4.1.2.1.1 Example 1

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

Applying the test

\begin{equation} \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 \tag {1}\end{equation}

Therefore

\begin{align*} \frac {\partial F}{\partial y} & =-\frac {2}{y^{3}}xy^{\prime }+\frac {1}{y^{2}}\\ \frac {\partial F}{\partial y^{\prime }} & =\frac {x}{y^{2}}\\ \frac {\partial F}{\partial y^{\prime \prime }} & =1 \end{align*}

Hence (1) becomes

\begin{align*} \left ( -\frac {2}{y^{3}}xy^{\prime }+\frac {1}{y^{2}}\right ) -\frac {d}{dx}\left ( \frac {x}{y^{2}}\right ) +\frac {d^{2}}{dx^{2}}\left ( 1\right ) & =0\\ \left ( -\frac {2}{y^{3}}xy^{\prime }+\frac {1}{y^{2}}\right ) -\left ( \frac {1}{y^{2}}-\frac {2xy^{\prime }}{y^{3}}\right ) & =0\\ 0 & =0 \end{align*}

Therefore this exact. We see that \(\left ( y^{\prime }-\frac {x}{y}\right ) ^{\prime }=y^{\prime \prime }-\left ( \frac {1}{y}+\frac {xy^{2}}{y^{\prime }}\right ) \). Which implies the ode is integrable as is. Which means

\begin{align} \int \left ( y^{\prime }-\frac {x}{y}\right ) ^{\prime }dx & =0\nonumber \\ y^{\prime }-\frac {x}{y} & =c \tag {2}\end{align}

Which can now be solved. In the above \(R\left ( x,y,y^{\prime }\right ) =\left ( y^{\prime }-\frac {x}{y}\right ) \). In other words \(F=\frac {d}{dx}R\). Hence

\[ \frac {d}{dx}R=0 \]

Integrating gives

\begin{align*} \int \frac {d}{dx}Rdx & =c\\ \int dR & =c\\ R & =c\\ y^{\prime }-\frac {x}{y} & =c \end{align*}

Which is the same as (2) above but shows how it came about more clearly.