Prove that $\frac{1}{N^2 + M^2}$ is an integrating factor for $M(x, y) dx + N(x, y) dy = 0$ when $N_x$ = -$M_y$ and $N_y$ = $M_x$

Question

Prove that $\frac{1}{N^2 + M^2}$ is an integrating factor for $M(x, y) dx + N(x, y) dy = 0$ when $N_x = -M_y$ and $N_y = M_x$

I tried to show that the equation $uMdx+uNdx=0$ is exact by showing that $\frac{\partial}{\partial y}[uM] = \frac{\partial}{\partial x}[uN]$, where $u = \frac{1}{N^2 + M^2}$ is the integrating factor

This is how I tried to do it:

$\frac{\partial}{\partial y}[uM] = \frac{\partial}{\partial y}[M]\cdot u+M\cdot\frac{\partial}{\partial y}[u]$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= \frac{\partial}{\partial y}[M]\cdot u+M \cdot \frac{\partial}{\partial y}[\frac{1}{N^2 + M^2}]$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= \frac{\partial}{\partial y}[M]\cdot u+M \cdot (-(M^2+N^2)^{(-2)}(2M\frac{\partial M}{\partial y}+2N\frac{\partial N}{\partial y})) $

Then I solved $M$ from $M dx + N dy = 0$

$\implies M = -N\frac{dy}{dx}$

and substitute that

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= \frac{\partial}{\partial y}[M]\cdot u-N\frac{dy}{dx} \cdot (-(M^2+N^2)^{(-2)}(2M\frac{\partial M}{\partial y}+2N\frac{\partial N}{\partial y})) $

then the $dy$ cancel out

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= \frac{\partial}{\partial y}[M]\cdot u-N \cdot (-(M^2+N^2)^{(-2)}(2M\frac{\partial M}{\partial x}+2N\frac{\partial N}{\partial x})) $

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= \frac{\partial}{\partial y}[M]\cdot u-N \frac{\partial}{\partial x}[u] $

then used the $N_x$ = -$M_y$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= -\frac{\partial}{\partial x}[N]\cdot u-N \frac{\partial}{\partial x}[u] $

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= -\frac{\partial}{\partial x}[uN]$

but I got $\frac{\partial}{\partial y}[uM] = -\frac{\partial}{\partial x}[uN]$ instead of $\frac{\partial}{\partial y}[uM] = \frac{\partial}{\partial x}[uN]$

Answer 1

The following are fairly clear: \begin{align} M_x&=N_y\,,\quad M_y=-N_x\\[2mm] u&=(M^2+N^2)^{-1}\,,\\[2mm] u_x&=-2(MM_x+NN_x)\,u^2=-2(MN_y-NM_y)\,u^2\,,\\[2mm] u_y&=-2(MM_y+NN_y)\,u^2\,,\\[2mm] (uM)_y&=uM_y+Mu_y\,,\\[2mm] (uN)_x&=uN_x+Nu_x\,. \end{align} Now use them patiently to transform the last identity to become $(uN)_x=(uM)_y\,:$

\begin{align}(uN)_x&=uN_x+Nu_x\,,\\[2mm]&=-uM_y+Nu_x=-uM_y-2N(MN_y-NM_y)\,u^2\\[2mm]&=-uM_y-2NMN_yu^2+2N^2M_yu^2+2M^2M_yu^2-2M^2M_yu^2\\[2mm]&=-uM_y-2NMN_yu^2+2M_yu-2M^2M_yu^2\\[2mm]&=uM_y-2M(MM_y+NN_y)u^2\\[2mm]&=uM_y+Mu_y=(uM)_y\,.\end{align}