§7.5 Modeling with differential equations

Introduction

Developing a differential equation

To model plastic accumulating in a lake, the natural differential equation for the amount $A(t)$ takes the form

An account earns $5\%$ continuous interest, so with $A(t)$ the balance the differential equation is

Adding a continual withdrawal of \10000$per year to the$5\%$ account changes the equation to

For the account $\frac{dA}{dt} = 0.05A-10000$, the equilibrium balance equals

With $A(0) = 100000$ the account $\frac{dA}{dt} = 0.05A-10000$ depletes after approximately

With an initial deposit of \300{,}000$earning$5\%$, the largest withdrawal rate that never depletes the account is

The body absorbs morphine at a rate proportional to the amount present. With $M(t)$ the amount and $k$ positive, the differential equation is

Morphine enters intravenously at $3$ milligrams per hour while the body absorbs at rate $kM$. The combined equation is

For the drip model $\frac{dM}{dt} = 3-kM$, the equilibrium $M = \frac{3}{k}$ is

A bank account with \10{,}000$grows at a continuous annual rate of$3\%$, so$\frac{dA}{dt} = 0.03A$. Find the growth rate at that balance, in dollars per year.

Morphine absorbs with half-life $2$ hours, so $M = M_0e^{-kt}$. Find $k$, rounded to three decimals.

With $k \approx 0.34657$ per hour and intravenous rate $q$, the equilibrium morphine level is $\frac{q}{k}$. Find the rate $q$ that yields an eventual level of $7$ milligrams, rounded to three decimals.

In the account $\frac{dA}{dt} = 0.05A-10000$, any initial balance of \200{,}000$ or more never depletes.

For the intravenous model $\frac{dM}{dt} = 3-kM$ starting from $M(0) = 0$, the solution $M = \frac{3}{k}(1-e^{-kt})$ approaches $\frac{3}{k}$ as $t \to \infty$.