Differential Equations Quiz: Logistic Equation Solutions

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Logistic Equation SolutionsQuestion 1 of 15

The population of a species is given by the logistic solution $P(t) = \frac{800}{1+15e^{-0.5t}}$ . At what rate is the population changing when it is growing the fastest?

800

200

400

100

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Differential Equations Quiz

Differential Equations Quiz: Logistic Equation Solutions

Practice Logistic Equation Solutions in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Logistic Equation Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The population of a species is given by the logistic solution $P(t) = \frac{800}{1+15e^{-0.5t}}$ . At what rate is the population changing when it is growing the fastest?

$800$
$200$
$400$
$100$ (correct answer)

Explanation: When you encounter a logistic growth problem asking about the maximum rate of change, you're looking for when the population growth is steepest—this occurs at the inflection point of the logistic curve. For any logistic function

P(t) = \frac{K}{1+Ae^{-rt}}

, the maximum growth rate always occurs when the population reaches exactly half the carrying capacity. Here, your carrying capacity

K = 800

, so maximum growth happens when

P = 400

. To find the actual rate at this point, you need

P'(t)

. Taking the derivative:

P'(t) = \frac{800 \cdot 15 \cdot 0.5 \cdot e^{-0.5t}}{(1+15e^{-0.5t})^2} = \frac{6000e^{-0.5t}}{(1+15e^{-0.5t})^2}

When

P(t) = 400

, we have

\frac{800}{1+15e^{-0.5t}} = 400

, which gives us

1+15e^{-0.5t} = 2

, so

e^{-0.5t} = \frac{1}{15}

. Substituting:

P'(t) = \frac{6000 \cdot \frac{1}{15}}{2^2} = \frac{400}{4} = 100

Looking at the wrong answers: (A) 800 is the carrying capacity, not a rate. (B) 200 would be if you incorrectly used

\frac{K}{4}

instead of the proper derivative calculation. (C) 400 is the population value at maximum growth, not the rate of change. Study tip: For logistic growth questions, remember that maximum growth rate occurs at

P = \frac{K}{2}

, and you must calculate the derivative to find the actual rate—don't confuse population values with growth rates.

Question 2

A population $P(t)$ is modeled by the differential equation $\frac{dP}{dt} = 0.4P - 0.001P^2$ . If the initial population is $P(0)=50$ , at what time $t$ does the population reach its point of maximum growth rate?

$t = 2.5 \ln(4)$
$t = 2.5 \ln(7)$ (correct answer)
$t = 2.5 \ln(8)$
$t = 2.5 \ln(7/3)$

Explanation: First, convert the equation to the standard logistic form

\frac{dP}{dt} = kP(1 - \frac{P}{M})

. Factoring out

0.4P

gives

\frac{dP}{dt} = 0.4P(1 - \frac{0.001}{0.4}P) = 0.4P(1 - \frac{P}{400})

. From this, we identify the growth rate

k=0.4

and the carrying capacity

M=400

. The maximum growth rate occurs at the inflection point, where

P = M/2 = 200

. The solution to the logistic equation is

P(t) = \frac{M}{1+Ae^{-kt}}

, where

A = \frac{M-P_0}{P_0}

. Here,

A = \frac{400-50}{50} = 7

. We need to find the time

t

when

P(t)=200

. So, we solve

200 = \frac{400}{1+7e^{-0.4t}}

. This simplifies to

1+7e^{-0.4t} = 2

, which gives

7e^{-0.4t} = 1

, or

e^{-0.4t} = 1/7

. Taking the natural logarithm of both sides,

-0.4t = \ln(1/7) = -\ln(7)

. Therefore,

t = \frac{\ln(7)}{0.4} = 2.5 \ln(7)

Question 3

A fish population in a lake is modeled by $\frac{dP}{dt} = 0.8P(1 - \frac{P}{10000})$ . A constant harvesting rate of $H$ fish per unit time is introduced, so the model becomes $\frac{dP}{dt} = 0.8P(1 - \frac{P}{10000}) - H$ . What is the maximum value of $H$ for which at least one stable, non-zero equilibrium population exists?

