4.4.2.3 Example 3

Solve

\[ yy^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}=0 \]

This ode can also be solved using the method of missing \(x\). Comparing the above to (1)

\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}

shows that

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

Then, we first verify the ode is exact using the conditions

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

This gives

\begin{align} 1 & =1\nonumber \\ 0 & =0\tag {2}\\ 0 & =0\nonumber \end{align}

Hence it is exact.. Since no initial conditions are given, then we will use (4). This gives

\begin{align*} \int _{0}^{x}a_{0}\left ( \alpha ,y,y^{\prime }\right ) d\alpha +\int _{0}^{y}a_{1}\left ( 0,\beta ,y^{\prime }\right ) d\beta +\int _{0}^{y^{\prime }}a_{2}\left ( 0,0,\gamma \right ) d\gamma & =c_{1}\\ 0+y^{\prime }\int _{0}^{y}d\beta +y\int _{0}^{y^{\prime }}d\gamma & =c_{1}\\ y^{\prime }y+yy^{\prime } & =c_{1}\\ 2y^{\prime }y & =c_{1}\end{align*}

Solving gives

\begin{align*} \int 2ydy & =\int c_{1}dx\\ y^{2} & =c_{1}x+c_{2}\end{align*}

Or

\begin{align*} y_{1} & =\sqrt {c_{1}x+c_{2}}\\ y_{2} & =-\sqrt {c_{1}x+c_{2}}\end{align*}

And this is the solution to the original ode.