4.9.2.1.9 Example 8  \(y^{\prime \prime }=\left ( y^{\prime }\right ) ^{3}-\left ( y^{\prime }\right ) ^{2}\)

\begin{align*} y^{\prime \prime } & =\left ( y^{\prime }\right ) ^{3}-\left ( y^{\prime }\right ) ^{2}\\ y\left ( 0\right ) & =3\\ y^{\prime }\left ( 0\right ) & =1 \end{align*}

This was solved also as missing \(x\) in the above section. let \(p=y^{\prime },y^{\prime \prime }=\frac {dp}{dx}\), the ode becomes

\begin{align*} \frac {dp}{dx} & =p^{3}-p^{2}\\ \int \frac {dp}{p^{3}-p^{2}} & =\int dx\\ \frac {1}{p}+\ln \left ( \frac {1-p}{p}\right ) & =x+c_{1}\\ p & =\frac {1}{\operatorname {LambertW}\left ( -c_{1}e^{x-1}\right ) +1}\end{align*}

When \(x=0,p\left ( 0\right ) =y^{\prime }\left ( 0\right ) =1\) and the above gives

\[ 1=\frac {1}{\operatorname {LambertW}\left ( -c_{1}e^{-1}\right ) +1}\]

Solving for \(c_{1}\) gives \(c_{1}=0\). Therefore the solution is

\begin{align*} p & =\frac {1}{\operatorname {LambertW}\left ( 0\right ) +1}\\ & =\frac {1}{0+1}\\ & =1 \end{align*}

But \(p=y^{\prime }\left ( x\right ) \). Hence the above gives \(y^{\prime }=1\), or \(y=x+c_{1}\). Using the other IC \(y\left ( 0\right ) =3\) gives \(c_{1}=3\). Hence solution is \(y=x+3\).  Which is same as was done in one of the above examples in the section of missing \(x\).