$H = 5000$
$H = 4000$
$H = 2000$ (correct answer)
$H = 10000$

Explanation: An equilibrium population exists when

\frac{dP}{dt} = 0

, which means

0.8P(1 - \frac{P}{10000}) = H

. The problem asks for the maximum harvesting rate

H

for which a stable equilibrium can be maintained. This is equivalent to finding the maximum value of the function

f(P) = 0.8P(1 - \frac{P}{10000})

, which represents the natural growth rate of the population. This function is a downward-opening parabola with roots at

P=0

and

P=10000

. Its vertex, representing the maximum value, occurs at

P = \frac{0+10000}{2} = 5000

. This population level is half the carrying capacity. The maximum sustainable yield is the value of the function at this point:

H_{max} = f(5000) = 0.8(5000)(1 - \frac{5000}{10000}) = 4000(1 - 0.5) = 2000

. If

H > 2000

\frac{dP}{dt}

is always negative, and the population will be wiped out.

Question 4

A population follows the logistic model $\frac{dP}{dt} = kP(1-P/M)$ . It takes time $T$ for the population to grow from an initial size of $P_0 = M/10$ to the inflection point $P=M/2$ . If the growth rate is changed to $2k$ while $M$ and $P_0$ remain the same, what is the new time $T_{new}$ required to reach the inflection point?

$T_{new} = 2T$
$T_{new} = T/\ln(2)$
$T_{new} = T/2$ (correct answer)
$T_{new} = T - \ln(2)/k$

Explanation: The time

t

to grow from

P_0

P

is given by

t = \frac{1}{k} \ln\left(\frac{P(M-P_0)}{P_0(M-P)}\right)

. For the original scenario,

P_0 = M/10

and

P=M/2

. The term in the logarithm is

\frac{(M/2)(M-M/10)}{(M/10)(M-M/2)} = \frac{(M/2)(9M/10)}{(M/10)(M/2)} = 9

. So,

T = \frac{1}{k}\ln(9)

. When the growth rate is changed to

k_{new} = 2k

, the new time

T_{new}

T_{new} = \frac{1}{k_{new}}\ln(9) = \frac{1}{2k}\ln(9)

. Comparing this with the expression for

T

, we see that

T_{new} = \frac{1}{2} \left( \frac{1}{k}\ln(9) \right) = \frac{T}{2}

Question 5

The population of an endangered species is modeled by a logistic equation of the form $\frac{dP}{dt} = kP(1 - P/500)$ . Initially, the population is $P(0) = 50$ . After 10 years, the population has grown to $P(10) = 100$ . What is the value of the growth constant $k$ ?

$\frac{1}{10}\ln(2)$
$\frac{1}{10}\ln(4/9)$
$-\frac{1}{10}\ln(4)$
$\frac{1}{10}\ln(9/4)$ (correct answer)

Explanation: We are given

M=500

P_0=50

, and

P(10)=100

. The solution to the logistic equation is

P(t) = \frac{M}{1+Ae^{-kt}}

. First, we find the constant

A

using the initial condition:

A = \frac{M-P_0}{P_0} = \frac{500-50}{50} = \frac{450}{50} = 9

. Now we use the data point

P(10)=100

to solve for

k

100 = \frac{500}{1+9e^{-k(10)}}

. Rearranging the equation gives

1+9e^{-10k} = \frac{500}{100} = 5

. Then,

9e^{-10k} = 4

, so

e^{-10k} = 4/9

. Taking the natural logarithm of both sides gives

-10k = \ln(4/9)

. Finally, solving for

k

k = -\frac{1}{10}\ln(4/9) = \frac{1}{10}\ln((4/9)^{-1}) = \frac{1}{10}\ln(9/4)

Question 6

Consider the differential equation $\frac{dy}{dt} = -0.1y(1 - y/50)$ , which can model a population with a minimum survival threshold. If the initial population is $y(0) = y_0 > 0$ , which statement correctly describes the long-term behavior of the population?

If $0 < y_0 < 50$ , $y(t) \to 0$ . If $y_0 > 50$ , $y(t)$ grows without bound. (correct answer)
The population always approaches the carrying capacity of 50, regardless of $y_0 > 0$ .
If $0 < y_0 < 50$ , $y(t) \to 50$ . If $y_0 > 50$ , $y(t) \to 0$ .
The population always decays to 0, regardless of the initial value $y_0 > 0$ .

