An item is initially sold at a price of $\$ p$ per unit. Over time, market forces push the price toward the equilibrium price, $\$ p^*$, at which supply balances demand. The Evans Price Adjustment model says that the rate of change in the market price, $\$ p$, is proportional to the difference between the market price and the equilibrium price.

(a) Write a differential equation for $p$ as a function of $t$.

Is the statement true or false? Give an explanation for your answer.

Euler’s method gives the arc length of a solution curve.

Solve the boundary value problem. \

$$p^{\prime \prime}+4 p^{\prime}+5 p=0 . \quad p(0)=1, \quad p(\pi / 2)=5$$

For the differential equation, find a solution which passes through the given point.

$$e^{-\cos \theta} \frac{d z}{d \theta}=\sqrt{1-z^2} \sin \theta, \quad z(0)=\frac{1}{2}$$

Suppose that $y=f(x)$ is a solution of the differential equation $d y / d x=g(x)$. Are the statements in the following Problems true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample.
If $g(x)$ is even, then $f(x)$ is odd.

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 1 million barrels of oil in the well; six years later 500,000 barrels remain. At what rate was the amount of oil in the well decreasing when there were 600,000 barrels remaining?

The amount of land in use for growing crops increases as the world's population increases. Suppose $A(t)$ represents the total number of hectares of land in use in year t. (A hectare is about $2 \dfrac{1}{2}$ acres.)
In $1950$ about $1 \cdot 10^9$ hectares of land were in use; in $1980$ the figure was $2 \cdot 10^{\circ}$. If the total amount of land available for growing crops is thought to be $3.2 \cdot 10^{17}$ hectares, when does this model predict it is exhausted? (Let $t=0$ in $1950$ .)

As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, a. Describe the practical meaning (in terms of the amount remembered) of the constants in the solution y=f(t).

In Figure $11.3$, the height, $y$, of the hanging cable above the horizontal line satisfies

$$\frac{d^{2}y}{dx^{2}}=k\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}.$$

(a) Show that $y=\frac{e^{x}+e^{-x}}{2}$ satisfies this differential equation if $k=1.$

Consider a conflict between two armies of x and y soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and $t$ represents time since the start of the battle, then $x$ and $y$ obey the differential equations

$$\begin{aligned} & \frac{d x}{d t}=-a y \\ & \frac{d y}{d t}=-b x \quad a, b>0 . \end{aligned}$$

Consider the battle of Iwo Jima, described in Problem 21. Take $a=0.05, b=0.01$ and assume the initial strength of the US troops to be 54,000 and that of the Japanese troops to be 21,500. (Again, ignore reinforcements.)
Assuming that the Japanese fought without surrendering until they had all been killed, as was the case, how many US troops does this model predict would be left when the battle ended?

Find the value(s) of $\omega$ for which $y=\cos\omega t$ satisfies

$$\frac{d^{2}y}{dt^{2}} + 9y=0.$$

Look at the pendulum motion in the previous problem. What effect does it have on $x$ as a function of time if:
$l$ is increased?

As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, a.

Solve the differential equation.

Use separation of variables to find the solutions to the differential equations subject to the given initial conditions.

$\frac{d y}{d x}=\frac{5 y}{x}, \quad y=3$ where $x=1$