Chapter 19 — Self-Assessment Quiz

Q: A first-order ODE is separable when it can be written in the form: (any right-hand side) only

B). The right side factors into a function of times a function of , which lets you rewrite it as and integrate each side separately. (§19.3)

Q: The initial-value problem , , has solution:

C) . Separating gives ; exponentiating and applying fixes the constant. This is the master equation of the chapter (decay when ). (§19.3, Worked Example 19.3.1)

Q: For the first-order linear equation , the integrating factor is:

A) . Multiplying through by this makes the left side an exact derivative, , which then integrates directly. (§19.4)

Q: A slope (direction) field for is useful primarily because it: gives the exact closed-form solution shows the slope a solution must have at every point, revealing qualitative behavior without solving only works for linear equations replaces the need for an initial condition

B). At each grid point it draws a short segment of slope ; solution curves flow tangent to those segments, so you can read off equilibria and long-term behavior even for equations with no closed form. (§19.5)

Q: Euler's method advances the numerical solution by:

A). Step forward by along the local slope — the tangent-line (linearization) idea of Chapter 11 iterated. Local error is , global error : first-order accurate. (§19.6)

Q: The logistic equation produces a population that: grows linearly forever grows purely exponentially forever follows an S-curve, growing fastest at and saturating at decays to zero

C). Near-exponential early, an inflection (fastest growth) at half capacity , then leveling at the carrying capacity . Solution: . (§19.7)

Q: Newton's law of cooling has solution:

B). The temperature gap decays exponentially, so relaxes toward ambient — fast at first, ever slower near equilibrium. This is the engine behind the forensic time-of-death estimate. (§19.8, Worked Example 19.8.1)

Q: In the SIR model, the basic reproduction number is:

B) . It comes from linearizing at the outbreak's start (): the epidemic grows when , i.e. . (§19.9)

Q: An infection has . The herd-immunity threshold — the immune fraction needed to stop sustained transmission — is:

C) . Transmission collapses once , i.e. . Note this is the threshold for control, not the final attack rate. (§19.9)

Q: Why is the SIR system solved numerically (e.g., with `solve_ivp`) rather than by a formula? The equations are linear but tedious It is a nonlinear system — the terms are products of unknowns — with no closed-form solution The Fundamental Theorem of Calculus does not apply Computers are simply faster

B). The mass-action infection terms multiply two unknown functions, making the system nonlinear; like most nonlinear ODEs it has no closed form, so we integrate numerically (RK45, the production-grade successor to the hand-coded Euler method of §19.6). (§19.9)

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Chapter 19 — Self-Assessment Quiz

Ten questions covering separable and linear equations, slope fields, Euler's method, the standard growth and cooling models, and the SIR model. Work each before opening the answer. Target time: ~20 minutes.

1. A first-order ODE is separable when it can be written in the form:

A) $y' = f(x) + g(y)$
B) $y' = g(x)\,h(y)$
C) $y' = f(x, y)$ (any right-hand side)
D) $y' = h(y)$ only

Answer

**B).** The right side factors into a function of $x$ times a function of $y$, which lets you rewrite it as $\dfrac{dy}{h(y)} = g(x)\,dx$ and integrate each side separately. *(§19.3)*

2. The initial-value problem $y' = ky$, $y(0) = y_0$, has solution:

A) $y = kt + y_0$
B) $y = e^{kt} + y_0$
C) $y = y_0\,e^{kt}$
D) $y = y_0\,k^{t}$

Answer

**C) $y = y_0 e^{kt}$.** Separating $dy/y = k\,dt$ gives $\ln|y| = kt + C$; exponentiating and applying $y(0) = y_0$ fixes the constant. This is the master equation of the chapter (decay when $k<0$). *(§19.3, Worked Example 19.3.1)*

3. For the first-order linear equation $y' + P(x)\,y = Q(x)$, the integrating factor is:

A) $\mu = e^{\int P\,dx}$
B) $\mu = e^{\int Q\,dx}$
C) $\mu = e^{\int (P - Q)\,dx}$
D) $\mu = \ln P$

Answer

**A) $\mu = e^{\int P\,dx}$.** Multiplying through by this $\mu$ makes the left side an exact derivative, $\dfrac{d}{dx}(\mu y) = \mu Q$, which then integrates directly. *(§19.4)*

