📝 Spread of disease differential equations (39 MCQs)

📖 Explanation: The term $(L - y)$ represents the susceptible population who have not yet contracted the disease but can become infected. The product $y(L-y)$ models the rate of new infections as encounters between infected and susceptible individuals. Option A confuses susceptibility with immunity; option C confuses this term with recovery dynamics; option D misidentifies it as the environmental carrying capacity from logistic growth models.

📖 Explanation: The maximum rate of spread occurs when $y = L/2 = 250$. Substituting: $dy/dt = 0.002 \times 250 \times (500 - 250) = 0.002 \times 250 \times 250 = 125$ students per day. Students often incorrectly substitute $y = L$ or $y = 0$, yielding zero, or use $y = 500$ giving 0, or use $L$ instead of $L/2$.

📖 Explanation: In the logistic model $dy/dt = k y (L - y)$, the equilibrium solutions are $y = 0$ and $y = L$. Since $y = 0$ is unstable and $y = L$ is stable, any positive initial population tends to $L = 100$ as $t \to \infty$. This represents the entire population becoming infected if no recovery or intervention occurs. Students might confuse this with carrying capacity models where the population stabilizes at L, but here L is the total susceptible population.

📖 Explanation: The student confuses the logistic model $dy/dt = k y (L - y)$ with the exponential growth model $dy/dt = k y$. The presence of the $(L - y)$ term represents limiting factors (such as susceptible population depletion). As $y$ approaches $L$, the growth rate $dy/dt$ approaches zero, so growth is not indefinite. This misunderstanding arises from ignoring the saturation effect of limited susceptible individuals.

📖 Explanation: The function $f(y) = y(L-y)$ is a downward parabola with vertex at $y = L/2$, where the product is maximized. At this point, the spread rate is fastest because there is a balance between many infected individuals and many susceptible individuals. Option A incorrectly states the derivative is zero (the derivative of $y$ is not zero); options C and D describe the behavior of the components, not the maximum of their product.

📖 Explanation: Using $dy/dt = 0.0001 \times 1 \times (1,000,000 - 1) \approx 0.0001 \times 1,000,000 = 100$ people per day. At the very beginning, $L - y \approx L$, so the rate is approximately $k L y_0$. Students might use $y = 900,000$ but the initial rate uses $y = 1$. Option A is the rate at $y = L/2$; C is from using $k y (L - y)$ at a later stage; D would occur at equilibrium.

📖 Explanation: The SIR model often includes a recovery term $-r y$, where $r$ is the recovery rate. The standard logistic model assumes a closed population with no recovery, so the correct model for disease with recovery and immunity is $dy/dt = k y (L - y) - r y$. Option C would imply recovery increases infection, which is unrealistic; D gives a negative spread rate.

📖 Explanation: When $y$ is small, $L - y \approx L$, so $dy/dt \approx k L y$. This is exponential growth with rate $kL = 0.01 \times 1000 = 10$ per day. Doubling time $T = \ln 2 / (kL) = 0.693 / 10 = 0.0693$ days, or about 99.8 minutes. The value 69.3 days would occur if $k = 0.01$ without multiplying by L. The 'no doubling' option confuses logistic vs exponential behavior at small y.

📖 Explanation: The logistic equation $dy/dt = y(1-y)$ has equilibrium solutions at $y = 0$ (unstable) and $y = 1$ (stable). For $0 < y < 1$, $dy/dt > 0$, so the solution increases toward $y = 1$. The slope field would show arrows pointing upward toward $y = 1$. Option A would be true for $y < 0$; C would require periodic coefficients; D would be true only if the initial condition were exactly at equilibrium (y=0 or y=1).

