2.24   ODE No. 24

\[ a y(x)^2-b x^{\nu }+y'(x)=0 \]

✓ Mathematica : cpu = 0.123047 (sec), leaf count = 733

DSolve[-(b*x^nu) + a*y[x]^2 + Derivative[1][y][x] == 0,y[x],x]
 

\[\left \{\left \{y(x)\to \frac {\frac {1}{2} (-1)^{\frac {1}{\nu +2}} (\nu +2)^{-\frac {1}{\nu +2}} a^{\frac {1}{2 (\nu +2)}+\frac {1}{2}} b^{\frac {1}{2 (\nu +2)}+\frac {1}{2}} x^{\frac {\nu +2}{2}-\frac {\frac {\nu }{2}+1}{\nu +2}} \Gamma \left (1+\frac {1}{\nu +2}\right ) \left (I_{\frac {1}{\nu +2}-1}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )+I_{1+\frac {1}{\nu +2}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )\right )+(-1)^{\frac {1}{\nu +2}} \left (1-\frac {\frac {\nu }{2}+1}{\nu +2}\right ) (\nu +2)^{-\frac {1}{\nu +2}} a^{\frac {1}{2 (\nu +2)}} b^{\frac {1}{2 (\nu +2)}} x^{-\frac {\frac {\nu }{2}+1}{\nu +2}} \Gamma \left (1+\frac {1}{\nu +2}\right ) I_{\frac {1}{\nu +2}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )+c_1 \left (\frac {1}{2} (\nu +2)^{-\frac {1}{\nu +2}} a^{\frac {1}{2 (\nu +2)}+\frac {1}{2}} b^{\frac {1}{2 (\nu +2)}+\frac {1}{2}} x^{\frac {\frac {\nu }{2}+1}{\nu +2}+\frac {\nu +2}{2}-1} \Gamma \left (1-\frac {1}{\nu +2}\right ) \left (I_{-1-\frac {1}{\nu +2}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )+I_{1-\frac {1}{\nu +2}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )\right )+\left (\frac {\nu }{2}+1\right ) (\nu +2)^{-\frac {1}{\nu +2}-1} a^{\frac {1}{2 (\nu +2)}} b^{\frac {1}{2 (\nu +2)}} x^{\frac {\frac {\nu }{2}+1}{\nu +2}-1} \Gamma \left (1-\frac {1}{\nu +2}\right ) I_{-\frac {1}{\nu +2}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )\right )}{a \left ((-1)^{\frac {1}{\nu +2}} (\nu +2)^{-\frac {1}{\nu +2}} a^{\frac {1}{2 (\nu +2)}} b^{\frac {1}{2 (\nu +2)}} x^{1-\frac {\frac {\nu }{2}+1}{\nu +2}} \Gamma \left (1+\frac {1}{\nu +2}\right ) I_{\frac {1}{\nu +2}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )+c_1 (\nu +2)^{-\frac {1}{\nu +2}} a^{\frac {1}{2 (\nu +2)}} b^{\frac {1}{2 (\nu +2)}} x^{\frac {\frac {\nu }{2}+1}{\nu +2}} \Gamma \left (1-\frac {1}{\nu +2}\right ) I_{-\frac {1}{\nu +2}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\frac {\nu +2}{2}}}{\nu +2}\right )\right )}\right \}\right \}\]

✓ Maple : cpu = 0.049 (sec), leaf count = 214

dsolve(diff(y(x),x)+a*y(x)^2-b*x^nu = 0,y(x))
 

\[y \left (x \right ) = \frac {-\sqrt {-a b}\, x^{\frac {\nu }{2}+1} \operatorname {BesselJ}\left (\frac {3+\nu }{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right ) c_{1} -\operatorname {BesselY}\left (\frac {3+\nu }{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right ) \sqrt {-a b}\, x^{\frac {\nu }{2}+1}+c_{1} \operatorname {BesselJ}\left (\frac {1}{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right )+\operatorname {BesselY}\left (\frac {1}{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right )}{x a \left (c_{1} \operatorname {BesselJ}\left (\frac {1}{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right )+\operatorname {BesselY}\left (\frac {1}{\nu +2}, \frac {2 \sqrt {-a b}\, x^{\frac {\nu }{2}+1}}{\nu +2}\right )\right )}\]

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 \(ab>0\)