1. ## [SOLVED] Problem

The following is an Exact Differential Equation :-

$\displaystyle (\frac{x+y}{y-1}) dx -\frac{1}{2}(\frac{x+1}{y-1})^2dy$

Multiplied both sides with $\displaystyle 2(y-1)^2$.

Then opened $\displaystyle -(x+1)^2$ and $\displaystyle (y-1)(x+y)$.

After partial differentiation the two sides doesn't come the same whereas this whole exercise is composed of questions which are Exact. And the end answer also doesn't come correct.

Is there a problem with my approach or what ?

2. Hello, Altair!

Check your work . . .

The following is an Exact Differential Equation:

. . $\displaystyle \left(\frac{x+y}{y-1}\right) dx -\frac{1}{2}\left(\frac{x+1}{y-1}\right)^2dy \;=\;0$

We have: .$\displaystyle M \;=\;\frac{x+y}{y-1}$

. . Then: .$\displaystyle M_y \:=\:\frac{(y-1)\!\cdot\!1 - (x+y)\!\cdot\!1}{(y-1)^2} \;=\;\boxed{-\frac{x+1}{(y-1)^2}}$

We have: .$\displaystyle N \;=\;-\frac{1}{2}\frac{(x+1)^2}{(y-1)^2} \;=\;-\frac{1}{2(y-1)^2}(x+1)^2$

. . Then: .$\displaystyle N_x\;=\;-\frac{1}{2(y-1)^2}\cdot2(x+1) \;=\;\boxed{-\frac{x+1}{(y-1)^2}}$