# How to show this Non-Exact Differential equation

Solve the following differential equation :

$$x \sin(y)dx+(x+1)\cos(y) dy$$

Here's where i'm stuck. I've let M=x siny and N=(x+1)cosy

Differentiating partially for both $M_y$=x cosy and $N_x$=cosy

$\frac{M_y-N_x}{N}$=$\frac{x-1}{x+1}$

I used a trick such that instead of x-1 I used x+1-2 and integrated to get my integrating factor as $e^x - \frac{1}{(x+1)^2}$

Now what do I do? I've tried multiplying my original equation by my integrating factor but it's long and disgusting and seems wrong. I have no idea how to proceed from here can anyone advise me

• do you meant $$x\sin(y)dx+(x+1)\cos(y)dy=0$$? – Dr. Sonnhard Graubner Mar 1 '18 at 17:03

if so then write your equation in the form $$\frac{x}{x+1}dx=-\frac{\cos(y)}{\sin(y)}dy$$