Particular Solution from Initial Condition

1. Rearrange the Differential Equation

Start with the given differential equation:

$$3x^2 y - 3x^3 \frac{dy}{dx} = \frac{6y^4}{x}$$

Divide every single term by $$y^4$$ to start separating the variables:

$$\frac{3x^2}{y^3} - \frac{3x^3}{y^4} \frac{dy}{dx} = \frac{6}{x}$$

Notice that the left side of the equation represents the exact result of applying the quotient rule derivative to the fraction $$\frac{x^3}{y^3}$$ with respect to $$x$$:

$$\frac{d}{dx}\left(\frac{x^3}{y^3}\right) = \frac{6}{x}$$

2. Integrate Both Sides

Integrate both sides of the equation with respect to $$x$$:

$$\int \frac{d}{dx}\left(\frac{x^3}{y^3}\right) \, dx = \int \frac{6}{x} \, dx$$

This clears the derivative on the left side and produces a natural logarithm on the right side:

$$\frac{x^3}{y^3} = 6\ln x + C$$

3. Find the Integration Constant C

We are given that the solution curve passes through the initial boundary point $$(1, 1)$$. Substitute $$x = 1$$ and $$y = 1$$ into our integrated equation:

$$\frac{1^3}{1^3} = 6\ln(1) + C$$

Since the natural logarithm of $$1$$ is zero ($$\ln 1 = 0$$), the equation simplifies directly to:

$$1 = 0 + C \implies C = 1$$

Now, substitute $$C = 1$$ back into the formula to get the final particular equation of the curve:

$$\frac{x^3}{y^3} = 6\ln x + 1$$

4. Evaluate the Curve at x = 2

To find the value of $$y(2)$$, substitute $$x = 2$$ directly into the particular equation:

$$\frac{2^3}{y^3} = 6\ln 2 + 1$$

$$\frac{8}{y^3} = 6\ln 2 + 1$$

Invert both sides of the fraction to isolate the $$y^3$$ term:

$$y^3 = \frac{8}{6\ln 2 + 1}$$

Take the cube root of both sides to get the final isolated value for $$y$$:

$$y = \frac{\sqrt[3]{8}}{\sqrt[3]{6\ln 2 + 1}}$$

$$y = \frac{2}{\sqrt[3]{6\ln 2 + 1}}$$

Final Answer

The value of $$y(2)$$ is equal to $$\frac{2}{\sqrt[3]{6\ln 2 + 1}}$$. Therefore, the correct option is (A).

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