📖 Explanation: Using Euler's method: Step 1: $y_1 = 2 + 0.5 \times 2 \times (10-2) \times 1 = 2 + 8 = 10$. Step 2: $y_2 = 10 + 0.5 \times 10 \times (10-10) \times 1 = 10 + 0 = 10$. Wait, that gives 10. But the exact solution approaches 10; however, Euler's method with $\Delta t = 1$ gives $y_1 = 10, y_2 = 10$. The correct answer should be 10.0. But the options include 6.0, 8.0, 9.0, 4.5. Let's recalc: $y_1 = 2 + 0.5*2*8*1 = 2 + 8 = 10$. $y_2 = 10 + 0.5*10*0*1 = 10$. So 10.0 is not in options. Let's adjust question to have answer 10.0. Actually, let's make it $\Delta t = 0.5$: $y_1 = 2 + 0.5*2*8*0.5 = 6$, $y_2 = 6 + 0.5*6*4*0.5 = 6 + 6 = 12$ (but L=10, so impossible). Let's use $\Delta t = 1$ but with $y_0 = 1$: $y_1 = 1 + 0.5*1*9 = 5.5$, $y_2 = 5.5 + 0.5*5.5*4.5 = 5.5 + 12.375 = 17.875$ (exceeds L). So the method is unstable if step too large. The correct answer is 10.0, which is not an option. I'll adjust the options to include 10.0. Let's change options: A) 6.0, B) 8.0, C) 9.0, D) 10.0. Correct: D.

📖 Explanation: The standard logistic model $dy/dt = k y (L - y)$ does NOT include recovery or immunity; it simply models the spread until everyone is infected. Thus, the assumption about permanent immunity is not part of this model; that belongs to SIR-type models. Options A, B, and D are actual limitations: A ignores demographics, B ignores heterogeneity, D assumes homogeneous mixing. The model with recovery would be $dy/dt = k y (L - y) - r y$, so the absence of recovery is a limitation, not immunity.

📖 Explanation: Substitute $y = 750$, $L = 1000$, $k = 0.001$: $dy/dt = 0.001 \times 750 \times (1000 - 750) = 0.001 \times 750 \times 250 = 0.001 \times 187500 = 187.5$ individuals per time unit. Students often forget to multiply by $(L-y)$ or use $L - y$ incorrectly. Option A is the product without k; C is if k=0.0001; D is if L=100.

📖 Explanation: The constant $k$ is a proportionality constant that depends on the nature of the disease and behavior patterns. It effectively combines the contact rate and transmission probability. Option A is $L$; option B would be a recovery term; option D is the carrying capacity $L$. The misconception that $k$ is the contact rate alone is common, but it actually is the constant of proportionality in the model.

📖 Explanation: The rate is maximized when $y = L/2 = 50$. The logistic solution is $y(t) = \frac{L y_0}{y_0 + (L - y_0)e^{-kLt}} = \frac{100 \times 5}{5 + 95 e^{-50t}} = \frac{500}{5 + 95 e^{-50t}}$. Set y=50: $50 = \frac{500}{5 + 95 e^{-50t}} \rightarrow 5 + 95 e^{-50t} = 10 \rightarrow 95 e^{-50t} = 5 \rightarrow e^{-50t} = 1/19 \rightarrow t = -\ln(1/19)/50 \approx 2.944/50 \approx 0.0589$ days. So about 0.06 days, not 6 or 10. The question needs to be adjusted. Let's change k to 0.01: $kL = 0.01 \times 100 = 1$. Then $t = -\ln(1/19)/1 \approx 2.94$ days. So answer C (approximately 3 days). Let's make options: A) 0.5 days, B) 1.5 days, C) 3 days, D) 5 days. Correct: C. Explanation: Solve for time to reach y=50 using logistic growth formula.

📖 Explanation: The model assumes the rate of new infections is proportional to the number of encounters between infected and susceptible individuals, which is proportional to the product $y(L-y)$. This is a fundamental assumption of the mass-action principle in epidemiology. Option A would give exponential growth; B would give linear growth; D would give faster-than-logistic growth. This is a direct recall of the modeling assumptions.

📖 Explanation: When $y = 900$, the number of susceptible individuals is $L - y = 1000 - 900 = 100$. This is a straightforward substitution from the model definition. Option B is the infected count; C would be if everyone was infected; D is the initial population. This tests direct recall of the meaning of $L - y$ as the susceptible population.

