The following shows how to determine if \(F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) =0\) is exact or not (without having to find the first integral \(R\)). This is based on page 164 in Murphy book. The second order ode must be of degree one. If it is, it can not be exact. The ode is exact iff\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} This turns out to be the same thing as using \(p^{\prime \prime }-q^{\prime }+r=0\) on the ode \(py^{\prime \prime }-q^{\prime }+r=0\). Let us apply the above test on second order ode which is known to be exact to see how it works. The ode is\begin {align*} F\left ( x,y,y^{\prime },y^{\prime \prime }\right ) & =0\\ xy^{\prime \prime }+\left ( y-1\right ) y^{\prime } & =0 \end {align*}
Hence the above test gives\begin {align*} y^{\prime }-\frac {d}{dx}\left ( y-1\right ) +\frac {d^{2}}{dx^{2}}\left ( x\right ) & =0\\ y^{\prime }-y^{\prime } & =0\\ 0 & =0 \end {align*}
Confirmed. Since the ode is linear, we could also apply \(p^{\prime \prime }-q^{\prime }+r=0\) to check, which is simpler. Here \(p=x,q=\left ( y-1\right ) ,r=0\). Therefore \begin {align*} p^{\prime \prime }-q^{\prime }+r & =0\\ 0-0+0 & =0 \end {align*}
The form (1) is given in Murphy book which is more general since it works on nonlinear and linear odes while \(p^{\prime \prime }-q^{\prime }+r=0\) is meant to be used for linear second order odes.
In implementation of the solver this is the same type of ode as "second order integrable as is" ode which is described below. I should merge these together. if a second order ode is exact, then it is also integrable ode as is. This is by definition of exactness above.