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JEE Advanced 2017

JEE Main 2017 Mathematics question on Differential Equations

If y=y(x)y = y(x) satisfies the differential equation 8x(9+x)dy=(4+9+x)−1dx8\sqrt{x}\left(\sqrt{9+\sqrt{x}}\right)dy = \left(\sqrt{4+\sqrt{9+\sqrt{x}}}\right)^{-1}dx, x>0x > 0 and y(0)=7y(0) = \sqrt{7}, then y(256)=y(256) =
  1. A.
    16
  2. B.
    3
  3. C.
    9
  4. D.
    80
Show answer and solution

Correct option: B

Solution

Substitute u = √x, then v = √(9+u), then w = √(4+v) to simplify the integral.

From dy = dx / [8√x · √(9+√x) · √(4+√(9+√x))]:

Let t = √(4+√(9+√x)). Then:
- √(9+√x) = t²-4, so 9+√x = (t²-4)², giving √x = (t²-4)²-9
- Differentiating: (1/2√x)dx = 2(t²-4)·2t dt, so dx = 8√x·(t²-4)t dt
- Also 8√x·√(9+√x) = 8√x·(t²-4)

So dy = [8√x·(t²-4)t dt] / [8√x·(t²-4)·t] = dt

Therefore y = t + C = √(4+√(9+√x)) + C

Applying y(0) = √7: √(4+√(9+0)) + C = √(4+3) + C = √7 + C = √7, so C = 0.

Thus y(x) = √(4+√(9+√x)).

At x = 256: √256 = 16, √(9+16) = √25 = 5, √(4+5) = √9 = 3.

Answer: 3

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