4.4.2.7 Example 7

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

Comparing to

\begin{equation} a_{2}\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+a_{1}\left ( x,y,y^{\prime }\right ) y^{\prime }+a_{0}\left ( x,y,y^{\prime }\right ) =0 \tag {1}\end{equation}

Then possible matches are

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

Or

\begin{align*} a_{2} & =y\\ a_{1} & =y^{\prime }\\ a_{0} & =1 \end{align*}

Or

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

We just need one match that satisfies the exactness conditions

\begin{align*} \frac {\partial a_{2}}{\partial y} & =\frac {\partial a_{1}}{\partial y^{\prime }}\\ \frac {\partial a_{2}}{\partial x} & =\frac {\partial a_{0}}{\partial y^{\prime }}\\ \frac {\partial a_{1}}{\partial x} & =\frac {\partial a_{0}}{\partial y}\end{align*}

Looking at the first match, then the conditions become

\[ 1=2y^{\prime }\]

Hence it fails. Looking at the second match

\begin{align*} 1 & =1\\ 0 & =0\\ 0 & =0 \end{align*}

This works, Therefore we will use \(a_{2}=y,a_{1}=y^{\prime },a_{0}=1\).  Since we are given initial conditions (even if partial), we will use Eq (3) which is

\[ \int _{x_{0}}^{x}a_{0}\left ( \alpha ,y,y^{\prime }\right ) d\alpha +\int _{y_{0}}^{y}a_{1}\left ( x_{0},\beta ,y^{\prime }\right ) d\beta +\int _{y_{0}^{\prime }}^{y^{\prime }}a_{2}\left ( x_{0},y_{0},\gamma \right ) d\gamma =0 \]

We are given \(x_{0},y_{0}^{\prime }\) but not \(y_{0}\). Hence in the above we will replace \(y_{0}\) by \(y\left ( 0\right ) \) and use the actual values for \(x_{0},y_{0}^{\prime }\) given. The above becomes, now using \(x_{0}=0,y_{0}^{\prime }=1,y_{0}=y\left ( 0\right ) \)

\begin{align*} \int _{x_{0}}^{x}\left ( 1\right ) d\alpha +\int _{y\left ( 0\right ) }^{y}y^{\prime }d\beta +\int _{y_{0}^{\prime }}^{y^{\prime }}y_{0}d\gamma & =0\\ \int _{0}^{x}\left ( 1\right ) d\alpha +\int _{y\left ( 0\right ) }^{y}y^{\prime }d\beta +\int _{1}^{y^{\prime }}y\left ( 0\right ) d\gamma & =0\\ x+y^{\prime }\left ( y-y\left ( 0\right ) \right ) +y\left ( 0\right ) \left ( y^{\prime }-1\right ) & =0\\ x+yy^{\prime }-y\left ( 0\right ) y^{\prime }+y\left ( 0\right ) y^{\prime }-y\left ( 0\right ) & =0\\ x+yy^{\prime }-y\left ( 0\right ) & =0 \end{align*}

Solving gives

\[ x^{2}-2xy\left ( 0\right ) +y^{2}-c_{1}=0 \]