Question

Let a, K and 0 N be positive real numbers such that 0 < N0 <K. Find the solution of the logistic equation

\frac{dN}{dt} = aN \left ( 1-\frac{N}{K} \right ),N(0) = N_{0}

If N0, N1, and N2 be the values of N at t = 0, t = T and t = 2T respectively, then show that 

\frac{N_{1}}{N_{0}}e^{-aT} = \frac{K-N_{1}}{K-N_{0}}

\frac{N_{2}}{N_{0}}e^{-2aT} = \frac{K-N_{2}}{K-N_{0}}

Also if at time t = t , the population N reaches half its carrying capacity then what is the form of the solution of the logistic equation?

15 Feb 2018
Answer :
Word Count : 478
The logistic equation given is: \[ \frac{dN}{dt} = aN \left(1 - \frac{N}{K} \right) \] with initial condition \( N(0) = N_0 \). ### Step 1: Solve the Logistic Equation This is a separable differential equation. We rewrite it as: \[ \frac{dN}{N(1 - N/K)} = a dt \] Using partial fraction decomposition, \[ \frac{1}{N(1 - N/K)} = \frac{A}{N} + \frac{B}{K-N} \] Multiplying both sides by \( N(K-N) \), \[ 1 = A(K-N) + BN \] Setting \( N = K \), we get \( 1 = A(0) + BK \Rightarrow B = \frac{1}{K} \). Setting \( N = 0 \), we get \( 1 = A(K-0) + B(0) \Rightarrow A = \frac{1}{K} \). Thus, \[ \frac{1}{N(1 - N/K)} = \frac{1}{K} \left(
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