We decompose the fraction (K)/(P(K-P)) into partial fractions (A)/(P) + (B)/(K-P). By equating the numerators, K = A(K-P) + BP. Setting P=0 gives K=AK, so A=1. Setting P=K gives K=BK, so B=1.

Now we integrate the decomposed fractions. The integral of (1)/(P) is ln|P| and the integral of (1)/(K-P) is -ln|K-P|. Combining the logarithms and exponentiating both sides, we get (P)/(K-P) = A e^(rt), where A = e^C.

Now we integrate the decomposed fractions. The integral of (1)/(P) is ln|P| and the integral of (1)/(K-P) is -ln|K-P|. Combining the logarithms and exponentiating both sides, we get (P)/(K-P) = A e^(rt), where A = e^C.

We use the initial condition P(0) = P_0 to find the constant A. Substituting t=0 and P=P_0 into the equation (P)/(K-P) = A e^(rt), we find A = (P_0)/(K-P_0). Substituting A back and solving for P(t) gives the logistic growth function.

We use the initial condition P(0) = P_0 to find the constant A. Substituting t=0 and P=P_0 into the equation (P)/(K-P) = A e^(rt), we find A = (P_0)/(K-P_0). Substituting A back and solving for P(t) gives the logistic growth function.

As t → ∞, e^(-rt) → 0, so P(t) → (K P_0)/(P_0) = K. This means the population approaches the carrying capacity K. As t → -∞, e^(-rt) → ∞, so P(t) → 0. The population starts from P_0 and grows towards K if 0 K, the population decreases towards K. If P_0 = K, the population remains constant at K. If P_0 = 0, the population remains 0.

As t → ∞, e^(-rt) → 0, so P(t) → (K P_0)/(P_0) = K. This means the population approaches the carrying capacity K. As t → -∞, e^(-rt) → ∞, so P(t) → 0. The population starts from P_0 and grows towards K if 0 K, the population decreases towards K. If P_0 = K, the population remains constant at K. If P_0 = 0, the population remains 0.

A population grows according to the logistic equation dP/dt = rP(1-P/K). If P(0) = P₀, find P(t) and analyze its behavior.

Answer: $P(t) = \frac{K P_0}{P_0 + (K-P_0)e^{-rt}}$ The population $P(t)$ starts at $P_0$ and approaches the carrying capacity $K$ as $t \to \infty$ .

Step-by-step solution

Step 1: Separate variables

The given logistic differential equation is $\frac{dP}{dt} = rP\left(1-\frac{P}{K}\right)$ . To solve this, we first separate the variables $P$ and $t$ to prepare for integration. We move all terms involving $P$ to one side and all terms involving $t$ to the other side.

Step 2: Integrate both sides using partial fractions

Before integrating, we can simplify the left side by multiplying the numerator and denominator by $K$ . Then, we use partial fraction decomposition for the integral on the left side. The integral of the right side is straightforward.

Step 3: Perform partial fraction decomposition

We decompose the fraction $\frac{K}{P(K-P)}$ into partial fractions $\frac{A}{P} + \frac{B}{K-P}$ . By equating the numerators, $K = A(K-P) + BP$ . Setting $P=0$ gives $K=AK$ , so $A=1$ . Setting $P=K$ gives $K=BK$ , so $B=1$ .

Step 4: Integrate and solve for P

Now we integrate the decomposed fractions. The integral of $\frac{1}{P}$ is $\ln|P|$ and the integral of $\frac{1}{K-P}$ is $-\ln|K-P|$ . Combining the logarithms and exponentiating both sides, we get $\frac{P}{K-P} = A e^{rt}$ , where $A = e^C$ .

Step 5: Apply initial condition and solve for P(t)

We use the initial condition $P(0) = P_0$ to find the constant $A$ . Substituting $t=0$ and $P=P_0$ into the equation $\frac{P}{K-P} = A e^{rt}$ , we find $A = \frac{P_0}{K-P_0}$ . Substituting $A$ back and solving for $P(t)$ gives the logistic growth function.

Step 6: Analyze the behavior of P(t)

As $t \to \infty$ , $e^{-rt} \to 0$ , so $P(t) \to \frac{K P_0}{P_0} = K$ . This means the population approaches the carrying capacity $K$ . As $t \to -\infty$ , $e^{-rt} \to \infty$ , so $P(t) \to 0$ . The population starts from $P_0$ and grows towards $K$ if $0 < P_0 < K$ . If $P_0 > K$ , the population decreases towards $K$ . If $P_0 = K$ , the population remains constant at $K$ . If $P_0 = 0$ , the population remains 0.

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