1.3.3.4.1 Example \(xy^{\prime }=2x^{5}+6y-5y^{2}\) Comparing to

\begin{equation} xy^{\prime }=cx^{n}+ay-by^{2} \tag {1}\end{equation}

Shows that \(c=2,n=5,a=6,b=5\). Checking case (i), where condition is \(n=2a\). We see this does not apply. Check case (ii) where condition is \(\frac {\left ( n-2a\right ) }{2n}=k\) with \(k\) positive integer. We see that \(\frac {\left ( n-2a\right ) }{2n}=\frac {\left ( 5-12\right ) }{10}=\frac {-7}{10}\) which is not positive integer. Finally checking case (iii) where condition is \(\frac {\left ( n+2a\right ) }{2n}\) being positive integer. But \(\frac {\left ( n+2a\right ) }{2n}=\frac {\left ( 5+12\right ) }{10}=\frac {17}{10}\ \)which is not positive integer. Since all three cases failed, we convert the ode to reduced Riccati using

\begin{align*} y & =zu\left ( z\right ) \\ z & =x^{a}\end{align*}

Or \(x=z^{\frac {1}{a}}\). Hence

\begin{align*} y^{\prime }\left ( x\right ) & =z^{\prime }u+z\frac {du}{dz}z^{\prime }\\ & =ax^{a-1}u+z\frac {du}{dz}ax^{a-1}\\ & =a\left ( z^{\frac {1}{a}}\right ) ^{a-1}u+z\frac {du}{dz}a\left ( z^{\frac {1}{a}}\right ) ^{a-1}\\ & =az^{\frac {a-1}{a}}u+z\frac {du}{dz}az^{\frac {a-1}{a}}\\ & =az^{\frac {a-1}{a}}u+az^{1+\frac {a-1}{a}}\frac {du}{dz}\\ & =az^{\frac {a-1}{a}}u+az^{\frac {a+a-1}{a}}\frac {du}{dz}\\ & =az^{\frac {a-1}{a}}u+az^{\frac {2a-1}{a}}\frac {du}{dz}\end{align*}

The original ode \(xy^{\prime }=cx^{n}+ay-by^{2}\) now becomes

\begin{align} z^{\frac {1}{a}}\left ( az^{\frac {a-1}{a}}u+az^{\frac {2a-1}{a}}\frac {du}{dz}\right ) & =c\left ( z^{\frac {1}{a}}\right ) ^{n}+azu-bz^{2}u^{2}\nonumber \\ azu+az^{\frac {2a-1}{a}+\frac {1}{a}}\frac {du}{dz} & =cz^{\frac {n}{a}}+azu-bz^{2}u^{2}\nonumber \\ az^{\frac {2a-1+1}{a}}\frac {du}{dz} & =cz^{\frac {n}{a}}-bz^{2}u^{2}\nonumber \\ az^{2}\frac {du}{dz} & =cz^{\frac {n}{a}}-bz^{2}u^{2}\nonumber \\ a\frac {du}{dz} & =cz^{\frac {n}{a}-2}-bu^{2}\nonumber \\ \frac {du}{dz} & =\frac {c}{a}z^{\frac {n}{a}-2}-\frac {b}{a}u^{2} \tag {2}\end{align}

Which is reduced Riccati form

\[ y^{\prime }=Ax^{N}+By^{2}\]

Where here \(A=\frac {c}{a},N=\frac {n}{a}-2,B=-\frac {b}{a}\) which can be solved as shown in the section of reduced Riccati ode above. Using \(c=2,n=5,a=6,b=5\) ode (2) becomes

\begin{align*} \frac {du}{dz} & =\frac {2}{6}z^{\frac {5}{6}-2}-\frac {5}{6}u^{2}\\ & =\frac {1}{3}z^{-\frac {7}{6}}-\frac {5}{6}u^{2}\end{align*}

Hence \(A=\frac {1}{3},B=-\frac {5}{6}\) and \(N=-\frac {7}{6}\). Since \(N\) is not the special case \(-2\). we use Reduced Riccati for the case where \(N\neq -2\), which gives

\begin{align} w & =\sqrt {z}\left \{ \begin {array} [c]{cc}c_{1}\operatorname {BesselJ}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {AB}z^{k}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {AB}z^{k}\right ) & AB>0\\ c_{1}\operatorname {BesselI}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-AB}z^{k}\right ) +c_{2}\operatorname {BesselK}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-AB}z^{k}\right ) & AB<0 \end {array} \right . \tag {3}\\ u & =-\frac {1}{B}\frac {w^{\prime }}{w}\nonumber \\ k & =1+\frac {N}{2}\nonumber \end{align}

