Answer
$C=x\sin y+y\sin x$
Work Step by Step
We are given:
$$(\sin y+y\cos x)dx+(x \cos y+\sin x)dy=0 (1)$$
Here,
$M(x,y)=\frac{\partial}{\partial y}(\sin y+y\cos x)=\cos y + \cos x$
$N(x,y)=\frac{\partial}{\partial x}(x \cos y+\sin x)=\cos y + \cos x$
$M_y=\cos y + \cos x \;\;\;\;,\;\;\;\;N_x=\cos y + \cos x$
$M_y=N_x$
$\Rightarrow$ (1) is exact differential equation
Therefore there exists a potential function $\phi$ such that
$\frac{\partial\phi}{\partial x}=\sin y+y\cos x$ ___(2)
Integrating (2) with respect to $x$ holding $y$ fixed
$\phi =x\sin y+y \sin x+f(y)$ ___(3)
Where $f(y)$ is arbitrary function of $y$
Hence we get:
$\frac{\partial\phi}{\partial y}=x\cos y+\sin x+f'(y)=x\cos y+\sin x$ (4)
We have
$f'(y)= 0\rightarrow f(y)=C$
$C$ is constant
The solution is:
$\phi(x,y)=x\sin y+y\sin x+C$
and $\phi(x,y)=0$
so:
$C=x\sin y+y\sin x$