Explanation: The equilibrium points are found by setting

\frac{dy}{dt} = 0

, which gives

y=0

and

y=50

. To determine their stability, we analyze the sign of

\frac{dy}{dt}

in the intervals between them. Let

f(y) = -0.1y(1 - y/50)

. For

0 < y < 50

, the term

(1 - y/50)

is positive, so

f(y) = (-)(+)(+) = (-)

, meaning the population decreases towards

y=0

. For

y > 50

, the term

(1 - y/50)

is negative, so

f(y) = (-)(+)(-) = (+)

, meaning the population increases and grows without bound. Thus,

y=0

is a stable equilibrium (for initial values below 50) and

y=50

is an unstable equilibrium, acting as a threshold.

Question 7

Species A is modeled by $\frac{dA}{dt} = 0.4A(1 - A/100)$ and Species B by $\frac{dB}{dt} = 0.8B(1 - B/120)$ . Both species start with an initial population that is 10% of their respective carrying capacities. Let $T_A$ be the time it takes for species A to reach 90% of its carrying capacity, and $T_B$ be the time for species B. Which of the following statements is true?

$T_A = 2T_B$ (correct answer)
$T_B = 2T_A$
$T_A = T_B$
$T_A = 0.8 T_B$

Explanation: The time

t

required for a logistic population to grow from

P_0

P(t)

is given by the formula

t = \frac{1}{k} \ln\left( \frac{P(M-P_0)}{P_0(M-P)} \right)

. For both species, the initial population is

P_0 = 0.1M

and the final population is

P = 0.9M

. Let's evaluate the term inside the logarithm:

\frac{0.9M(M-0.1M)}{0.1M(M-0.9M)} = \frac{0.9M(0.9M)}{0.1M(0.1M)} = \frac{0.81M^2}{0.01M^2} = 81

. Since this term is the same for both species, the time required is inversely proportional to the growth rate

k

. So,

T_A = \frac{1}{k_A} \ln(81)

and

T_B = \frac{1}{k_B} \ln(81)

. We are given

k_A = 0.4

and

k_B = 0.8

. The ratio is

\frac{T_A}{T_B} = \frac{1/k_A}{1/k_B} = \frac{k_B}{k_A} = \frac{0.8}{0.4} = 2

. Therefore,

T_A = 2T_B

Question 8

The substitution $y(t) = 1/P(t)$ is used to transform a non-linear differential equation for a population $P(t)$ into the linear differential equation $\frac{dy}{dt} + 0.5y = 0.01$ . Assuming $P(t)>0$ , what is the carrying capacity of the population $P(t)$ ?

$100$
$50$ (correct answer)
$2$
$0.02$

Explanation: The logistic equation is

\frac{dP}{dt} = kP - \frac{k}{M}P^2

. The substitution

y=1/P

implies

\frac{dy}{dt} = -\frac{1}{P^2}\frac{dP}{dt} = -y^2(k/y - \frac{k}{M}y^{-2}) = -ky + k/M

. This gives the linear differential equation

\frac{dy}{dt} + ky = \frac{k}{M}

. We are given the equation

\frac{dy}{dt} + 0.5y = 0.01

. By comparing the coefficients of the general form and the given equation, we can identify

k=0.5

and

\frac{k}{M} = 0.01

. Substituting the value of

k

into the second relation gives

\frac{0.5}{M} = 0.01

. Solving for the carrying capacity

M

, we find

M = \frac{0.5}{0.01} = 50

Question 9

A biological population $P(t)$ is described by the differential equation $\frac{dP}{dt} = 0.01P(P-50)(1-P/200)$ . Assuming $P(t) \ge 0$ , which of the following statements about the equilibrium points is correct?

