Example

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

Hence \(f\left ( x,y,y^{\prime }\right ) =2yy^{\prime }\). At \(x=0\) then \(f=4\) which is continuous. And \(f_{y}=2y^{\prime }\) which at \(x_{0}\) becomes \(4\). This is also continuous. And \(f_{y^{\prime }}=2y\) which at \(x_{0}\) becomes \(4\) which is also continuous. Hence solution exists and is unique on interval that contains \(x=0\). The solution can be found as follows

Let \(y^{\prime }=p\left ( y\right ) \) then \(y^{\prime \prime }=\frac {dp}{dx}=\frac {dp}{dy}\frac {dy}{dx}=\frac {dp}{dy}p\). The ode becomes

\begin{align*} \frac {dp}{dy}p & =2yp\\ \frac {dp}{dy} & =2y \end{align*}

But at \(x=0\) we have \(y\left ( 0\right ) =1\) and \(y^{\prime }\left ( 0\right ) =p\left ( y\left ( 0\right ) \right ) =p\left ( 1\right ) =2\). This is the initial condition used for solving the above quadrature ode. Integrating the above gives

\[ p=y^{2}+c_{1}\]

Applying IC \(p\left ( 1\right ) =2\) gives 

\begin{align*} 2 & =1+c_{1}\\ c_{1} & =1 \end{align*}

Hence \(p=y^{2}+1\). But \(y^{\prime }=p\) or \(y^{\prime }=y^{2}+1\). This is separable with initial conditions \(y\left ( 0\right ) =1\). Integrating gives

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

Applying IC

\[ \arctan \left ( 1\right ) =c_{2}\]

So \(c_{2}=\frac {\pi }{4}\). Hence the solution becomes

\begin{align*} \arctan \left ( y\right ) & =x+\frac {\pi }{4}\\ y\left ( x\right ) & =\tan \left ( x+\frac {\pi }{4}\right ) \end{align*}