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JEE Main 2019 · 8 Apr · Shift I

JEE Main 2019 Mathematics question on Differential Equations

Let y=y(x)y = y(x) be the solution of the differential equation, (x2+1)2dydx+2x(x2+1)y=1(x^2 + 1)^2 \frac{dy}{dx} + 2x(x^2 + 1)y = 1 such that y(0)=0y(0) = 0. If a y(1)=π32\sqrt{a}\, y(1) = \frac{\pi}{32}, then the value of 'aa' is
  1. A.
    4
  2. B.
    12\frac{1}{2}
  3. C.
    1
  4. D.
    116\frac{1}{16}
Show answer and solution

Correct option: D

Solution

Divide by (x²+1)², giving dy/dx + 2x/(x²+1) · y = 1/(x²+1)².

Integrating factor: e^(∫2x/(x²+1)dx) = e^(ln(x²+1)) = x²+1.

Multiply through: d/dx[(x²+1)y] = 1/(x²+1).

Integrate: (x²+1)y = ∫dx/(x²+1) = arctan(x) + C.

With y(0)=0: C=0, so y = arctan(x)/(x²+1).

At x=1: y(1) = arctan(1)/2 = (π/4)/2 = π/8.

From √a · y(1) = π/32: √a · (π/8) = π/32, so √a = 1/4, giving a = 1/16.

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