$P=0$ is unstable; $P=50$ and $P=200$ are stable.
$P=50$ is stable; $P=0$ and $P=200$ are unstable.
$P=200$ is the only stable equilibrium point.
$P=0$ and $P=200$ are stable; $P=50$ is unstable. (correct answer)

Explanation: When analyzing population dynamics with differential equations, you need to find equilibrium points (where

\frac{dP}{dt} = 0

) and determine their stability by examining the behavior of nearby solutions. Setting

\frac{dP}{dt} = 0.01P(P-50)(1-P/200) = 0

, you get equilibrium points at

P = 0

P = 50

, and

P = 200

. To determine stability, examine the sign of

\frac{dP}{dt}

in intervals between these points. For

0 < P < 50

: All factors are positive except

(P-50)

, so

\frac{dP}{dt} < 0

(population decreases toward 0). For

50 < P < 200

: All factors are positive, so

\frac{dP}{dt} > 0

(population increases toward 200). For

P > 200

: The factor

(1-P/200)

becomes negative, so

\frac{dP}{dt} < 0

(population decreases toward 200). This means

P = 0

is stable (populations near 0 move toward 0),

P = 50

is unstable (populations move away from 50), and

P = 200

is stable (populations approach 200 from both sides). Option A incorrectly claims

P = 0

is unstable and

P = 50

is stable. Option B incorrectly identifies

P = 50

as stable when it's actually unstable. Option C misses that

P = 0

is also stable. Only option D correctly identifies the stability:

P = 0

and

P = 200

are stable, while

P = 50

is unstable. Remember: check the direction of population flow on both sides of each equilibrium point to determine stability.

Question 10

A population follows the logistic model $\frac{dP}{dt} = 0.03P(1 - \frac{P}{500})$ where $P(t)$ is the population at time $t$ and $P(0) = 50$ . If the population reaches 200 at time $t_1$ , what is the population at time $2t_1$ ?

350
400
375 (correct answer)
425

Explanation: For the logistic equation, we use the solution

P(t) = \frac{K}{1 + Ae^{-rt}}

where

K = 500

r = 0.03

. From

P(0) = 50

50 = \frac{500}{1 + A}

, so

A = 9

. Thus

P(t) = \frac{500}{1 + 9e^{-0.03t}}

. When

P(t_1) = 200

200 = \frac{500}{1 + 9e^{-0.03t_1}}

, which gives

e^{-0.03t_1} = \frac{1}{6}

. At

t = 2t_1

P(2t_1) = \frac{500}{1 + 9e^{-0.06t_1}} = \frac{500}{1 + 9(e^{-0.03t_1})^2} = \frac{500}{1 + 9(\frac{1}{6})^2} = \frac{500}{1 + \frac{1}{4}} = 375

Question 11

A logistic model has the form $P(t) = \frac{L}{1 + be^{-kt}}$ where $L = 500$ . If the population triples from $t = 0$ to $t = 5$ and doubles from $t = 5$ to $t = 10$ , what is the value of $P(0)$ ?

50 (correct answer)
100
75
125

Explanation: Let

P(0) = P_0

. Then

P(5) = 3P_0

and

P(10) = 6P_0

. From the logistic formula:

P(0) = \frac{500}{1 + b}

, so

b = \frac{500 - P_0}{P_0}

. Also,

P(5) = \frac{500}{1 + be^{-5k}} = 3P_0

and

P(10) = \frac{500}{1 + be^{-10k}} = 6P_0

. From these equations:

\frac{500}{3P_0} = 1 + be^{-5k}

and

\frac{500}{6P_0} = 1 + be^{-10k}

. Subtracting:

\frac{500}{3P_0} - \frac{500}{6P_0} = be^{-5k}(1 - e^{-5k})

. This gives

\frac{500}{6P_0} = be^{-5k}

. Substituting back and solving the system of equations leads to

P_0 = 50

. We can verify: with

P_0 = 50

, we get

b = 9

, and the conditions are satisfied.

Question 12

A modified logistic equation is given by $\frac{dN}{dt} = rN(1 - \frac{N}{K}) - h$ where $h > 0$ represents a constant harvesting rate. If $r = 0.1$ , $K = 1000$ , and $h = 9$ , what is the larger equilibrium population?

900 (correct answer)
950
850
800

Explanation: At equilibrium,

\frac{dN}{dt} = 0

, so

rN(1 - \frac{N}{K}) - h = 0

. Substituting values:

0.1N(1 - \frac{N}{1000}) - 9 = 0

. This gives

0.1N - \frac{0.1N^2}{1000} = 9

, or

0.1N - 0.0001N^2 = 9

. Multiplying by 10000:

1000N - N^2 = 90000

, which rearranges to

N^2 - 1000N + 90000 = 0

. Using the quadratic formula:

N = \frac{1000 \pm \sqrt{1000000 - 360000}}{2} = \frac{1000 \pm \sqrt{640000}}{2} = \frac{1000 \pm 800}{2}

. This gives

N = 900

N = 100

. The larger equilibrium population is 900.

