Value of Function Satisfying Given Radical Differential Equation

To find the value of $y(256)$, we start by rewriting the given differential equation in a separable form:

$8\sqrt{x}\left(\sqrt{9 + \sqrt{x}}\right) dy = \left(\sqrt{4 + \sqrt{9 + \sqrt{x}}}\right)^{-1} dx$

$dy = \frac{dx}{8\sqrt{x}\left(\sqrt{9 + \sqrt{x}}\right)\left(\sqrt{4 + \sqrt{9 + \sqrt{x}}}\right)}$

To integrate the right-hand side, we make the substitution: $u = \sqrt{4 + \sqrt{9 + \sqrt{x}}}$

Differentiating $u$ with respect to $x$ using the chain rule: $du = \frac{1}{2\sqrt{4 + \sqrt{9 + \sqrt{x}}}} \cdot \frac{d}{dx}\left(4 + \sqrt{9 + \sqrt{x}}\right)$ $du = \frac{1}{2\sqrt{4 + \sqrt{9 + \sqrt{x}}}} \cdot \frac{1}{2\sqrt{9 + \sqrt{x}}} \cdot \frac{d}{dx}\left(9 + \sqrt{x}\right)$ $du = \frac{1}{2\sqrt{4 + \sqrt{9 + \sqrt{x}}}} \cdot \frac{1}{2\sqrt{9 + \sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} dx$ $du = \frac{dx}{8\sqrt{x}\sqrt{9 + \sqrt{x}}\sqrt{4 + \sqrt{9 + \sqrt{x}}}}$

Thus, the differential equation simplifies to: $dy = du$

Integrating both sides with respect to their variables: $y(x) = u + C$ $y(x) = \sqrt{4 + \sqrt{9 + \sqrt{x}}} + C$

Using the initial condition $y(0) = \sqrt{7}$: $y(0) = \sqrt{4 + \sqrt{9 + \sqrt{0}}} + C = \sqrt{4 + 3} + C = \sqrt{7} + C$ Given $y(0) = \sqrt{7}$, we have: $\sqrt{7} + C = \sqrt{7} \implies C = 0$

So, the function $y(x)$ is given by: $y(x) = \sqrt{4 + \sqrt{9 + \sqrt{x}}}$

Now, we evaluate $y(256)$: $y(256) = \sqrt{4 + \sqrt{9 + \sqrt{256}}}$ Since $\sqrt{256} = 16$: $y(256) = \sqrt{4 + \sqrt{9 + 16}} = \sqrt{4 + \sqrt{25}} = \sqrt{4 + 5} = \sqrt{9} = 3$

Thus, the value of $y(256)$ is $3$.

Correct Option: [A]