I just need someone to point out my mistake, I'm trying to find the integrating factor to make the following equation an exact ODE.
$$\frac{x+y}{y-1}dx-\frac{1}{2}(\frac{x+y}{y-1})^2dy=0$$
I multiplied the original equation by -2 for my own ease to make it
$$\frac{-2x-2y}{y-1}dx+(\frac{x+y}{y-1})^2dy=0$$
so it matches the form
\begin{equation*} M(x,y)dx+N(x,y)dy=0 \end{equation*}
So far I've worked out that
\begin{equation*} \frac {\partial M} {\partial y}=\frac{2x+2}{(y-1)^2} \end{equation*}
\begin{equation*} \frac {\partial N} {\partial x}=\frac{2(x+y)}{(y-1)^2} \end{equation*}
leading me to my frustration
Test Case 1 \begin{equation*} \frac {1} {N}*(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})=\frac{2-2y}{(x+y)^2} \end{equation*}
Test Case 2 \begin{equation*} \frac {1} {M}*(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial Y})=\frac{-1}{y+x} \end{equation*}
both are failing for me and I am at wits end on where am I wrong any form of guidance will be much MUCH appreciated
$$\frac{x+y}{y-1}dx-\frac{1}{2}(\frac{x+y}{y-1})^2dy=0$$ $$2({y-1})dx-({x+y})dy=0$$ I substituted $u=y-1$ and $v=x+1$ for simplicity: $$2udv-({u+v})du=0$$ $$(2udv-vdu)-udu=0$$ Divide by $2\sqrt u$: $$(\sqrt udv-\dfrac 1{2\sqrt u}vdu)-\dfrac {\sqrt u} 2du=0$$ $$d(\dfrac v {\sqrt u})-\dfrac 1{2\sqrt u}du=0$$ Integration gives: $$v(u)=u+c\sqrt u$$ Finally: $$x(y)=y-2+c\sqrt {y-1}$$