In this exercise, let y(t) be the solution of dy/dt=0.3y(2-y) with y(0)=1. Determine $\lim \limits_{t \to \infty}$ y(t) without solving for y explicitly.

Solve the Initial Value Problem.

$$\frac { d y } { d t } = t e ^ { - y },$$

y(1) = 0

Find the solution of the logistic equation

$$\frac { d y } { d t } = 0.4 y ( 4 - y )$$

satisfying y(0) = 8.

Assume that $R=110 \Omega, L=10 \mathrm{H}$, and $V(t)=e^{-t}$.

Calculate $t_m$ and $I\left(t_m\right)$, where $t_m$ is the time at which I(t) has a maximum value.

In this exercise, find the solutions of the logistic equation dy/dt=y(4-y) satisfying the initial conditions:

(a) y(0)=1

(b) y(0)=4

(c) y(0)=6

Assume that $R=110 \Omega, L=10 \mathrm{H}$, and $V(t)=e^{-t}$.

Use a computer algebra system to sketch the graph of the solution for $0 \leq t \leq 3$.

Continuing with the previous exercise, let tank 2 be another tank filled with V$_2$ gallons of water. Assume that the dye solution from tank 1 empties into tank 2 as in the previous figure, mixes instantaneously, and leaves tank 2 at the same rate R. Let c$_2$(t) be the dye concentration in tank 2 at time t.

(a) Explain why c$_2$ satisfies the differential equation

$$\frac{d c_2}{d t}=\frac{R}{V_2}\left(c_1-c_2\right)$$

(b) Use the solution to the previous exercise to solve for $c_2(t)$ if $V_1=300$, $V_2=200, R=50$, and $c_0=10$.

(c) Find the maximum concentration in tank 2.

(d) Plot the solution.

Assume that $R=110 \Omega, L=10 \mathrm{H}$, and $V(t)=e^{-t}$.

Solve Eq. $\frac{d I}{d t}+\frac{R}{L} I=\frac{1}{L} V(t)$ with initial condition I(0)=0.

Assume that V(t) = V is constant and I(0) = 0. (a) Solve for I(t). (b) Show that

$$\lim _ { t \rightarrow \infty } I ( t ) = V / R$$

and that I(t) reaches approximately 63% of its limiting value after L/R seconds. (c) How long does it take for I(t) to reach 90% of its limiting value if R = 500 ohms, L = 4 henries, and V = 20 volts?

In the earlier example, we calculated the thickness of a glacier assuming the friction at the base of the glacier is constant. In this exercise, we consider cases where the friction varies along the length of the glacier.

(a) Solve the glacier thickness differential equation, and use the equation

$$T \frac{d T}{d x}=\frac{\tau}{\rho g}$$

for T(x) with $\tau(x)=0.3 x(1000-x)$ N/m$^2$ and initial condition T(0)=0.

(b) Sketch the graph of T for $0 \leq x \leq 1000$ m.

Solve for l(t), assuming that R = 500 ohms, L = 4 henries, and V = 20 cos(80) volts.

Solve the Initial Value Problem.

$$\frac { d y } { d t } = y e ^ { - t },$$

y(0) = 1

In the earlier example, we calculated the thickness of a glacier assuming the friction at the base of the glacier is constant. In this exercise, we consider cases where the friction varies along the length of the glacier.

(a) Solve the glacier thickness differential equation, and use the equation below

$$T \frac{d T}{d x}=\frac{\tau}{\rho g}$$

for T(x) with $\tau(x)=75 x$ N/m$^2$ and initial condition T(0)=0.

(b) Sketch the graph of T for $0 \leq x \leq 1000$ m.

The trough (dimensions in centimeters) is filled with water. At time t=0 (in seconds), water begins leaking through a hole at the bottom of area 4$cm^2$ . Let y(t) be the water height at time t . Find a differential equation for y(t) and solve it to determine when the water level decreases to 60 cm .