4.1.2.3.5 Example 5
\begin{align*} y^{\prime \prime }-y^{\prime }e^{y} & =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\\ -y^{\prime }e^{y}-\frac {d}{dx}\left ( -e^{y}\right ) +\frac {d^{2}}{dx^{2}}\left ( 1\right ) & =0\\ -y^{\prime }e^{y}+y^{\prime }e^{y} & =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 & =-y^{\prime }e^{y}\\ & =-pe^{y}\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\\ -pe^{y} & =\left ( \frac {\partial }{\partial x}p+\psi _{x}\right ) +\left ( \frac {\partial }{\partial y}p+\psi _{y}\right ) p\\ -pe^{y} & =\psi _{x}+\psi _{y}p \end{align*}
Comparing terms shows that
\begin{align*} -e^{y} & =\psi _{y}\\ 0 & =\psi _{x}\end{align*}
Integrating the first equation gives \(\psi =-e^{y}+h\left ( x\right ) \). Partial differentiating this w.r.t. \(x\) gives \(\psi _{x}=h^{\prime }\left ( x\right ) \). comparing
this to the second equation above gives \(h^{\prime }\left ( x\right ) =0\), hence \(h\left ( x\right ) =c\). Hence \(h=-x^{2}+c\). Therefore\(\ \psi =-e^{y}+c\). Eq (1A) becomes
\begin{align*} R & =p-e^{y}+c\\ & =y^{\prime }-e^{y}+c \end{align*}
Therefore, since \(\phi =c_{1}\) a constant, then the above becomes (by merging the constants)
\[ y^{\prime }-e^{y}=c_{2}\]
Which is
the reduced ode to solve.