Example xy′ = 2x4 − 18y + 5y2

Shows that \(c=2,n=4,a=-18,b=-5\). Hence \(\frac {\left ( n-2a\right ) }{2n}=\frac {\left ( 4-2\left ( -18\right ) \right ) }{8}=\frac {40}{8}=5\). Therefore \(k=5\) which is positive integer.

\begin{align*} y & =\frac {a}{b}+\frac {x^{n}}{y_{1}}\\ y_{1} & =\frac {a+n}{c}+\frac {x^{n}}{y_{2}}\\ y_{2} & =\frac {a+2n}{b}+\frac {x^{n}}{y_{3}}\\ y_{3} & =\frac {a+3n}{c}+\frac {x^{n}}{y_{4}}\\ y_{4} & =\frac {a+4n}{b}+\frac {x^{n}}{y_{5}}\\ & \vdots \\ y_{k-1} & =\frac {a+\left ( k-1\right ) n}{\Delta }+\frac {x^{n}}{y_{k}}\end{align*}

Where \(\Delta =c\) when index \(i\) on \(y_{i}\) is odd and \(\Delta =b\) when index \(i\) on \(y_{i}\) is even. In this example, we found that \(k=5\). Therefore we have (we stop at \(k-1=4\))

\begin{align} y & =\frac {a}{b}+\frac {x^{n}}{y_{1}}\nonumber \\ y_{1} & =\frac {a+n}{c}+\frac {x^{n}}{y_{2}}\nonumber \\ y_{2} & =\frac {a+2n}{b}+\frac {x^{n}}{y_{3}}\nonumber \\ y_{3} & =\frac {a+3n}{c}+\frac {x^{n}}{y_{4}}\nonumber \\ y_{4} & =\frac {a+4n}{b}+\frac {x^{n}}{y_{5}} \tag {1}\end{align}

Now, we just need to ﬁnd \(y_{5}=y_{k}\) to ﬁnish. To ﬁnd \(y_{5}\) we use either

\begin{equation} xy_{5}^{\prime }=bx^{n}+\left ( a+nk\right ) y_{5}-cy_{5}^{2} \tag {A}\end{equation}

\begin{equation} xy_{5}^{\prime }=cx^{n}+\left ( a+nk\right ) y_{5}-by_{5}^{2} \tag {B}\end{equation}

ODE (A) is used when \(k\) is odd and ode (B) is used when \(k\) is even. Since \(k=5\) is odd, then we use (A). Hence the ode to solve for \(y_{5}\) is

\begin{align*} xy_{5}^{\prime } & =-5x^{4}+\left ( -18+20\right ) y_{5}-2y_{5}^{2}\\ & =-5x^{4}+2y_{5}-2y_{5}^{2}\end{align*}

(used upper case letters now, so not to confuse with lower case letters used for the original ode).

Now we see that \(N=4,A=2,C=-5,B=2\) which means \(N=2A\). But this is case (i). This always happens with case (ii) that we end up having to solve one ode using case (i) to ﬁnish. If we do not end up with case(i) ode, then we have made a mistake

But we know how to solve case (i) which we did above. This gives us \(y_{2}\). We plug this into (1) and now we have the solution for \(y\).

Now we solve (B) as we did in part (i). Let \(y_{2}=x^{A}u\). Then ode (B) becomes (as was shown in case (i) example)

\begin{equation} u^{\prime }=x^{A-1}\left ( C-Bu^{2}\right ) \tag {3}\end{equation}

\[ u=\frac {1}{B}\tanh \left ( \frac {\sqrt {CB}\left ( C_{1}A+x^{A}\right ) }{A}\right ) \sqrt {CB}\]

\begin{align*} y_{5}x^{-A} & =\frac {1}{B}\tanh \left ( \frac {\sqrt {CB}\left ( C_{1}A+x^{A}\right ) }{A}\right ) \sqrt {CB}\\ y_{5} & =\sqrt {CB}\frac {x^{A}}{B}\tanh \left ( \frac {\sqrt {CB}\left ( C_{1}A+x^{A}\right ) }{A}\right ) \end{align*}

\begin{align} y_{5} & =\sqrt {-10}\frac {x^{2}}{2}\tanh \left ( \frac {\sqrt {-10}\left ( 2C_{1}+x^{2}\right ) }{2}\right ) \nonumber \\ & =\frac {\sqrt {-10}}{2}x^{2}\tanh \left ( \frac {\sqrt {-10}x^{2}}{2}+C_{2}\right ) \nonumber \\ & =\frac {i\sqrt {10}}{2}x^{2}\tanh \left ( \frac {i\sqrt {10}x^{2}}{2}+C_{2}\right ) \tag {4}\end{align}

But \(\tanh \left ( iz\right ) =i\tan \left ( z\right ) \), hence the above becomes

\begin{align*} y_{5} & =\frac {i^{2}\sqrt {10}}{2}x^{2}\tan \left ( \frac {\sqrt {10}x^{2}}{2}+C_{2}\right ) \\ & =\frac {-\sqrt {10}}{2}x^{2}\tan \left ( \frac {\sqrt {10}x^{2}}{2}+C_{2}\right ) \end{align*}

\[ y=\frac {a}{b}+\frac {x^{n}}{\frac {a+n}{c}+\frac {x^{n}}{\frac {a+2n}{b}+\frac {x^{n}}{\frac {a+3n}{c}+\frac {x^{n}}{\frac {a+4n}{b}+\frac {x^{n}}{y_{5}}}}}}\]

Substituting all parameters into the above and using \(y_{2}\) from (4) gives, using \(c=2,n=4,a=-18,b=-5\)

\[ y=\frac {-18}{-5}+\frac {x^{4}}{\frac {-18+4}{2}+\frac {x^{4}}{\frac {-18+8}{-5}+\frac {x^{4}}{\frac {-18+12}{2}+\frac {x^{4}}{\frac {-18+16}{-5}+\frac {x^{4}}{\frac {-\sqrt {10}}{2}x^{2}\tan \left ( \frac {\sqrt {10}x^{2}}{2}+C_{2}\right ) }}}}}\]

\[ y=\frac {18}{5}+\frac {x^{4}}{-7+\frac {x^{4}}{2+\frac {x^{4}}{-3+\frac {x^{4}}{\frac {2}{5}-\frac {2x^{2}}{\sqrt {10}\tan \left ( \frac {\sqrt {10}x^{2}}{2}+C_{2}\right ) }}}}}\]