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JEE Main 2014

JEE Main 2014 Mathematics question on Differential Equations

Let the population of rabbits surviving at a time tt be governed by the differential equation dp(t)dt=12p(t)−200\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200. If p(0)=100p(0) = 100, then p(t)p(t) is equal to
  1. A.
    400−300et2400 - 300e^{\frac{t}{2}}
  2. B.
    300−200et2300 - 200e^{\frac{t}{2}}
  3. C.
    600−500et2600 - 500e^{\frac{t}{2}}
  4. D.
    400−300et2400 - 300e^{\frac{t}{2}}
Show answer and solution

Correct option: A

Solution

The ODE dp/dt - (1/2)p = -200 is linear. The integrating factor is e^(-t/2), giving d/dt[p·e^(-t/2)] = -200e^(-t/2). Integrating: p·e^(-t/2) = 400e^(-t/2) + C, so p(t) = 400 + Ce^(t/2). Applying p(0) = 100: 100 = 400 + C, so C = -300. Therefore p(t) = 400 - 300e^(t/2).

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