Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Calculus

Question:

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

Options:

3

9

16

80

Correct Answer:

3

Explanation:

The correct answer is option (1) : 3

$8\sqrt{x} (\sqrt{9+\sqrt{x}}dy = (\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1}dx$

$⇒dy=\frac{1}{\begin{Bmatrix}8\sqrt{x}\left(\sqrt{9+\sqrt{x}}\right)\end{Bmatrix}\begin{Bmatrix}\sqrt{4+\sqrt{9+\sqrt{x}}}\end{Bmatrix}}dx$

Let $ 4+\sqrt{9+\sqrt{x}}=t.$ Then, $\frac{1}{2\sqrt{9+\sqrt{x}}}×\frac{1}{2\sqrt{x}}dx=dt$

Substituting $ 4 + \sqrt{9+\sqrt{x}}=t$ in the given differential equation,

we obtain $dy=\frac{1}{2\sqrt{t}}dt.$ Integrating, we obtain

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

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, $y (256) = 3.$