7.1 Modeling Situations with Differential Equations

Learning objectives

—FUN-7.A—Interpret verbal statements of problems as differential equations involving a derivative expression• FUN-7.A Interpret verbal statements of problems as differential equations involving a derivative expression.• FUN-7.A.1 Differential equations relate a function of an independent variable and the function’s derivatives.• Enduring understanding FUN-7: Solving differential equations allows us to determine functions and develop models.

A differential equation relates an unknown function to one or more of its derivatives. If a quantity $P$ depends on time $t$ , then $dP/dt$ represents its instantaneous rate of change; the words describing that rate determine the right-hand side of the equation.

Verbal relationship	Differential equation
changes at the constant rate $r$	$\frac{dP}{dt}=r$
grows at a rate proportional to $P$	$\frac{dP}{dt}=kP$ , $k>0$
decreases at a rate proportional to $P$	$\frac{dP}{dt}=-kP$ , $k>0$
rate equals a stated function $g(t,P)$	$\frac{dP}{dt}=g(t,P)$

Suppose a population $P(t)$ grows at a rate proportional to its current size. “Rate” gives $dP/dt$ , “proportional to its current size” gives $kP$ , and “grows” requires a positive sign. The model is $dP/dt=kP$ with $k>0$ . If $t$ is measured in years, $dP/dt$ has units of individuals per year and $k$ has units of $1/$ year.

The differential equation describes the rate rule; it is not yet the function $P(t)$ . A statement such as $P(0)=P_0$ is an initial condition that can select one solution later. Preserve the wording carefully: “decreases by $r$ per unit time” gives $dP/dt=-r$ , whereas “decreases proportionally to $P$ ” gives $dP/dt=-kP$ .

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—7.2 Verifying Solutions for Differential EquationsRead