separable differential equations using logarithmic and exponential functions

Question

The population of a certain community is known to increase at a rate proportional to the number of people present at any time. Find the population P(t) at any time. If the population has doubled in 5 years, how long will it take to triple? To quadruple?

First, solve the differential equation dP/dt=kP, where k is a constant and subject to P(0)=Po. "Po" reads "P not."

Write the equation as dP/P=k dt and integrate which gives ln(P)=kt+C1 or P=e^kt+C1.

Relabeling e^C1 as C, then yields P(t)=Ce^kt.

At t=0, it follows that Po=Ce^0=C and so P(t)=Po(e^kt).

Now at t=5, I have P(5)=2Po=Po(e^5k) or 2=e^5k; ln2=ln(e^5k), ln(2)=5k, k=ln(2)/5=0.1386.

Thus, for anytime t≥0,  P(t)=Po(e^0.1386t).

To find the time at which the population triples, solve 3Po=Po(e^0.1386t) for t: 0.1386t=ln(3), t=ln(3)/0.1386≈7.9 yrs.

To find the time at which the population quadruples (increase fourfold), solve 4Po=Po(e^0.1386t) for t: t=ln(4)/0.1386=10 yrs. Are those the correct answers?   

 

 

Answer 1

Hi Anthony!

Yes, your calculations and final answers are correct! You've done a great job setting up and solving the differential equation. There's just one small notation issue in your integration step that I'd like to clarify.

Step 1: Solve the differential equation

We start with:

dP/dt = kP

Separate the variables:

dP/P = k dt

Integrate both sides:

ln|P| = kt + C₁

Since population is positive, we can write:

ln(P) = kt + C₁

Exponentiate both sides:

P = e^(kt + C₁)

Using the properties of exponents:

P = e^(C₁) × e^(kt)

Let C = e^(C₁).

Therefore:

P(t) = Ce^(kt)

Using the initial condition P(0) = P₀:

P₀ = Ce⁰ = C

So our population growth model is:

P(t) = P₀e^(kt)

Your approach here was correct! Just remember that when exponentiating, the entire expression kt + C₁ must be in the exponent.

Step 2: Find the growth constant

We know the population doubles in 5 years.

Therefore:

P(5) = 2P₀

Substitute into the population model:

2P₀ = P₀e^(5k)

Divide both sides by P₀:

2 = e^(5k)

Take the natural logarithm:

ln(2) = 5k

Solve for k:

k = ln(2)/5

k ≈ 0.138629

Therefore, the population model is:

P(t) = P₀e^[(ln(2)/5)t]

This can also be written as:

P(t) = P₀ × 2^(t/5)

The second form is especially convenient because it shows that the population doubles every 5 years.

Step 3: Find when the population triples

We want:

P(t) = 3P₀

Substitute:

3P₀ = P₀e^(kt)

Divide by P₀:

3 = e^(kt)

Take the natural logarithm:

ln(3) = kt

Solve for t:

t = ln(3)/k

Substitute k = ln(2)/5:

t = 5ln(3)/ln(2)

t ≈ 7.9248 years

Therefore, the population triples after approximately 7.9 years.

Your answer is correct!

Step 4: Find when the population quadruples

We want:

P(t) = 4P₀

Substitute:

4P₀ = P₀e^(kt)

Divide by P₀:

4 = e^(kt)

Take the natural logarithm:

ln(4) = kt

Solve for t:

t = ln(4)/k

Substitute:

t = 5ln(4)/ln(2)

Since ln(4) = 2ln(2):

t = 5 × 2

t = 10 years

Your answer is exactly correct!

Final Answers

1. Population model: P(t) = P₀e^[(ln(2)/5)t]
2. Growth constant: k ≈ 0.138629
3. Time to triple: Approximately 7.92 years
4. Time to quadruple: Exactly 10 years

A Helpful Shortcut

Since the population doubles every 5 years, it must quadruple after two doubling periods.

After 5 years: 2P₀

After 10 years: 4P₀

This confirms our answer without needing additional calculus!

The time to triple requires logarithms because 3 is not an integer power of 2.

Overall, your reasoning and final answers are correct. The only adjustment needed is writing P = e^(kt + C₁), rather than P = e^kt + C₁, immediately after exponentiating.

Excellent work, and I hope this helps!