1.3.1.2.3 Example \(y^{\prime }=x^{3}+y^{2}\)

\[ y^{\prime }=x^{3}+y^{2}\]

Comparing this to \(y^{\prime }=ax^{n}+by^{2}\) shows that \(a=1,b=1,n=3\). We see that \(n\) do not satisfy that \(\frac {n}{2n+4}=\frac {3}{6+4}=\frac {3}{10}\) being an integer. Hence we expect that applying (2) will give solution in cylindrical functions and not elementary functions. Since \(ab>0\) then applying

\begin{align*} w & =\sqrt {x}c_{1}\operatorname {BesselJ}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) \\ k & =1+\frac {3}{2}=\frac {5}{2}\end{align*}

Hence

\[ w=\sqrt {x}c_{1}\operatorname {BesselJ}\left ( \frac {1}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) \]

Hence

\[ y=-\frac {w^{\prime }}{w}\]

Simplifying the above gives

\[ y=\frac {x^{\frac {3}{2}}\left ( -c_{1}\operatorname {BesselJ}\left ( \frac {-4}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) -\operatorname {BesselY}\left ( \frac {-4}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) \right ) }{c_{1}\operatorname {BesselJ}\left ( \frac {1}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) +\operatorname {BesselY}\left ( \frac {1}{5},\frac {2}{5}x^{\frac {5}{2}}\right ) }\]

We see that the solution is in terms of cylindrical functions. Because \(n\) did not satisfy that \(\frac {n}{2n+4}\) is integer. But the main point is that (2) can still be used to solve the special Riccati ode.