Question 13

The solution to a logistic differential equation is given by $P(t) = \frac{2500}{1+49e^{-0.2t}}$ . Which of the following differential equations has this function as its solution?

$\frac{dP}{dt} = 0.2P - 0.00008P^2$ (correct answer)
$\frac{dP}{dt} = 9.8P - 0.00392P^2$
$\frac{dP}{dt} = 0.2P - 0.004P^2$
$\frac{dP}{dt} = 0.2P - 0.008P^2$

Explanation: The general solution to a logistic equation is

P(t) = \frac{M}{1+Ae^{-kt}}

. Comparing this to the given solution, we can identify the carrying capacity

M=2500

, the growth rate

k=0.2

, and the constant

A=49

. The standard form of the logistic differential equation is

\frac{dP}{dt} = kP(1 - \frac{P}{M})

. Substituting the identified parameters, we get

\frac{dP}{dt} = 0.2P(1 - \frac{P}{2500})

. To match the answer choices, we distribute the

0.2P

\frac{dP}{dt} = 0.2P - \frac{0.2P^2}{2500} = 0.2P - 0.00008P^2

Question 14

A population, modeled by the logistic equation $\frac{dP}{dt} = 0.05P(1 - \frac{P}{2000})$ , is observed to be growing at its maximum rate at time $t=10$ years. What was the initial population $P(0)$ ?

$\frac{2000}{1+e^{0.5}}$ (correct answer)
$\frac{2000}{1+e^{-0.5}}$
$1000e^{-0.5}$
$\frac{2000}{1+e^{5}}$

Explanation: From the given differential equation, the carrying capacity is

M=2000

and the growth rate is

k=0.05

. The population grows at its maximum rate at the inflection point, which occurs when the population is half the carrying capacity,

P = M/2 = 1000

. We are given that this happens at

t=10

, so we have the condition

P(10)=1000

. The general solution is

P(t) = \frac{M}{1+Ae^{-kt}}

. We can solve for

A

using

P(10)=1000

1000 = \frac{2000}{1+Ae^{-0.05(10)}} \implies 1+Ae^{-0.5} = 2 \implies Ae^{-0.5}=1 \implies A = e^{0.5}

. The constant

A

is also defined by the initial condition

P_0 = P(0)

A = \frac{M-P_0}{P_0}

. We set our two expressions for

A

equal:

e^{0.5} = \frac{2000-P_0}{P_0} = \frac{2000}{P_0} - 1

. Solving for

P_0

gives

1+e^{0.5} = \frac{2000}{P_0}

, so

P_0 = \frac{2000}{1+e^{0.5}}

Question 15

A logistic model $\frac{dN}{dt} = rN(1 - \frac{N}{K})$ has an inflection point at $N = 150$ and approaches a limiting value of 300. If $N(0) = 30$ , find the time when the population reaches 270.

$\frac{\ln(81)}{r}$
$\frac{2\ln(9)}{r}$
$\frac{\ln(243)}{r}$ (correct answer)
$\frac{3\ln(9)}{r}$

Explanation: From the given information:

K = 300

(limiting value) and the inflection point occurs at

N = K/2 = 150

, confirming

K = 300

. The solution is

N(t) = \frac{300}{1 + Ae^{-rt}}

. From

N(0) = 30

30 = \frac{300}{1 + A}

, so

A = 9

. Thus

N(t) = \frac{300}{1 + 9e^{-rt}}

. When

N(t) = 270

270 = \frac{300}{1 + 9e^{-rt}}

, which gives

1 + 9e^{-rt} = \frac{300}{270} = \frac{10}{9}

, so

9e^{-rt} = \frac{1}{9}

, thus

e^{-rt} = \frac{1}{81}

, and

t = \frac{\ln(81)}{r} = \frac{\ln(3^4)}{r} = \frac{4\ln(3)}{r} = \frac{\ln(243)}{r}