Exponential models | Differential equations

What you’ll learn:

How to derive the general solution to exponential differential equations
How the initial condition leads to the specific form $y = y_{0} e^{k t}$

In section 7.1, we saw that when a quantity changes at a rate proportional to its current size, it can be modeled by the differential equation

$\frac{d y}{d t} = k y$

To solve this, use separation of variables:

$\frac{1}{y} d y = k d t$

$\int \frac{1}{y} d y = \int k d t$

$ln ∣ y ∣ = k t + C$

$y = e^{k t + C}$

$y = e^{k t} \cdot e^{C}$

Since $e^{C}$ is just a constant, it can be replaced with the general constant term $C$ .

If the initial condition for this model is $y (0) = y_{0}$ , then

$y_{0} = C e^{0}$

$y_{0} = C$

So we can replace $C$ in the general solution with the initial value $y_{0}$ .

$y = y_{0} e^{k t}$

This equation models any situation where a quantity grows or decays exponentially, starting from an initial value/amount $y_{0}$ at time $t = 0$ .

Examples

1. A population grows at a rate proportional to its current size. If it doubles every 5 years, how many years will it take to triple?

Solution

(spoiler)

The differential equation that models this situation is

$\frac{d P}{d t} = k P$

and the general solution for the differential equation is

$P (t) = P_{0} e^{k t}$

where $P_{0}$ is the initial population and $P (t)$ is the size of the population at time $t$ (in years).

Since we know the population doubles after 5 years, $P = 2 P_{0}$ when $t = 5$ . Setting up the equation to find $k$ ,

$2 P_{0} = P_{0} e^{5 k}$

$2 = e^{5 k}$

$ln (2) = 5 k$

$k = \frac{ln ( 2 )}{5}$

Now we can substitute this into the equation to find when the population $P$ will be $3 P_{0}$ (triple the initial population.

$3 P_{0} = P_{0} e^{(l n (2) /5) t}$

$3 = e^{(l n (2) /5) t}$

$ln (3) = \frac{ln ( 2 )}{5} t$

$t = \frac{5 ln ( 3 )}{ln ( 2 )}$

$\approx 7.925$

It will take roughly 8 years for the population to triple.

2. A radioactive substance decays according to the model $A (t) = A_{0} e^{k t}$ . The half-life, or the time it takes for the substance to decay to half its original amount, is $10$ hours.

a) What is the value of $k$ ?
b) If there is initially $100$ grams of the substance, how long will it take to decay to $20$ grams?

Solutions

a) Value of $k$

(spoiler)

The half-life of $10$ hours gives us more than just 1 piece of information: at $t = 10$ , the amount $A$ is half of the original amount of $A_{0}$ , or $A = \frac{1}{2} A_{0}$ . Then

$\frac{1}{2} A_{0} = A_{0} e^{10 k}$

Solving for $k$ ,

$\frac{1}{2} = e^{10 k}$

$ln (1/2) = 10 k$

$k = \frac{ln ( 1/2 )}{10}$

b) Time to decay from $100$ to $20$ grams

(spoiler)

We’ve found $k$ to be $f r a c ln (0.5) 10$ so the exponential model is

$A (t) = A_{0} e^{l n (0.5) t /10}$

We want to find $t$ given $A_{0} = 100$ and $A (t) = 20$ , so

$20 = 100 e^{l n (0.5) t /10}$

$\frac{20}{100} = e^{l n (0.5) t /10}$

$ln (20/100) = \frac{ln ( 0.5 )}{10} t$

$t = ln (20/100) \cdot \frac{10}{ln ( 0.5 )}$

$= 23.219$

It will take a little over $23$ hours to decay from $100$ to $20$ grams.

Here is a question similar to #5 on the FRQ portion of the 2012 AB exam:

A tank is being filled with water, and the rate at which the water level rises is proportional to the difference between the tank’s maximum height and the current water level. At time $t = 0$ , the water is $10$ cm deep. $H (t)$ is the height of the water, in centimeters, at time $t$ . minutes after filling begins. The tank has a maximum height of $120$ cm. The rate of change of the water height is modeled by the differential equation

$\frac{d H}{d t} = \frac{1}{4} (120 - H)$

a) Is the water rising faster when the water level is at $30$ cm or when it is at $90$ cm?

b) Use separation of variables to write an expression for $H (t)$ , the particular solution to the differential equation with initial condition $H (0) = 10$ .

Solutions

a) $30$ vs. $90$ cm

(spoiler)

To find at which height the water is rising faster, we compare the rate $\frac{d H}{d t}$ when $H = 30$ vs. $H = 90$ .

b) Expression for $H (t)$

(spoiler)

Separate the variables:

$\frac{1}{120 - H} d H = \frac{1}{4} d t$

Integrate both sides:

$\int \frac{1}{120 - H} d H = \int \frac{1}{4} d t$

$- ln ∣120 - H ∣ = \frac{t}{4} + C$

Plug in initial condition $H = 10, t = 0$ to find $C$ :

$- ln (120 - 10) = \frac{1}{4} (0) + C$

$C = - ln (110)$

Solve for $H$ :

$- ln ∣120 - H ∣ = \frac{t}{4} - ln (110)$

$ln ∣120 - H ∣ = - \frac{t}{4} + ln (110)$

$ln ∣120 - H ∣ - ln (110) = - \frac{t}{4}$

$ln (\frac{120 - H}{110}) = - \frac{t}{4}$

$\frac{120 - H}{110} = e^{- t /4}$

$120 - H = 110 e^{- t /4}$

$H (t) = 120 - 110 e^{- t /4}$

What you’ll learn:

How to derive the general solution to exponential differential equations
How the initial condition leads to the specific form $y = y_{0} e^{k t}$

In section 7.1, we saw that when a quantity changes at a rate proportional to its current size, it can be modeled by the differential equation

$\frac{d y}{d t} = k y$

To solve this, use separation of variables:

$\frac{1}{y} d y = k d t$

$\int \frac{1}{y} d y = \int k d t$

$ln ∣ y ∣ = k t + C$

$y = e^{k t + C}$

$y = e^{k t} \cdot e^{C}$

Since $e^{C}$ is just a constant, it can be replaced with the general constant term $C$ .

