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AIEEE 2003

JEE Main 2003 Mathematics question on Differential Equations

If y(t)y(t) is a solution of (1+t)dydt−ty=1(1+t)\frac{dy}{dt} - ty = 1 and y(0)=−1y(0) = -1, then y(1)y(1) is equal to
  1. A.
    −1/2-1/2
  2. B.
    e+1/2e + 1/2
  3. C.
    e−1/2e - 1/2
  4. D.
    1/21/2
Show answer and solution

Correct option: A

Solution

The integrating factor for dy/dt - t/(1+t) · y = 1/(1+t) is e^(∫-t/(1+t)dt) = e^(-(t - ln(1+t))) = e^(ln(1+t) - t) = (1+t)e^(-t).

Multiplying through: d/dt[y·(1+t)e^(-t)] = e^(-t), so y·(1+t)e^(-t) = -e^(-t) + C.

Applying y(0) = -1: (-1)(1)(1) = -1 + C → C = 0.

So y = -1/(1+t). At t = 1: y(1) = -1/2.

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