Studying for a test and I am a bit confused with this problem.
$$y'+\frac y{(2xy-e^{-2y})}=0$$
It says that you can rewrite it as the following:
$${(2xy-e^{-2x})}dy+ydx=0$$
I understand that you multiply both sides by the denominator but I am confused as to how it causes the change of variable in the exponential?
And on the answer sheet instead of doing $M_y$ and $N_x$, it does $M_y$ and $N_y$, giving this as the value for $N_y$:
$$N_y=2y$$
I understand the method but it almost seems like the answer sheet is wrong. Because then it says:
$$\frac {\partial \mu}{\partial x}=\frac {M_y-N_x}M=\frac {1-2y}y$$
It just doesn't make any sense to me whatsoever how you get these values?
First question it seems it's just a typo $$y'+\frac y{(2xy-e^{-2y})}=0$$ $$\frac {dy}{dx}+\frac y{(2xy-e^{-2y})}=0$$ $${dy}+\frac y{(2xy-e^{-2y})}dx=0$$ $${(2xy-e^{-2y})}{dy}+ ydx=0$$ For the integrating factor you only need a function of y
To make it exact... $${(2xy-e^{-2y})}{dy}+ ydx=0$$ $$2y{(2xy-e^{-2y})}{dy}+ 2y^2dx=0$$ $$Qdy+Pdx=0$$ We need for exactness that : $$\frac {\partial P}{\partial y}-\frac {\partial Q}{\partial x}=0$$ $$4y-4y=0$$