* Which of the following statements are TRUE for the differential equation $$ \frac{d y}{d x}=\frac{2 x^{3}}{y^{3}} $$. I The differential equation can be separated as $$ y^{3} d y=2 x^{3} d x $$ II The general solution is $$ \frac{y^{4}}{4}=\frac{x^{4}}{2}+C $$ III The particular solution at the point $$ (1,0) $$ is $$ \frac{y^{4}}{4}=\frac{x^{4}}{2}+\frac{1}{2} $$ A I and II B I and III C II and III D I, II and III

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**Step 1. Separation of Variables**

We are given the differential equation  
$$
\frac{dy}{dx} = \frac{2x^3}{y^3}.
$$  
Multiplying both sides by $$y^3$$ and $$dx$$ gives:  
$$
y^3\, dy = 2x^3\, dx.
$$  
Thus, the equation is separable as indicated in Statement I.

**Step 2. Finding the General Solution**

Integrate both sides:  
$$
\int y^3 \, dy = \int 2x^3 \, dx.
$$

The left-hand side becomes:  
$$
\frac{y^4}{4},
$$  
and the right-hand side becomes:  
$$
\frac{2x^4}{4} = \frac{x^4}{2}.
$$

So, the general solution is:  
$$
\frac{y^4}{4} = \frac{x^4}{2} + C.
$$  
This confirms Statement II.

**Step 3. Examining the Particular Solution**

We are asked to verify the particular solution at the point $$(1, 0)$$. Substitute $$x = 1$$ and $$y = 0$$ into the general solution:  
$$
\frac{0^4}{4} = \frac{1^4}{2} + C \quad \Rightarrow \quad 0 = \frac{1}{2} + C.
$$

Solving for $$C$$ gives:  
$$
C = -\frac{1}{2}.
$$

Thus, the particular solution is:  
$$
\frac{y^4}{4} = \frac{x^4}{2} - \frac{1}{2}.
$$

Statement III claims that the particular solution is  
$$
\frac{y^4}{4} = \frac{x^4}{2} + \frac{1}{2},
$$  
which is incorrect since the constant should be $$-\frac{1}{2}$$ instead of $$+\frac{1}{2}$$.

**Conclusion**

- Statement I is true.
- Statement II is true.
- Statement III is false.

The correct answer is: A I and II.
A I and II

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