Since \(AB=-\frac {5}{18}<0\) then the solution is

\[ w=\sqrt {z}c_{1}\operatorname {BesselI}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-AB}z^{k}\right ) +c_{2}\operatorname {BesselK}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-AB}z^{k}\right ) \]

But \(k=1+\frac {-\frac {7}{6}}{2}=\frac {5}{12}\) and the above becomes

\begin{align*} w & =\sqrt {z}c_{1}\operatorname {BesselI}\left ( \frac {1}{2\frac {5}{12}},\frac {1}{\frac {5}{12}}\sqrt {\frac {5}{18}}z^{\frac {5}{12}}\right ) +c_{2}\operatorname {BesselK}\left ( \frac {1}{2\frac {5}{12}},\frac {1}{\frac {5}{12}}\sqrt {\frac {5}{18}}z^{\frac {5}{12}}\right ) \\ & =\sqrt {z}c_{1}\operatorname {BesselI}\left ( \frac {5}{6},\frac {12}{5}\sqrt {\frac {5}{18}}z^{\frac {5}{12}}\right ) +c_{2}\operatorname {BesselK}\left ( \frac {5}{6},\frac {12}{5}\sqrt {\frac {5}{18}}z^{\frac {5}{12}}\right ) \end{align*}

Hence

\begin{align*} u\left ( z\right ) & =-\frac {1}{B}\frac {w^{\prime }\left ( z\right ) }{w\left ( z\right ) }\\ & =-\frac {1}{-\frac {5}{6}}\frac {w^{\prime }\left ( z\right ) }{w\left ( z\right ) }\\ & =\frac {6}{5}\frac {w^{\prime }\left ( z\right ) }{w\left ( z\right ) }\end{align*}

Carrying out the simplification above gives

\[ u\left ( z\right ) =\frac {1}{z^{\frac {1}{6}}}\frac {2\sqrt {10}\left ( \operatorname {BesselI}\left ( \frac {1}{5},\frac {2\sqrt {10}}{5}z^{\frac {5}{12}}\right ) c_{1}-\operatorname {BesselK}\left ( \frac {1}{5},\frac {2\sqrt {10}}{5}z^{\frac {5}{12}}\right ) \right ) }{-\operatorname {BesselI}\left ( \frac {1}{5},\frac {2\sqrt {10}}{5}z^{\frac {5}{12}}\right ) \sqrt {10}c_{1}+\sqrt {10}\operatorname {BesselK}\left ( \frac {1}{5},\frac {2\sqrt {10}}{5}z^{\frac {5}{12}}\right ) +10z^{\frac {5}{12}}\left ( \operatorname {BesselK}\left ( \frac {4}{5},\frac {2\sqrt {10}}{5}z^{\frac {5}{12}}\right ) +\operatorname {BesselI}\left ( -\frac {4}{5},\frac {2\sqrt {10}}{5}z^{\frac {5}{12}}\right ) c_{1}\right ) }\]

Now that we found \(u\), then \(y\) for the original ode is found, since

\[ y=zu\left ( z\right ) \]

This completes the solution. We need to also change all \(z\) in the solution to \(x\) using \(z=x^{a}\) after applying the above. This results in

\[ y\left ( x\right ) =-x^{5}\frac {2\sqrt {10}\left ( \operatorname {BesselI}\left ( \frac {1}{5},\frac {2\sqrt {10}}{5}x^{\frac {5}{2}}\right ) c_{1}-\operatorname {BesselK}\left ( \frac {1}{5},\frac {2\sqrt {10}}{5}x^{\frac {5}{2}}\right ) \right ) }{\operatorname {BesselI}\left ( \frac {1}{5},\frac {2\sqrt {10}}{5}x^{\frac {5}{2}}\right ) \sqrt {10}c_{1}+\sqrt {10}\operatorname {BesselK}\left ( \frac {1}{5},\frac {2\sqrt {10}}{5}x^{\frac {5}{2}}\right ) +10x^{\frac {5}{2}}\left ( \operatorname {BesselK}\left ( \frac {4}{5},\frac {2\sqrt {10}}{5}x^{\frac {5}{2}}\right ) +\operatorname {BesselI}\left ( -\frac {4}{5},\frac {2\sqrt {10}}{5}x^{\frac {5}{2}}\right ) c_{1}\right ) }\]