If y=y(x) satisfies the differential equation | vedclass

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(A) Given the differential equation: $8 \sqrt{x}(\sqrt{9+\sqrt{x}}) dy = \frac{1}{\sqrt{4+\sqrt{9+\sqrt{x}}}} dx$. Rearranging the terms,we get: $dy =

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$3$

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(A) Given the differential equation: $8 \sqrt{x}(\sqrt{9+\sqrt{x}}) dy = \frac{1}{\sqrt{4+\sqrt{9+\sqrt{x}}}} dx$. Rearranging the terms,we get: $dy = \frac{dx}{8 \sqrt{x} \sqrt{9+\sqrt{x}} \sqrt{4+\sqrt{9+\sqrt{x}}}}$. Let $u = 4+\sqrt{9+\sqrt{x}}$. Then $du = \frac{1}{2\sqrt{9+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} dx = \frac{dx}{4\sqrt{x}\sqrt{9+\sqrt{x}}}$. Substituting this into the differential equation: $dy = \frac{1}{2} \cdot \frac{du}{\sqrt{u}}$. Integrating both sides: $y = \int \frac{1}{2\sqrt{u}} du = \sqrt{u} + C$. Substituting back $u$: $y = \sqrt{4+\sqrt{9+\sqrt{x}}} + C$. Given $y(0) = \sqrt{7}$,we have $\sqrt{4+\sqrt{9+0}} + C = \sqrt{7} \Rightarrow \sqrt{4+3} + C = \sqrt{7} \Rightarrow \sqrt{7} + C = \sqrt{7} \Rightarrow C = 0$. Thus,$y = \sqrt{4+\sqrt{9+\sqrt{x}}}$. For $x=256$,$y(256) = \sqrt{4+\sqrt{9+\sqrt{256}}} = \sqrt{4+\sqrt{9+16}} = \sqrt{4+\sqrt{25}} = \sqrt{4+5} = \sqrt{9} = 3$.

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

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