Applications of Differential Equations

The population growth can be modeled by the differential equation $\frac{dP}{dt} = kP$, where $P$ is the population size and $k$ is a constant. Solving this equation with the given initial condition gives $P(t) = 100e^{kt}$. Since the population doubles in 10 years, we have $200 = 100e^{10k}$, which implies $k = \frac{\ln 2}{10}$. Therefore, the population after 20 years is $P(20) = 100e^{20k} = 100e^{2\ln 2} = 1600$ rabbits.

The radioactive decay can be modeled by the differential equation $\frac{dQ}{dt} = -kQ$, where $Q$ is the amount of the substance and $k$ is a constant. Solving this equation with the given half-life gives $Q(t) = Q_0e^{-kt}$, where $Q_0$ is the initial amount. Since the half-life is 10 years, we have $\frac{Q_0}{2} = Q_0e^{-10k}$, which implies $k = \frac{\ln 2}{10}$. Therefore, the amount remaining after 20 years is $Q(20) = Q_0e^{-20k} = Q_0e^{-2\ln 2} = \frac{Q_0}{4} = 25%$ of the original amount.

The general solution to the differential equation is $x(t) = A\cos(\omega t + \phi)$, where $A$ is the amplitude, $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency, and $\phi$ is the phase angle. Using the initial conditions, we have $x(0) = A\cos\phi = 1$ and $\frac{dx}{dt}(0) = -A\omega\sin\phi = 0$. This implies that $\phi = 0$ and $A = 1$. Therefore, the amplitude of the motion is 1 meter.

Let $Q(t)$ be the amount of salt in the tank at time $t$. The rate of change of $Q(t)$ is given by $\frac{dQ}{dt} = (0.5\cdot 10) - \frac{Q}{100\cdot 10}$, which simplifies to $\frac{dQ}{dt} = 5 - \frac{Q}{1000}$. Solving this equation with the initial condition $Q(0) = 0$ gives $Q(t) = 50(1 - e^{-t/200})$. Therefore, the amount of salt in the tank after 10 minutes is $Q(10) = 50(1 - e^{-1/20}) \approx 50$ pounds.

The equilibrium point is the point where $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$. Setting $\frac{dx}{dt} = 0$ and solving for $x$ gives $x = \frac{K}{1 + \alpha y}$. Setting $\frac{dy}{dt} = 0$ and solving for $y$ gives $y = \frac{N}{1 + \beta x}$. Substituting the expression for $x$ into the expression for $y$, we get $y = \frac{N}{1 + \beta\frac{K}{1 + \alpha y}}$. Solving this equation for $y$ gives $y = 375$. Substituting this value of $y$ into the expression for $x$, we get $x = 750$. Therefore, the equilibrium point is (750, 375).