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Question Number 129949 by bemath last updated on 21/Jan/21

$$\mathrm{Given}8\sqrt{\mathrm{x}}\left(\sqrt{9+\sqrt{\mathrm{x}}}\right)\mathrm{dy}=\frac{\mathrm{dx}}{\sqrt{4+\sqrt{9+\sqrt{\mathrm{x}}}}},\mathrm{x}>1$$ $$\mathrm{and}\mathrm{y}\left(0\right)=\sqrt{7}.\mathrm{Find}\mathrm{y}\left(256\right).$$

Answered by liberty last updated on 21/Jan/21

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{8\sqrt{\mathrm{x}}\left(\sqrt{9+\sqrt{\mathrm{x}}}\right)\left(\sqrt{4+\sqrt{9+\sqrt{\mathrm{x}}}}\right)}$$ $$\mathrm{let}\sqrt{4+\sqrt{9+\sqrt{\mathrm{x}}}}=\mathrm{z};\sqrt{9+\sqrt{\mathrm{x}}}={\mathrm{z}}^2-4$$ $$\frac{\frac{1}{2\sqrt{\mathrm{x}}}}{2\sqrt{9+\sqrt{\mathrm{x}}}}\mathrm{dx}=2\mathrm{z}\mathrm{dz}$$ $$\frac{\mathrm{dx}}{4\sqrt{\mathrm{x}}\sqrt{9+\sqrt{\mathrm{x}}}}=2\mathrm{z}\mathrm{dz};\frac{1}{8\sqrt{\mathrm{x}}\sqrt{9+\sqrt{\mathrm{x}}}}\mathrm{dx}=\mathrm{z}\mathrm{dz}$$ $$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{z}\mathrm{dz}}{\mathrm{z}}\Rightarrow \mathrm{y}=\mathrm{z}+\mathrm{c}$$ $$\iff \mathrm{y}\left(\mathrm{x}\right)=\sqrt{4+\sqrt{9+\sqrt{\mathrm{x}}}}+\mathrm{c}$$ $$\mathrm{y}\left(0\right)=\sqrt{7}+\mathrm{c}\Rightarrow \mathrm{c}=0$$ $$\therefore \mathrm{y}\left(\mathrm{x}\right)=\sqrt{4+\sqrt{9+\sqrt{\mathrm{x}}}}$$ $$\mathrm{y}\left(256\right)=\sqrt{4+\sqrt{9+\sqrt{256}}}$$ $$\mathrm{y}\left(256\right)=\sqrt{4+\sqrt{25}}=\sqrt{9}=3$$

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