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