4.9.1.1.7 Example 7 \(yy^{\prime \prime }-\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{3}=0,y\left ( 0\right ) =-1,y^{\prime }\left ( 0\right ) =0\)

\begin{equation} yy^{\prime \prime }-\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{3}=0 \tag {1}\end{equation}

With IC

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

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

\begin{align} y\frac {dp}{dy}p-p^{2}+p^{3} & =0\tag {2}\\ p^{\prime } & =\frac {p^{2}-p^{3}}{yp}\nonumber \\ p^{\prime } & =\frac {p-p^{2}}{y}\nonumber \end{align}

This is separable. Solving

\[ \int \frac {dp}{p^{2}-p}=-\int \frac {1}{y}dy\qquad p-p^{2}\neq 0 \]

This gives

\[ \frac {p-1}{p}=\frac {c_{1}}{y}\]

Applying IC \(p=0\) at \(y=-1\) show there is no solution as we obtain \(-1=0\). Hence no general solution exists. Let look for singular solution. This happens when \(p-p^{2}=0\) or \(p=0\) and \(p=1\). Looking at \(p=0\) means \(y^{\prime }=0\) or \(y=c\). At IC this gives \(c=-1\). Hence \(y=-1\). This also satisfies \(y^{\prime }\left ( 0\right ) =0\). So \(y=-1\) is valid singular solution. Let look at \(p=1\) which means \(y^{\prime }=1\) or \(y=x+c_{1}\). At first IC this gives \(c_{1}=-1\). Hence solution now becomes \(y=x-1\). But this does not satisfy \(y^{\prime }\left ( 0\right ) =0\). Therefore only

\[ y=-1 \]

Is solution (singular).