4.1.2.3.3 Example 3

\begin{align*} y^{\prime \prime }-2yy^{\prime } & =0\\ F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) & =0 \end{align*}

First we need to determine if it is exact or not.  Applying the test

\begin{align*} \frac {\partial F}{\partial y}-\frac {d}{dx}\left ( \frac {\partial F}{\partial y^{\prime }}\right ) +\frac {d^{2}}{dx^{2}}\left ( \frac {\partial F}{\partial y^{\prime \prime }}\right ) & =0\\ -2y^{\prime }-\frac {d}{dx}\left ( -2y\right ) +\frac {d^{2}}{dx^{2}}\left ( 1\right ) & =0\\ -2y^{\prime }+2\frac {d}{dx}\left ( y\right ) & =0\\ -2y^{\prime }+2y^{\prime } & =0\\ 0 & =0 \end{align*}

So it exact.  Comparing this ode to  \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =f\left ( x,y,y^{\prime }\right ) y^{\prime \prime }+g\left ( x,y,y^{\prime }\right ) \) shows that

\begin{align*} f & =1\\ g & =-2yy^{\prime }\\ & =-2yp \end{align*}

Therefore (4) becomes

\begin{align} R & =\int fdp+\psi \left ( x,y\right ) \nonumber \\ & =p+\psi \left ( x,y\right ) \tag {1A}\end{align}

Hence (5) becomes

\begin{align*} g & =\left ( \frac {\partial }{\partial x}\left ( \int fdp\right ) +\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}\left ( \int fdp\right ) +\psi _{y}\right ) p\\ -2yp & =\left ( \frac {\partial p}{\partial x}+\psi _{x}\right ) +\left ( \frac {\partial p}{\partial y}+\psi _{y}\right ) p\\ -2yp & =\psi _{x}+\psi _{y}p \end{align*}

Comparing terms shows that

\begin{align*} -2y & =\psi _{y}\\ 0 & =\psi _{x}\end{align*}

Integrating the first equation gives \(\psi =-y^{2}+h\left ( x\right ) \). Differentiating this w.r.t. \(x\) gives \(\psi _{x}=h^{\prime }\left ( x\right ) \). comparing this to the second equation above gives \(0=h^{\prime }\left ( x\right ) \), hence \(h\left ( x\right ) =c\). Hence \(\psi =-y^{2}+c\).  Therefore (1A) becomes

\[ R=p-y^{2}+c \]

Therefore, since \(R=c_{1}\) a constant, then the above becomes (by merging the constants)

\begin{align*} p-y^{2} & =c_{2}\\ y^{\prime }-y^{2} & =c_{2}\end{align*}

This is the reduced ode.