Exams › JEE Main › Maths
Let the population of rabbits surviving at a time t be governed by the differential equation dp(t)/dt = 1/2 p(t) − 200. If p(0) = 100, then p(t) equals:
- 400 − 300 e^(−t/2)
- 400 − 300 e^(t/2)
- 300 − 200 e^(−t/2)
- 600 − 500 e^(t/2)
Correct answer: 400 − 300 e^(t/2)
Solution
The correct option represents the solution to the differential equation, which describes exponential growth with a carrying capacity. By solving the equation with the initial condition p(0) = 100, we find that the population approaches a stable equilibrium of 400 as time progresses, while the term involving e^(t/2) captures the transient behavior of the population growth.
Related JEE Main Maths questions
- If √(1-x²ⁿ)+√(1-y²ⁿ)=a(xⁿ-yⁿ), then the value of (√(1-x²ⁿ) dy)/(√(1-y²ⁿ) dx) is
- For a curve that passes through the point (4, 0), the slope is governed by
dy/dx = y/x + 5x/((x + 2)(x − 3)).
If the point (5, a) lies on this curve, what is the value of a?
- Which differential equation represents the family of all conics whose axes are aligned with the coordinate axes?
- Find the equation of the curve that satisfies (xy - x²) (dy)/(dx) = y² and passes through the point (-1, 1).
- For the differential equation y = y/x + x/y, if its general solution is written as y = x / log|Cx|, then the function φ(x/y) is
- For the differential equation dy/dx = [y f'(x) − y²]/f(x), where f(x) is a specified function, the solution is
⚔️ Practice JEE Main Maths free + battle 1v1 →