4.9.2.1.2 Example 2 \(y^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}=1\)

\begin{align} y^{\prime \prime }+\left ( y^{\prime }\right ) ^{2} & =1\tag {1}\\ y\left ( 0\right ) & =0\nonumber \\ y^{\prime }\left ( 0\right ) & =1\nonumber \end{align}

Let \(p\left ( x\right ) =y^{\prime }\) then \(y^{\prime \prime }=p^{\prime }\) and the ode becomes

\begin{align*} p^{\prime }+p^{2} & =1\\ p^{\prime } & =1-p^{2}\\ \frac {dp}{dx} & =1-p^{2}\\ \int \frac {dp}{1-p^{2}} & =\int dx\\ \operatorname {arctanh}\left ( p\right ) & =x+c_{1}\\ p & =\tanh \left ( x+c_{1}\right ) \end{align*}

At \(x=0,p=1\) hence

\begin{equation} 1=\tanh \left ( c_{1}\right ) \nonumber \end{equation}

There is no solution. Hence no general solution exist. Now we look for singular solution. This happens when \(1-p^{2}=0\) or \(p^{2}=1\) or \(p=\pm 1\). For \(p=1\) this means \(y^{\prime }=1\) or \(y=x+c\) which at IC gives \(c=0\). Hence singular solution is

\[ y=x \]

This satisfies both IC’s. If we try \(p=-1\) it gives \(y=-x\) but this does not satisfy IC. So only solution is \(y=x\).