Answer
$y=\sqrt \frac{10-x^3}{x^2}$
Work Step by Step
We are given:
$$(3x^2+2xy^2)dx+(2x^2y)dy=0$$
We have that:
$$M(x,y)=3x^2+2xy^2 \wedge N(x,y)=2x^2y$$
then
$$M'(x,y)=4xy =N'(x,y)$$
So this equation is exist.
Let's set:
$$\phi(x,y)=0$$
There exists a potential function $\phi$ satisfies:
$$\frac{\partial \phi}{\partial x}=M \wedge \frac{\partial \phi}{\partial y}=N$$
$$\frac{\partial \phi}{\partial x}=3x^2+2xy^2 \wedge \frac{\partial \phi}{\partial y}=2x^2y$$
Integrating $\frac{\partial \phi}{\partial x}=3x^2+2xy^2 $ we get
$$\phi=x^3+x^2y^2+f(y)$$
Since $\frac{\partial \phi}{\partial y}=2x^2y+f'(y)$
We obtain:
$$f'(y)=0 \rightarrow f(y)=C$$
$C$ is constant of integration
The potential solution is:
$$\phi=x^3+x^2y^2$$
Hence general solution is $C=x^3+x^2y^2$
We are given $y(1)=3$
$$C=1+3^2=10$$
The final solution is:
$$y=\sqrt \frac{10-x^3}{x^2}$$
