Differential Equations Quiz: Logistic Equation Solutions

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.

Question 1

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

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 2

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

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 3

1. $\frac{2000}{1+e^{0.5}}$ (correct answer)
2. $\frac{2000}{1+e^{-0.5}}$
3. $1000e^{-0.5}$
4. $\frac{2000}{1+e^{5}}$

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 $P_0 = P(0)$ as $A = \frac{M-P_0}{P_0}$. 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 4

1. $H = 5000$
2. $H = 4000$
3. $H = 2000$ (correct answer)
4. $H = 10000$

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 5

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

Explanation: The time $t$ to grow from $P_0$ to $P$ 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 $T_{new}$ is $T_{new} = \frac{1}{k_{new}}\ln(9) = \frac{1}{2k}\ln(9)$. 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 6

1. $800$
2. $200$
3. $400$
4. $100$ (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 7

1. $\frac{1}{10}\ln(2)$
2. $\frac{1}{10}\ln(4/9)$
3. $-\frac{1}{10}\ln(4)$
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 8

1. If $0 < y_0 < 50$, $y(t) \to 0$. If $y_0 > 50$, $y(t)$ grows without bound. (correct answer)
2. The population always approaches the carrying capacity of 50, regardless of $y_0 > 0$.
3. If $0 < y_0 < 50$, $y(t) \to 50$. If $y_0 > 50$, $y(t) \to 0$.
4. 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 9

1. $T_A = 2T_B$ (correct answer)
2. $T_B = 2T_A$
3. $T_A = T_B$
4. $T_A = 0.8 T_B$

Explanation: The time $t$ required for a logistic population to grow from $P_0$ to $P(t)$ 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 10

1. $100$
2. $50$ (correct answer)
3. $2$
4. $0.02$

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 11

1. $P=0$ is unstable; $P=50$ and $P=200$ are stable.
2. $P=50$ is stable; $P=0$ and $P=200$ are unstable.
3. $P=200$ is the only stable equilibrium point.
4. $P=0$ 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 12

1. 350
2. 400
3. 375 (correct answer)
4. 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 13

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

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

Question 14

1. 50 (correct answer)
2. 100
3. 75
4. 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 15

1. 900 (correct answer)
2. 950
3. 850
4. 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 $N = 900$ or $N = 100$. The larger equilibrium population is 900.