Question

If $y=y\left(x\right)$ 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 $y\left(0\right)=\sqrt{7}$, then $y\left(256\right)$___.

• A. $16$
• B. $80$
• C. $3$
• D. $9$

Answer 1

The correct option is C $3$

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

Let $9+\sqrt{x}=t^2$

$\Rightarrow \frac{1}{2\sqrt{x}}dx=2tdt$

$\Rightarrow \frac{dx}{\sqrt{x}}=4tdt$

$\Rightarrow 8dy=\frac{4tdt}{t\sqrt{4+t}}$

$\Rightarrow 2dy=\frac{dt}{\sqrt{4+t}}$

Integrate both sides

$\Rightarrow \int 2dy=\int \frac{dt}{\sqrt{4+t}}$

$\Rightarrow 2y=2\sqrt{4+t}+c_1$

$\Rightarrow y=\sqrt{4+t}+C$

Resubstituting the value of $t$,

$\Rightarrow y=\sqrt{4+\sqrt{9+\sqrt{x}}}+C$

Now, for $x=0$, we have $y=\sqrt{7}$

$\Rightarrow \sqrt{7}=\sqrt{4+\sqrt{9+0}}+C$

$\Rightarrow \sqrt{7}=\sqrt{7}+C$

$\Rightarrow C=0$

Hence, the solution of the given differential equation is

$y=\sqrt{4+\sqrt{9+\sqrt{x}}}$

Now, for $x=256$

$y=\sqrt{4+\sqrt{9+\sqrt{256}}}$

$\Rightarrow y=\sqrt{4+\sqrt{9+16}}$

$\Rightarrow y=\sqrt{4+\sqrt{25}}$

$\Rightarrow y=\sqrt{4+5}=\sqrt{9}=3$

$\therefore y\left(256\right)=3$

Hence, option $C$ is correct.