4.9.1.1.12 Example 12 \(y^{\prime \prime }=\left ( y^{\prime }\right ) ^{3}-\left ( y^{\prime }\right ) ^{2},y\left ( 0\right ) =3,y^{\prime }\left ( 0\right ) =1\)

\begin{equation} y^{\prime \prime }=\left ( y^{\prime }\right ) ^{3}-\left ( y^{\prime }\right ) ^{2} \tag {1}\end{equation}

With IC

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

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=p\frac {dp}{dy}\). Hence the ode becomes

\begin{align} p\frac {dp}{dy} & =p^{3}-p^{2}\tag {2}\\ \frac {dp}{dy} & =p^{2}-p\hspace {0.5in}p\neq 0\nonumber \end{align}

Which is first order ode in \(p\left ( y\right ) \) of type quadrature. The solution is

\begin{align*} \int \frac {dp}{p^{2}-p} & =\int dy\\ \ln \left ( p-1\right ) -\ln p & =y+c_{1}\\ \ln \left ( \frac {p-1}{p}\right ) & =y+c_{1}\\ \frac {p-1}{p} & =e^{y+c_{1}}\\ p-1 & =pe^{y+c_{1}}\\ p\left ( 1-e^{y+c_{1}}\right ) & =1\\ p & =\frac {1}{\left ( 1-e^{y+c_{1}}\right ) }\end{align*}

But \(p=y^{\prime }\) hence the above becomes

\begin{equation} y^{\prime }=\frac {1}{\left ( 1-e^{y+c_{1}}\right ) } \tag {3A}\end{equation}

Using initial conditions at \(x=0\) the above becomes

\begin{align*} 1 & =\frac {1}{\left ( 1-e^{3+c_{1}}\right ) }\\ \left ( 1-e^{3+c_{1}}\right ) & =1\\ 1-e^{3+c_{1}} & =1\\ e^{3+c_{1}} & =0\\ c_{1} & =-\infty \end{align*}

Hence (3A) becomes

\[ y^{\prime }=1 \]

Or

\[ y=x+c_{2}\]

Using the initial conditions \(y\left ( 0\right ) =3\) the above becomes

\[ 3=c_{2}\]

Hence the solution is

\[ y=3+x \]