If $y = y(x)$ satisfies the differential equation $8\sqrt{x}\left(\sqrt{9+\sqrt{x}}\right)dy = \left(\sqrt{4+\sqrt{9+\sqrt{x}}}\right)^{-1}dx$, $x>0$ and $y(0)=\sqrt{7}$, then $y(256)=$

If y = y(x) satisfies the differential equation 8√x(√9+√x)dy = (√4+√9+√x)⁻¹dx, x>0 and y(0)=√7, then y(256)= $8\sqrt{x}\,\sqrt{9+\sqrt{x}}\,dy = \dfrac{dx}{\sqrt{4+\sqrt{9+\sqrt{x}}}},\qquad x>0$ $y(0) = \sqrt{7}$ $\text{To find: } y(256)$ $\text{Plan: substitute } s = \sqrt{x}, \text{ then } u = \sqrt{9+s}, \text{ and integrate}$ $8\sqrt{x}\,\sqrt{9+\sqrt{x}}\,dy = \dfrac{dx}{\sqrt{4+\sqrt{9+\sqrt{x}}}}$ $\dfrac{dy}{dx} = \dfrac{1}{8\sqrt{x}\,\sqrt{9+\sqrt{x}}\,\sqrt{4+\sqrt{9+\sqrt{x}}}}$ $s = \sqrt{x},\quad dx = 2s\,ds \Rightarrow \dfrac{dy}{ds} = \dfrac{1}{4\sqrt{9+s}\,\sqrt{4+\sqrt{9+s}}}$ $u = \sqrt{9+s},\quad ds = 2u\,du,\quad \sqrt{9+s}=u \Rightarrow \dfrac{dy}{du} = \dfrac{2u}{4u\sqrt{4+u}} = \dfrac{1}{2\sqrt{4+u}}$ $y = \int \dfrac{du}{2\sqrt{4+u}} = \sqrt{4+u} + C$ $y(x) = \sqrt{4 + \sqrt{9 + \sqrt{x}}} + C$ $y(0) = \sqrt{4+\sqrt{9+0}} + C = \sqrt{7} + C = \sqrt{7} \Rightarrow C = 0$ $\Rightarrow y(x) = \sqrt{4 + \sqrt{9 + \sqrt{x}}}$ $y(256) = \sqrt{4+\sqrt{9+\sqrt{256}}} = \sqrt{4+\sqrt{9+16}} = \sqrt{4+5} = \sqrt{9} = 3$ $\boxed{\text{option (1)}}$