If $y=y(x)$ satisfies the differential equation
$8 \sqrt{x}(\sqrt{9+\sqrt{x}}) d y=(\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} d x, x>0$
and $y(0)=\sqrt{7}$, then $y(256)=$
Answer & explanation
Correct answer: option 1
The given differential equation is
$8 \sqrt{x}(\sqrt{9+\sqrt{x}}) d y=(\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} d x$
$\Rightarrow \quad d y=\frac{1}{\{8 \sqrt{x}(\sqrt{9+\sqrt{x}})\}\{\sqrt{4+\sqrt{9+\sqrt{x}}}\}} d x $
Let $4+\sqrt{9+\sqrt{x}}=t$. Then, $\frac{1}{2 \sqrt{9+\sqrt{x}}} \times \frac{1}{2 \sqrt{x}} d x=d t$
Substituting $4+\sqrt{9+\sqrt{x}}=t$ in the given differential equation, we obtain $d y=\frac{1}{2 \sqrt{t}} d t$. Integrating, we obtain
$y=\sqrt{t}+C \Rightarrow y=\sqrt{4+\sqrt{9+\sqrt{x}}}+C$ ......(i)
It is given that $y=\sqrt{7}$ when $x=0$. Substituting these values in (i), we get $C=0$.
Putting $C=0$ in (i), we get : $y=\sqrt{4+\sqrt{9+\sqrt{x}}}$
Putting $x=256$, we get $y=3$.
Hence, $y(256)=3$.