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