JEE Main 2021 · 24 February · morning · Single correct · +4 / −1
The population P = P(t) at time 't' of a certain species follows the differential equation
dP/dt = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is :
- (a) log ₑ18
- (b) 1/2log ₑ18
- (c) 2log ₑ18
- (d) log ₑ9
Answer and solution
Answer (c) 2log ₑ18
Solution
Maths sir · 4 steps
Pull out the 0.5 so the right side becomes 0.5(P-900), then separate the variables and integrate.
- Factor the right side. Write 0.5P - 450 = 0.5(P - 900). Now the equation reads dP/(P-900) = 0.5 dt, with P on one side and t on the other.
- Integrate. This gives ln|P-900| = 0.5t + C, so P - 900 = K e^(t/2) for some constant K. At t = 0, P = 850, so K = 850 - 900 = -50.
P(t) = 900 - 50 e^(t/2)
- Set P to zero. 0 = 900 - 50e^(t/2) gives e^(t/2) = 18. Take the natural log: t/2 = logₑ 18.
- So. t = 2 logₑ 18, which is option (c).
The idea to keep. For dy/dt = ay + b, write it as a(y + b/a). The solution is y = -b/a + Ceᵃᵗ, and the constant a in the exponent decides the factor in the time.
Answer from the NTA final answer key. Solution by PrepWiser.