JEE Advanced 2017 Paper 2Question 1
If $y = y(x)$ satisfies the differential equation
$$8\sqrt{x}\left(\sqrt{9+\sqrt{x}}\right)dy = \left(\sqrt{4+\sqrt{9+\sqrt{x}}}\right)^{-1}dx, \quad x>0$$
and $y(0)=\sqrt{7}$, then $y(256)=$
Differential Equation
Solution
Concept. The right side is exactly the derivative of $\sqrt{4+\sqrt{9+\sqrt x}}$. Formulas used. (1) Chain rule: $\dfrac{d}{dx}\sqrt{u}=\dfrac{u'}{2\sqrt u}$Step 1 - Differentiate the guess. Let $w=\sqrt{4+\sqrt{9+\sqrt x}}$. Applying (1) three times: $$\frac{dw}{dx}=\frac{1}{2w}\cdot\frac{1}{2\sqrt{9+\sqrt x}}\cdot\frac{1}{2\sqrt x}=\frac{1}{8\sqrt x\,\sqrt{9+\sqrt x}\,w}.$$ This is exactly $\frac{dy}{dx}$ from the given equation. Step 2 - Constant. So $y=w+C$. At $x=0$: $w=\sqrt{4+3}=\sqrt7=y(0)$, so $C=0$. Step 3 - At $x=256$. $\sqrt{256}=16$, $\sqrt{25}=5$, $\sqrt9=3$. So $y(256)=3$. Answer: (A)