91Ó°ÊÓ

We are given the initial-value problem: \[P'=0.02P\left(1-\frac{P}{1000}\right), \quad P(0)=1000\]

To find the equilibrium points, we need to set \(P'=0\) and solve for \(P\): \begin{align*} 0 &= 0.02P\left(1-\frac{P}{1000}\right) \\ 0 &= P\left(1-\frac{P}{1000}\right) \end{align*} This equation is satisfied when \(P=0\) or \(P=1000\). These are our two equilibrium points.

The given differential equation is separable, so we can rearrange the terms and integrate both sides: \[\frac{dP}{P\left(1-\frac{P}{1000}\right)}=0.02\,dt\] Now, integrate both sides of the equation: \[\int\frac{dP}{P\left(1-\frac{P}{1000}\right)} = \int0.02\,dt\] Use the partial fraction decomposition for the left side of the equation: \[\frac{A}{P}+\frac{B}{1-\frac{P}{1000}} = \frac{A(1-\frac{P}{1000})+BP}{P\left(1-\frac{P}{1000}\right)}\] Comparing the numerators, we see that \(A=1\) and \(B=-1\). The left side now becomes: \[\frac{1}{P}-\frac{1}{1-\frac{P}{1000}}\] Integrate both sides: \[\int \left(\frac{1}{P}-\frac{1}{1-\frac{P}{1000}}\right) dP = \int 0.02\,dt\] \[\ln\left|P\right|-\ln\left|1-\frac{P}{1000}\right| =0.02t+C\] Applying exponentials to both sides the equation becomes: \[\frac{P}{1-\frac{P}{1000}}=k e^{0.02t}\] Where \(k=e^C\). Now, use the initial condition \(P(0) = 1000\) to find the value of \(k\): \[\frac{1000}{1-\frac{1000}{1000}}=ke^{0.02(0)} \Rightarrow \frac{1000}{0}=k\] This situation implies that the solution is not possible. This means that the initial condition lies exactly on the equilibrium point \(P=1000\). Therefore, the solution remains at the equilibrium point.

Since we found that the solution remains at the equilibrium point, we can conclude that: \[\lim_{t \to \infty} P(t) = 1000\] The given statement is true, and the explanation is that the initial condition is at the equilibrium point.