If the initial condition for this model is $y (0) = y_{0}$ , then

$y_{0} = C e^{0}$

$y_{0} = C$

So we can replace $C$ in the general solution with the initial value $y_{0}$ .

$y = y_{0} e^{k t}$

This equation models any situation where a quantity grows or decays exponentially, starting from an initial value/amount $y_{0}$ at time $t = 0$ .

Examples

1. A population grows at a rate proportional to its current size. If it doubles every 5 years, how many years will it take to triple?

Solution

(spoiler)

The differential equation that models this situation is

$\frac{d P}{d t} = k P$

and the general solution for the differential equation is

$P (t) = P_{0} e^{k t}$

where $P_{0}$ is the initial population and $P (t)$ is the size of the population at time $t$ (in years).

Since we know the population doubles after 5 years, $P = 2 P_{0}$ when $t = 5$ . Setting up the equation to find $k$ ,

$2 P_{0} = P_{0} e^{5 k}$

$2 = e^{5 k}$

$ln (2) = 5 k$

$k = \frac{ln ( 2 )}{5}$

Now we can substitute this into the equation to find when the population $P$ will be $3 P_{0}$ (triple the initial population.

$3 P_{0} = P_{0} e^{(l n (2) /5) t}$

$3 = e^{(l n (2) /5) t}$

$ln (3) = \frac{ln ( 2 )}{5} t$

$t = \frac{5 ln ( 3 )}{ln ( 2 )}$

$\approx 7.925$

It will take roughly 8 years for the population to triple.

2. A radioactive substance decays according to the model $A (t) = A_{0} e^{k t}$ . The half-life, or the time it takes for the substance to decay to half its original amount, is $10$ hours.

a) What is the value of $k$ ?
b) If there is initially $100$ grams of the substance, how long will it take to decay to $20$ grams?

Solutions

a) Value of $k$

(spoiler)

The half-life of $10$ hours gives us more than just 1 piece of information: at $t = 10$ , the amount $A$ is half of the original amount of $A_{0}$ , or $A = \frac{1}{2} A_{0}$ . Then

$\frac{1}{2} A_{0} = A_{0} e^{10 k}$

Solving for $k$ ,

$\frac{1}{2} = e^{10 k}$

$ln (1/2) = 10 k$

$k = \frac{ln ( 1/2 )}{10}$

b) Time to decay from $100$ to $20$ grams

(spoiler)

We’ve found $k$ to be $f r a c ln (0.5) 10$ so the exponential model is

$A (t) = A_{0} e^{l n (0.5) t /10}$

We want to find $t$ given $A_{0} = 100$ and $A (t) = 20$ , so

$20 = 100 e^{l n (0.5) t /10}$

$\frac{20}{100} = e^{l n (0.5) t /10}$

$ln (20/100) = \frac{ln ( 0.5 )}{10} t$

$t = ln (20/100) \cdot \frac{10}{ln ( 0.5 )}$

$= 23.219$

It will take a little over $23$ hours to decay from $100$ to $20$ grams.

Here is a question similar to #5 on the FRQ portion of the 2012 AB exam:

A tank is being filled with water, and the rate at which the water level rises is proportional to the difference between the tank’s maximum height and the current water level. At time $t = 0$ , the water is $10$ cm deep. $H (t)$ is the height of the water, in centimeters, at time $t$ . minutes after filling begins. The tank has a maximum height of $120$ cm. The rate of change of the water height is modeled by the differential equation

$\frac{d H}{d t} = \frac{1}{4} (120 - H)$

a) Is the water rising faster when the water level is at $30$ cm or when it is at $90$ cm?

b) Use separation of variables to write an expression for $H (t)$ , the particular solution to the differential equation with initial condition $H (0) = 10$ .

Solutions

a) $30$ vs. $90$ cm

(spoiler)

To find at which height the water is rising faster, we compare the rate $\frac{d H}{d t}$ when $H = 30$ vs. $H = 90$ .

b) Expression for $H (t)$

(spoiler)

Separate the variables:

$\frac{1}{120 - H} d H = \frac{1}{4} d t$

Integrate both sides:

$\int \frac{1}{120 - H} d H = \int \frac{1}{4} d t$

$- ln ∣120 - H ∣ = \frac{t}{4} + C$

Plug in initial condition $H = 10, t = 0$ to find $C$ :

$- ln (120 - 10) = \frac{1}{4} (0) + C$

$C = - ln (110)$

Solve for $H$ :

$- ln ∣120 - H ∣ = \frac{t}{4} - ln (110)$

$ln ∣120 - H ∣ = - \frac{t}{4} + ln (110)$

$ln ∣120 - H ∣ - ln (110) = - \frac{t}{4}$

$ln (\frac{120 - H}{110}) = - \frac{t}{4}$

$\frac{120 - H}{110} = e^{- t /4}$

$120 - H = 110 e^{- t /4}$

$H (t) = 120 - 110 e^{- t /4}$