📖 Explanation: The equilibrium solutions are $y = 0$ and $y = L$. Linearizing near $y = 0$, $dy/dt \approx kL y$, so $y$ grows exponentially (unstable). Near $y = L$, let $z = L - y$, then $dz/dt = -k L z$, so $z$ decays (stable). Thus, y=0 is unstable and y=L is stable. This is a classic result for the logistic equation. Students often confuse stability or forget to check the sign of the derivative near the equilibria.

📖 Explanation: The general solution is $y = \frac{1}{1 + C e^{-t}}$. Plugging $t=0, y=0.2$: $0.2 = \frac{1}{1 + C} \rightarrow 1 + C = 5 \rightarrow C = 4$. Option B is the negative of 4; option C is $L - y_0$; option D is $y_0$. This tests the ability to use initial conditions to find the integration constant, a common step in solving separable equations.

📖 Explanation: The logistic curve $y(t) = \frac{L}{1 + C e^{-kLt}}$ has an inflection point at $y = L/2$, where the curvature changes from concave up to concave down. This is the point of maximum growth rate, meaning the disease is spreading fastest at that time. Option A is the initial time; B describes the region after the inflection point; D is the asymptotic approach to L. Students often confuse the inflection point with the start of the epidemic.

📖 Explanation: In epidemiological modeling, $k$ typically represents the product of the contact rate and the transmission probability. This is a key parameter in the basic reproduction number $R_0$. Option B is $L$; option C is derived from the model but not $k$; option D would be a separate term in SIR models. This tests the student's ability to connect mathematical parameters to real-world meanings, which is essential for modeling.

📖 Explanation: If $y(0) = 0$, then $dy/dt = 0.5 \times 0 \times (100 - 0) = 0$. However, a disease cannot start with zero infected individuals because there would be no spread. The model is undefined or trivial at y=0. The correct initial condition must be $y(0) > 0$. This highlights a common error: assuming the model works for y=0. In reality, an initial infected population is needed for the disease to propagate. Option B is the rate at y=1; C is at y=50; D is at y=100.

📖 Explanation: As $t \to \infty$, $e^{-kLt} \to 0$, so $y(t) \to L$. This means the entire susceptible population will eventually become infected in this model, assuming no recovery or intervention. Option B is the opposite; C would require periodic forcing; D is the value at the inflection point. This is a direct consequence of the model's structure and is a key prediction of the simple logistic model for disease spread.

📖 Explanation: Euler's method: $y_1 = y_0 + f(y_0) \Delta t = 100 + 0.2 \times 100 \times (1000 - 100) \times 0.5 = 100 + 0.2 \times 100 \times 900 \times 0.5 = 100 + 9000 = 9100$. That's too large; the step size is too big. Let's reduce step size to 0.01: $y_1 = 100 + 0.2 \times 100 \times 900 \times 0.01 = 100 + 180 = 280$. Not in options. Let's use $\Delta t = 0.001$: $y_1 = 100 + 0.2 \times 100 \times 900 \times 0.001 = 100 + 18 = 118$. The exact value at t=1 is $y = \frac{1000 \times 100}{100 + 900 e^{-200}} \approx 1000/(1 + 9 e^{-200}) \approx 1000$. Actually, with k=0.2, L=1000, kL=200, after t=1, e^{-200} is essentially 0, so y ≈ 1000. So Euler with small step will approach 1000. The question needs to be realistic. Let's choose k=0.001, L=1000, so kL=1. Then exact at t=1: $y = 1000/(1 + 9 e^{-1}) = 1000/(1+3.31) = 232$. Euler with $\Delta t = 0.5$: $y_1 = 100 + 0.001 \times 100 \times 900 \times 0.5 = 100 + 45 = 145$. $y_2 = 145 + 0.001 \times 145 \times 855 \times 0.5 = 145 + 61.99 = 206.99 \approx 207$. So answer ~207. Options: A) 145, B) 207, C) 232 (exact), D) 300. Correct: B for Euler approximation.

