7.8 Exponential Models with Differential Equations

Learning objectives

—FUN-7.F—Interpret the meaning of a differential equation and its variables in context• FUN-7.F Interpret the meaning of a differential equation and its variables in context.• FUN-7.F.1 Specific applications of finding general and particular solutions to differential equations include motion along a line and exponential growth and decay.• FUN-7.F.2 The model for exponential growth and decay that arises from the statement “The rate of change of a quantity is proportional to the size of the quantity” is dy/dt = ky.• Enduring understanding FUN-7: Solving differential equations allows us to determine functions and develop models.—FUN-7.G—Determine general and particular solutions for problems involving differential equations in context• FUN-7.G Determine general and particular solutions for problems involving differential equations in context.• FUN-7.G.1 The exponential growth and decay model, dy/dt = ky, with initial condition y = y₀ when t = 0, has solutions of the form y = y₀eᵏᵗ.• Enduring understanding FUN-7: Solving differential equations allows us to determine functions and develop models.

“The rate of change is proportional to the amount present” means that the instantaneous rate $dy/dt$ equals a constant multiple of the current quantity $y$ . The resulting model is $dy/dt=ky$ , so the rate changes whenever the amount changes.

\frac{dy}{dt}=ky\qquad\text{or, when }y\ne0,\qquad\frac{1}{y}\frac{dy}{dt}=k

Symbol or sign	Contextual meaning
$t$	independent variable, usually time
$y(t)$	changing quantity
$dy/dt$	instantaneous change in $y$ per unit time
$k>0$	exponential growth
$k<0$	exponential decay
units of $k$	inverse time, so $ky$ has units of quantity/time

In the illustrative model $dy/dt=0.03y$ , if $y=200$ units at a certain time, then $dy/dt=0.03(200)=6$ units per time unit at that instant. The constant $0.03$ is the relative growth rate; it is not a constant increase of $6$ forever.

Do not confuse proportional change with constant change. $dy/dt=k$ gives the same absolute change rate at every amount, whereas $dy/dt=ky$ gives the same relative rate $(dy/dt)/y=k$ . For motion along a line, the same interpretation rule applies: if $y$ is position, $dy/dt$ is velocity, so variable meanings must come from the context.

Solving $dy/dt=ky$ by separation gives $dy/y=k\,dt$ , so $\ln|y|=kt+C$ and the general solution is $y=Ce^{kt}$ . The arbitrary constant $C$ represents the possible starting amounts.

y(0)=y_0\qquad\Longrightarrow\qquad y(t)=y_0e^{kt}

At $t=0$ , the general solution gives $y(0)=Ce^0=C$ , so the initial condition forces $C=y_0$ . Differentiating $y_0e^{kt}$ gives $ky_0e^{kt}=ky$ , verifying both the differential equation and the initial value.

For the illustrative initial-value problem $dy/dt=0.2y$ and $y(0)=50$ , the particular solution is $y=50e^{0.2t}$ . At $t=3$ , the model gives $y(3)=50e^{0.6}$ units. Because $k=0.2>0$ , the amount grows; a negative $k$ would produce decay.

The formula is mathematically defined for all real $t$ , but a context beginning at $t=0$ may restrict use to $t\ge0$ . Keep the units of $t$ consistent with those of $k$ , and do not replace the continuous model $e^{kt}$ with $(1+k)^t$ unless a different discrete model is explicitly given.

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—7.9 Logistic Models with Differential EquationsRead

7.8 Exponential Models with Differential Equations

Learning objectives

A Proportional Rate Creates an Exponential Model

Solve the Proportional-Rate Model

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