2.20.3.3.3 Case when \(\frac {\left ( n+2a\right ) }{2n}=k\) with k positive integer. (section 3-3 a(iii))

2.20.3.3.3.1 Example \(xy^{\prime }=2x^{4}+6y-5y^{2}\) Comparing to

\[ xy^{\prime }=cx^{n}+ay-by^{2}\]

Shows that \(n=4,c=2,a=6,b=5\). Hence \(\frac {\left ( n+2a\right ) }{2n}=\frac {\left ( 4+12\right ) }{8}=\frac {16}{8}=2\) a positive integer. Therefore \(k=2\). This is similar to case (ii) above, where here also the solution is finite continued fraction, but now \(-a\) is used in place of \(a\) as in case (ii) and first term \(y\) is different that in case (ii). But everything else is the same. Hence the solution now becomes

\begin{align*} y & =\frac {x^{n}}{y_{1}}\\ y_{1} & =\frac {n-a}{c}+\frac {x^{n}}{y_{2}}\\ y_{2} & =\frac {2n-a}{b}+\frac {x^{n}}{y_{3}}\\ y_{3} & =\frac {3n-a}{c}+\frac {x^{n}}{y_{4}}\\ y_{4} & =\frac {4n-a}{b}+\frac {x^{n}}{y_{5}}\\ & \vdots \\ y_{k-1} & =\frac {\left ( k-1\right ) n-a}{\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 \(k=2\). Therefore we have (we stop at \(k-1=1\))

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

Now, we just need to find \(y_{2}=y_{k}\) to finish. To find \(y_{2}\) we use either

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

Or

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

ODE (A) is used when \(k\) is odd and ode (B) is used when \(k\) is even. Since \(k=2\) is even, then we use (B). Hence the ode to solve for \(y_{2}\) is

\[ xy_{2}^{\prime }=cx^{n}+\left ( nk-a\right ) y_{2}-by_{2}^{2}\]

But in this problem, \(n=4,c=2,a=6,b=5\), hence the above becomes

\begin{align} xy_{2}^{\prime } & =2x^{4}+\left ( 8-6\right ) y_{2}-5y_{2}^{2}\nonumber \\ & =2x^{4}+2y_{2}-5y_{2}^{2} \tag {B1}\end{align}

Notice that this has form

\[ xy_{2}^{\prime }=Cx^{N}+Ay_{2}-By_{2}^{2}\]

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

Now we see that \(N=4,A=2,C=2,B=5\) which means \(N=2A\). Then we use case (i) to solve (B1). Let \(y_{2}=x^{A}u\). Then ode (B1)

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

We see that (3) is separable. Solving (3) gives

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

But \(y_{2}=x^{A}u\), hence \(u=y_{2}x^{-A}\). Therefore

\begin{align*} y_{2}x^{-A} & =\frac {1}{B}\tanh \left ( \frac {\sqrt {CB}\left ( C_{1}A+x^{A}\right ) }{A}\right ) \sqrt {CB}\\ y_{2} & =\sqrt {CB}\frac {x^{A}}{B}\tanh \left ( \frac {\sqrt {CB}\left ( C_{1}A+x^{A}\right ) }{A}\right ) \end{align*}

Plugging in values for \(N=4,A=2,C=2,B=5\) gives

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

Now we go back to (1) and find \(y\)

\begin{align} y & =\frac {x^{n}}{y_{1}}\nonumber \\ y_{1} & =\frac {n-a}{c}+\frac {x^{n}}{y_{2}}\nonumber \end{align}

Substituting all parameters into the above and using \(y_{2}\) from (4) gives, using \(n=4,c=2,a=6,b=5\),

\begin{align*} y & =\frac {x^{4}}{\frac {4-6}{2}+\frac {x^{4}}{\sqrt {10}\frac {x^{2}}{5}\tanh \left ( \frac {\sqrt {10}x^{2}}{2}+C_{2}\right ) }}\\ & =\frac {x^{4}}{-1+\frac {5x^{2}}{\sqrt {10}\tanh \left ( \frac {\sqrt {10}x^{2}}{2}+C_{2}\right ) }}\end{align*}