2.20.3.5.2 Example \(y^{\prime }=x^{5}+\left ( \frac {1}{x}-2x^{4}\right ) y+x^{3}y^{2}\)

Comparing to \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\) shows that

\begin{align*} f_{0} & =x^{5}\\ f_{1} & =\left ( \frac {1}{x}-2x^{4}\right ) \\ f_{2} & =x^{3}\end{align*}

We see that \(y_{1}=x\) is a solution. Hence let \(y=y_{1}+\frac {1}{u}\). The ode becomes

\[ u^{\prime }-\frac {u}{x}=-x^{3}\]

Which is linear. Solving using integrating factor gives

\[ u=\frac {1}{x}\left ( c_{1}-\frac {x^{5}}{5}\right ) \]

Hence

\begin{align} y & =y_{1}+\frac {1}{u}\nonumber \\ & =x+\frac {1}{\frac {1}{x}\left ( c_{1}-\frac {x^{5}}{5}\right ) }\nonumber \\ & =x+\frac {5x}{\left ( 5c_{1}-x^{5}\right ) }\nonumber \\ & =x+\frac {5x}{c_{2}-x^{5}}\nonumber \\ & =\frac {x\left ( x^{5}-c_{2}-5\right ) }{x^{5}-c_{2}} \tag {A}\end{align}

Using the direct formula (1) given earlier, we can write

\begin{align} y & =y_{1}+\Phi \frac {1}{c_{1}-\int \Phi f_{2}dx}\nonumber \\ & =x+\Phi \frac {1}{c_{1}-\int \Phi x^{3}dx} \tag {B}\end{align}

Where

\begin{align*} \Phi & =e^{\int 2f_{2}y_{1}+f_{1}dx}\\ & =e^{\int 2x^{4}+\left ( \frac {1}{x}-2x^{4}\right ) dx}\\ & =e^{\ln x}\\ & =x \end{align*}

Hence (B) becomes

\begin{align*} y & =x+\frac {x}{c_{1}-\int x^{4}dx}\\ & =x+\frac {x}{c_{1}-\frac {x^{5}}{5}}\\ & =x+\frac {5x}{5c_{1}-x^{5}}\\ & =\frac {x\left ( 5c_{1}-x^{5}\right ) +5x}{5c_{1}-x^{5}}\\ & =\frac {x\left ( 5c_{1}-x^{5}+5\right ) }{5c_{1}-x^{5}}\\ & =\frac {x\left ( c_{2}-x^{5}+5\right ) }{c_{2}-x^{5}}\\ & =\frac {x\left ( x^{5}-c_{2}-5\right ) }{x^{5}-c_{2}}\end{align*}

Which is the same solution in (A). The direct formula might be easier to use.