nu +2 \right ) ^{-1}}\left (2\,{\frac {\sqrt {-ab}{x}^{\nu /2+1}}{\nu +2}}\right )} \right ) \left ( {\it \_C1}\,{{\sl J}_{ \left ( \nu +2 \right ) ^{-1}}\left (2\,{\frac {\sqrt {-ab}{x}^{\nu /2+1}}{\nu +2}}\right )}+{{\sl Y}_{ \left ( \nu +2 \right ) ^{-1}}\left (2\,{\frac {\sqrt {-ab}{x}^{\nu /2+1}}{\nu +2}}\right )} \right ) ^{-1}} \right \}$

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 $$ab>0$$