Finding the Potential Function

Let

$$M(x,y)=3x^2y+2xy+y^3$$

and

$$N(x,y)=x^3+x^2+3xy^2$$

First check whether the differential equation is exact.

We have

$$\frac{\partial M}{\partial y}=3x^2+2x+3y^2$$

and

$$\frac{\partial N}{\partial x}=3x^2+2x+3y^2$$

Since

$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

the equation is exact.

Hence there exists a function

$$F(x,y)$$

such that

$$dF=M\,dx+N\,dy$$

Integrating

$$M$$

with respect to

$$x$$,

we get

$$F(x,y)=x^3y+x^2y+xy^3+\phi(y)$$

Differentiating with respect to

$$y$$,

$$\frac{\partial F}{\partial y}=x^3+x^2+3xy^2+\phi'(y)$$

Comparing with

$$N=x^3+x^2+3xy^2$$

gives

$$\phi'(y)=0$$

Hence

$$\phi(y)=C$$

and therefore

$$F(x,y)=x^3y+x^2y+xy^3$$

The implicit solution is

$$x^3y+x^2y+xy^3=C$$

Using the initial condition

$$y(1)=1$$,

$$C=1^3(1)+1^2(1)+1(1)^3$$

$$=1+1+1$$

$$=3$$

Therefore the required solution is

$$x^3y+x^2y+xy^3=3$$

Therefore, $$\text{Answer}=(B)$$.