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The ODE \begin{align} N(x,y)p+M(x,y)=0;\quad [p=y'_x] \end{align} can be tested for exactness by checking if $N_x=M_y$. This test is found by assuming a first integral of the form $U(x,y)=C$ with derivative $U_yp+U_x=0$. If $U$ is a nice function then $U_{yx}=U_{xy}$, and if $U$ is indeed a first integral it must be that $N_x=M_y$.

Consider now the second order ODE \begin{align} N(x,y,p)p'_x+M(x,y,p)=0. \end{align} To see if this equation is exact we assume a first integral of the form $U(x,y,p)=C$ with derivative $U_{p}p'+U_yp+U_x=0$. If $U$ is indeed a first integral we are left with the conditions \begin{align}\tag{1} U_{p}=N,\quad U_yp+U_x=M. \end{align} If I take two partial derivatives w.r.t $p$ to eliminate $U$ I find the equation \begin{align}\tag{2}\label{2} pN_{yp}+2N_y+N_{xp}=M_{pp}. \end{align} It seems to me, though, that this equation is necessary but not sufficient for arbitrary $N$ and $M$. Consider the ODE $p'_x+f(x,y)p+g(x,y)=0$, which is obviously exact only if $f_x=g_y$ but satisfies equation (2). Is there a correct way to eliminate $U$ in equations (1)? Or perhaps equation (2) works if we add some restrictions on $N$ and $M$? I imagine taking several derivatives of equation (1) is to blame for this.

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