Suppose that $P$ dollars is invested at an annual interest rate of $r \times 100 \%$. If the accumulated interest is credited to the account at the end of the year, then the interest is said to be compounded annually; if it is credited at the end of each 6-month period, then it is said to be compounded semiannually; and if it is credited at the end of each 3-month period, then it is said to be compounded quarterly. The more frequently the interest is compounded, the better it is for the investor since more of the interest is itself earning interest.

One can imagine interest to be compounded each day, each hour, each minute, and so forth. Carried to the limit one can conceive of interest compounded at each instant of time; this is called continuous compounding. Thus, from part (a), the value $A$ of $P$ dollars after $t$ years when invested at an annual rate of $r \times 100 \%$, compounded continuously, is

$$A=\lim _{n \rightarrow+\infty} P\left(1+\frac{r}{n}\right)^{n t}$$

Use the fact that $\lim _{x \rightarrow 0}(1+x)^{1 / x}=e$ to prove that $A=P e^{r t}$.

Use Euler's Method with a step size of $\Delta t=0.1$ to approximate the solution of the initial-value problem

$$y^{\prime}=1+5 t-y, \quad y(1)=5$$

over the interval $[1,2]$.

The equation

$$x^2 \frac{d^2 y}{d x^2}+p x \frac{d y}{d x}+q y=0 \quad(x>0)$$

where $p$ and $q$ are constants, is called Euler's equidimensional equation. Show that the substitution $x=e^2$ transforms this equation into the equation

$$\frac{d^2 y}{d z^2}+(p-1) \frac{d y}{d z}+q y=0$$

In a certain 1-hour period the number of bacteria in a colony increases by $25 \%$. Assuming an exponential growth model, what is the doubling time for the colony?

In each part, find the solution of the differential equation

$$x \frac{d y}{d x}+y=x$$

that satisfies the initial condition.

$$y(-1)=2$$

In each part, find the solution of the differential equation

$$x \frac{d y}{d x}+y=x$$

that satisfies the initial condition. (a) $y(1)=2$

Show that if a quantity $y=y(t)$ has an exponential model, and if $y\left(t_1\right)=y_1$ and $y\left(t_2\right)=y_2$, then the doubling time or the half-life $T$ is

$$T=\left|\frac{\left(t_2-t_1\right) \ln 2}{\ln \left(y_2 / y_1\right)}\right|$$

Consider the initial-value problem

$$y^{\prime}=\cos 2 \pi t, \quad y(0)=1$$

Use Euler's Method with five steps to approximate $y(1)$.

Find all values of $k$ for which the differential equation $y^{\prime \prime}+k y^{\prime}+k y=0$ has a general solution of the given form. (a) $y=c_1 e^{a x}+c_2 e^{b x}$ (b) $y=c_1 e^{a x}+c_2 x e^{a x}$ (c) $y=c_1 e^{a x} \cos b x+c_2 e^{a x} \sin b x$

Suppose that a quantity $y$ has an exponential growth model $y=y_0 e^{k t}$ or an exponential decay model $y=y_0 e^{-k t}$, and it is known that $y=y_1$ if $t=t_1$. In each case find a formula for $k$ in terms of $y_0, y_1$, and $t_1$, assuming that $t_1 \neq 0$.

Show that if $e^x$ and $e^{-x}$ are solutions of a second-order linear homogeneous differential equation, then so are $\cosh x$ and $\sinh x$.

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of $x$.

$y-\frac{d y}{d x} \sec x=0$

Use Euler's Method with the given step size $\Delta x$ to approximate the solution of the initial-value problem over the stated interval. Present your answer as a table and as a graph.

$$d y / d x=\sin y, y(0)=1,0 \leq x \leq 2, \Delta x=0.5$$

Express $y=4\left(0.5^t\right)$ in the form $y=y_0 e^{-k t}$.