📖 Explanation: The logistic model is a simplification that assumes homogeneous mixing (everyone contacts everyone equally), no latent period (immediate infectiousness), and no spatial or social structure. Real epidemics have contact heterogeneity, latent periods, and indirect transmission (e.g., airborne). Thus, all options are valid limitations. Option A is a key limitation; B is often true but not always (airborne diseases); C is true for many diseases. The comprehensive answer is D, as all are limitations.

📖 Explanation: First, $dy/dt = 0.1 y (200 - y)$. At t=0, y=10, so $dy/dt = 0.1 \times 10 \times 190 = 190$. Now, $d^2y/dt^2 = 0.1 (dy/dt)(200 - 2y) = 0.1 \times 190 \times (200 - 20) = 0.1 \times 190 \times 180 = 3420$. That's not in options. Let's check: $f(y) = 0.1 y (200 - y)$, so f&#039;(y) = 0.1 (200 - 2y). Then d^2y/dt^2 = f&#039;(y) \cdot dy/dt = 0.1 (200 - 2y) \times 0.1 y (200 - y). At y=10: $= 0.1 \times 180 \times 0.1 \times 10 \times 190 = 0.1 \times 180 \times 190 = 3420$. None of the options. Let's change to $k=0.01$, L=100, y0=10: $dy/dt = 0.01 \times 10 \times 90 = 9$. $d^2y/dt^2 = 0.01 (100 - 20) \times 9 = 0.01 \times 80 \times 9 = 7.2$. Options: A) 0, B) 7.2, C) 9, D) 20. Correct: B. So the answer is 7.2, which is not in original options. I'll adjust options: A) 0, B) 7.2, C) 9, D) 18. Correct: B.

📖 Explanation: When $y_0 \ll L$, $L - y \approx L$, so $dy/dt \approx kL y$. This is the exponential growth model with growth rate $kL$. This is the initial phase of the epidemic where the disease spreads exponentially. Option B is decay; C describes the full logistic curve but incorrectly emphasizes the lag phase; D would be if k=0. This is a key insight: the early spread of a disease is approximately exponential.

📖 Explanation: The standard logistic model predicts $y \to L$, i.e., everyone gets infected. Adding a recovery term $-r y$ (as in the SIR model) creates an equilibrium where the disease may not infect everyone, depending on the basic reproduction number $R_0 = kL/r$. Option B would change the interaction order; C would remove the carrying capacity; D would revert to exponential growth. Thus, A is the most realistic modification to prevent full infection.

📖 Explanation: If $y_0 = 0$, then $y(t) = \frac{L \times 0}{0 + (L - 0)e^{-kLt}} = 0$. The disease cannot start with zero infected individuals; the model predicts no spread. Option A suggests undefined, but mathematically it is well-defined as 0. Option C would be true only if $y_0 > 0$; D is the limit for $y_0 > 0$. This highlights the importance of initial conditions in logistic models and the trivial solution at y=0.

📖 Explanation: Given $dy/dt = k y (L-y)$, plug in $y=100$, $L=10,000$, $dy/dt=500$: $500 = k \times 100 \times (10,000 - 100) = k \times 100 \times 9,900 = k \times 990,000$. Thus $k = 500 / 990,000 \approx 0.000505$. So $k \approx 0.0005$. Option B is 0.005 (ten times too large); C is 0.05; D is 0.5. This tests the ability to solve for a parameter given a data point, a common task in calibration.

📖 Explanation: Setting $dy/dt = 0$: $y (k(L-y) - r) = 0$. The nontrivial equilibrium is $k(L-y) - r = 0 \rightarrow L - y = r/k \rightarrow y = L - r/k$. This is the endemic equilibrium in the SIR model with recovery. Option B is $r/k$, which would be the value if L=0; C is the logistic equilibrium without recovery; D is the trivial solution. This shows how adding recovery creates a non-zero equilibrium below L.

📖 Explanation: The logistic model assumes a closed population. If births and deaths occur, the population L changes over time, so the final number infected may not be exactly L. Option B would actually support the conclusion (no recovery means everyone gets infected); C is required for spread; D is a simplifying assumption but does not prevent full infection. This tests understanding of model assumptions and their impact on predictions.

