Answer
$C=x^3e^{3y}-3\cos y$
Work Step by Step
We are given
$x^2ydx+y(x^3+e^{-3y}\sin y )dy=0$ ___(1)
Here, $M(x,y)=x^2y$
$N(x,y)=y(x^3+e^{-3y}\sin y)$
We have:
$\frac{M_y-N_x}{N}=3-\frac{1}{y}=f(y)$
The integrating factor is:
$I(x)=e^{\int f(x)dy}=e^{\int 3-\frac{1}{y}dy}=e^{3y-\ln y }=\frac{e^{3y}}{y}$
Multiplying equation (1) by $I(x)=\frac{e^{3y}}{y}$
$\frac{e^{3y}}{y}(x^2ydx+y(x^3+e^{-3y}\sin y )dy)=0$
$x^2e^{3y}dx+(x^3e^{3y}+\sin y)dy=0$
Then:
$\frac{\partial \phi}{\partial x}=x^2e^{3y}$
Integrating both sides:
$\phi=\frac{x^3e^{3y}}{3})+f(y)$
We have:
$\frac{\partial phi}{\partial y}=x^3e^{3y}+f'(y)=x^2e^{3y}+\sin y$
$\rightarrow f'(y)=\sin y$
Since $f'(y)=\sin y \rightarrow f(y)=-\cos y + C$
C is a constant integration
The solution is:
$\phi (x,y)=\frac{x^3e^{3y}}{3}-\cos y+C=0$
or
$C=x^3e^{3y}-3\cos y$
