Logistic Equation Solutions - Differential Equations

1. 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 $k$?
   Answer: $\frac{1}{10}\ln(9/4)$ 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 $k$: $k = -\frac{1}{10}\ln(4/9) = \frac{1}{10}\ln((4/9)^{-1}) = \frac{1}{10}\ln(9/4)$.
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?
   Answer: $t = 2.5 \ln(7)$ 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, $t = \frac{\ln(7)}{0.4} = 2.5 \ln(7)$.
3. 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?
   Answer: $100$ 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.
4. 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 $P(0)$?
   Answer: $\frac{2000}{1+e^{0.5}}$ 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 $P_0 = \frac{2000}{1+e^{0.5}}$.
5. 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 $P(0)$?
   Answer: 50 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.
6. 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?
   Answer: $P=0$ and $P=200$ are stable; $P=50$ is unstable. 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.
7. 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?
   Answer: If $0 < y_0 < 50$, $y(t) \to 0$. If $y_0 > 50$, $y(t)$ grows without bound. 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.
8. 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?
   Answer: 900 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.
9. 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?
   Answer: $\frac{dP}{dt} = 0.2P - 0.00008P^2$ 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 $0.2P$: $\frac{dP}{dt} = 0.2P - \frac{0.2P^2}{2500} = 0.2P - 0.00008P^2$.
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 $2t_1$?
    Answer: 375 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 $t = 2t_1$: $P(2t_1) = \frac{500}{1 + 9e^{-0.06t_1}} = \frac{500}{1 + 9(e^{-0.03t_1})^2} = \frac{500}{1 + 9(\frac{1}{6})^2} = \frac{500}{1 + \frac{1}{4}} = 375$.
11. 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?
    Answer: $T_{new} = T/2$ 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 $T_{new} = \frac{1}{2} \left( \frac{1}{k}\ln(9) \right) = \frac{T}{2}$.
12. 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?
    Answer: $T_A = 2T_B$ 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, $T_A = 2T_B$.
13. 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.
    Answer: $\frac{\ln(243)}{r}$ 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 $t = \frac{\ln(81)}{r} = \frac{\ln(3^4)}{r} = \frac{4\ln(3)}{r} = \frac{\ln(243)}{r}$.
14. 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 $P(t)$?
    Answer: $50$ 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 $M = \frac{0.5}{0.01} = 50$.
15. 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?
    Answer: $H = 2000$ 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.

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