JEE Differential Equation

Calculus Level 4

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

If $y = y(x)$ satisfies the above differential equation and $y(0) = \sqrt 7$ , find $y(256)$ ?

16 3 9 80

1 solution

Chew-Seong Cheong
May 23, 2017

$\begin{aligned} 8\sqrt x \left(\sqrt{9+\sqrt x}\right) \ dy & = \left(\sqrt{4+\sqrt{9+\sqrt x}}\right)^{-1}dx \\ \implies y(x) & = \int \frac {dx}{8\sqrt x \left({\color{#3D99F6}\sqrt{9+\sqrt x}}\right)\left(\sqrt{4+{\color{#3D99F6}\sqrt{9+\sqrt x}}}\right)} & \small \color{#3D99F6} \text{Let }u = \sqrt{9+\sqrt x}, \ x = \left(u^2 - 9\right)^2, \ dx = 4u\left(u^2 - 9\right) \\ & = \int \frac {4u\left(u^2 - 9\right)\ du}{8\left(u^2 - 9\right)u\sqrt{4+u}} \\ & = \int \frac {du}{2\sqrt{4+u}} \\ & = \sqrt{4+u} + \color{#3D99F6}C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = \sqrt{4+{\color{#3D99F6}\sqrt{9+\sqrt x}}} + C \\ \implies y(0) & = \sqrt{4+\sqrt{9+\sqrt 0}} + C = \sqrt 7 \\ \implies C & = 0 \\ \implies y(x) & = \sqrt{4+\sqrt{9+\sqrt x}} \\ \implies y(256) & = \sqrt{4+\sqrt{9+\sqrt{256}}} = \boxed{3} \end{aligned}$

@Ravneet Singh , Have you seen that this year JEE Advanced and even Mains was easy, I am not telling that JEE Adv was totally easy. But it is easier than last year. Only Physics is twisted, Chemistry is also like maths, Level is OK . Isnt it?

Md Zuhair - 4 years ago

Yeah this year's JEE Was comparable to JEE 2013,14,15 . But certainly easier than jee 2016

Prakhar Bindal - 4 years ago

@Md Zuhair yes maths was on easier side this year. I think in 2011,maths part was this easy. Last year's maths part was not that hard, but was really lenthy.

Ravneet Singh - 4 years ago

JEE Differential Equation

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