📖 Explanation: The function $f(y) = y(L-y)$ is a parabola symmetric about $y = L/2$, meaning $f(L/2 + a) = f(L/2 - a)$. This implies the infection rate is symmetric about the midpoint, and the logistic curve $y(t)$ is symmetric about its inflection point if plotted against time (in the sense of odd symmetry after a transformation). Option A is a direct property; B is a geometric interpretation; C is a verbal interpretation. Thus, D is correct. This requires students to connect algebraic symmetry to epidemiological meaning.

📖 Explanation: The maximum rate is $k L_{eff}^2 / 4 = 0.001 \times 500^2 / 4 = 0.001 \times 250,000 / 4 = 250 / 4 = 62.5$ per day. Wait, the units: if k=0.001, L=500, then max rate = k * (L/2)*(L/2) = 0.001 * 250 * 250 = 62.5. Not in options. Let's change k=0.00001, L=1000: max = 0.00001 * 500 * 500 = 2.5. Let's use k=0.01, L=100: max = 0.01 * 50 * 50 = 25. Options: A) 0.25, B) 0.5, C) 1.0, D) 2.0. Not matching. Let's use k=0.01, L=10: max = 0.01 * 5 * 5 = 0.25. So answer A. This tests the effect of intervention on the peak infection rate.

📖 Explanation: The second derivative changes from positive to negative at the inflection point $y = L/2$, which is where the growth rate $dy/dt$ is maximized. This is the peak of the epidemic curve. Option B is true in the sense that after the peak the rate decreases, but the sign change itself indicates the maximum rate. Option C would be after the peak? No, the rate is still positive; it's just decreasing. Option D is at y=L. This tests understanding of the relationship between derivatives and epidemic dynamics.

📖 Explanation: Omitting $(L-y)$ gives $dy/dt = k y$, which is exponential growth. This predicts unlimited growth of infected individuals, which is unrealistic. Option B would occur with a negative rate; C would occur if $k$ is constant and y appears linearly? Actually $dy/dt = k y$ is exponential, not linear. This highlights a critical error: forgetting the limiting factor changes the model entirely, leading to unrealistic predictions. This is a common mistake in modeling.

📖 Explanation: The infection rate is maximized at $y = L/2 = 100$. This is a direct property of the logistic model: the product $y(L-y)$ is maximized at the midpoint. Option A is the initial condition; C is the total population; D is double the initial. This is a basic recall question about the logistic model's properties.

📖 Explanation: The key difference is the saturation term $(L-y)$ in the logistic model, which reduces the growth rate as y approaches L. This prevents unbounded growth. Option B is incorrect; the growth rate in exponential is constant (relative), but the absolute growth rate $dy/dt$ grows with y; in logistic, the absolute growth rate first increases then decreases. Option C is wrong because both are nonlinear if y appears in a product with itself? Actually exponential is linear in y, logistic is nonlinear in y. But the main point is the limiting factor. This tests understanding of model assumptions.

📖 Explanation: The time to reach $y = L/2$ is found from the logistic solution: $y = L/(1 + C e^{-kLt})$. Setting $y = L/2$: $1 + C e^{-kLt} = 2 \rightarrow e^{-kLt} = 1/C \rightarrow t = \frac{\ln C}{kL} = \frac{\ln((L-y_0)/y_0)}{kL} = \frac{\ln(L/y_0 - 1)}{kL}$. Option A is correct. Option B is the doubling time in exponential growth; C is not dimensionally consistent; D is missing L. This requires deriving the time to reach a specific population size, a multi-step problem.

📖 Explanation: The simple logistic model lacks recovery, immunity, demographics, and intervention. Real diseases often have recovery (reducing infected count), births/deaths (changing population), and vaccination (reducing susceptible). Any of these extensions can result in the disease not infecting everyone. Option A changes the population; B reduces L; C introduces recovery. Thus, D is the most comprehensive answer. This tests the student's ability to think about model improvements.