4. A slope (direction) field for $y' = F(x,y)$ is useful primarily because it:

A) gives the exact closed-form solution
B) shows the slope a solution must have at every point, revealing qualitative behavior without solving
C) only works for linear equations
D) replaces the need for an initial condition

Answer

**B).** At each grid point it draws a short segment of slope $F(x,y)$; solution curves flow tangent to those segments, so you can read off equilibria and long-term behavior even for equations with no closed form. *(§19.5)*

5. Euler's method advances the numerical solution by:

A) $y_{n+1} = y_n + h\,F(x_n, y_n)$
B) $y_{n+1} = y_n - h\,F(x_n, y_n)$
C) $y_{n+1} = y_n + F(x_n, y_n)/h$
D) $y_{n+1} = h\,y_n\,F(x_n, y_n)$

Answer

**A).** Step forward by $h$ along the local slope $F(x_n, y_n)$ — the tangent-line (linearization) idea of Chapter 11 iterated. Local error is $O(h^2)$, global error $O(h)$: first-order accurate. *(§19.6)*

6. The logistic equation $P' = rP\!\left(1 - \dfrac{P}{K}\right)$ produces a population that:

A) grows linearly forever
B) grows purely exponentially forever
C) follows an S-curve, growing fastest at $P = K/2$ and saturating at $K$
D) decays to zero

Answer

**C).** Near-exponential early, an inflection (fastest growth) at half capacity $P = K/2$, then leveling at the carrying capacity $K$. Solution: $P = K/(1 + Ae^{-rt})$. *(§19.7)*

7. Newton's law of cooling $\dfrac{dT}{dt} = -k(T - T_{\text{env}})$ has solution:

A) $T = T_{\text{env}}\,e^{-kt}$
B) $T = T_{\text{env}} + (T_0 - T_{\text{env}})\,e^{-kt}$
C) $T = T_0 + kt$
D) $T = T_0\,e^{-kt}$

Answer

**B).** The temperature gap $T - T_{\text{env}}$ decays exponentially, so $T$ relaxes toward ambient — fast at first, ever slower near equilibrium. This is the engine behind the forensic time-of-death estimate. *(§19.8, Worked Example 19.8.1)*

8. In the SIR model, the basic reproduction number is:

A) $R_0 = \beta / \gamma$
B) $R_0 = \beta N / \gamma$
C) $R_0 = \gamma N / \beta$
D) $R_0 = \beta + \gamma$

Answer

**B) $R_0 = \beta N / \gamma$.** It comes from linearizing $I' = \beta S I - \gamma I \approx (\beta N - \gamma)I$ at the outbreak's start ($S \approx N$): the epidemic grows when $\beta N > \gamma$, i.e. $R_0 > 1$. *(§19.9)*

9. An infection has $R_0 = 4$. The herd-immunity threshold — the immune fraction needed to stop sustained transmission — is:

A) $25\%$
B) $50\%$
C) $75\%$
D) $100\%$

Answer

**C) $75\%$.** Transmission collapses once $R_0(1-p) < 1$, i.e. $p > 1 - 1/R_0 = 1 - 1/4 = 3/4$. Note this is the *threshold for control*, not the final attack rate. *(§19.9)*

10. Why is the SIR system solved numerically (e.g., with solve_ivp) rather than by a formula?

A) The equations are linear but tedious
B) It is a nonlinear system — the $\beta SI$ terms are products of unknowns — with no closed-form solution
C) The Fundamental Theorem of Calculus does not apply
D) Computers are simply faster

Answer

**B).** The mass-action infection terms $\beta SI$ multiply two unknown functions, making the system nonlinear; like most nonlinear ODEs it has no closed form, so we integrate numerically (RK45, the production-grade successor to the hand-coded Euler method of §19.6). *(§19.9)*

Scoring

9–10 correct: Excellent — you have strong command of ODE methods and the SIR model. Move on to the exercises' ⭐⭐⭐⭐ challenges and the Chapter 39 capstone preview.
7–8 correct: Solid. Revisit whichever of separable (§19.3), integrating factor (§19.4), or Euler's method (§19.6) tripped you up.
5–6 correct: Re-derive the integrating factor and re-implement Euler's method by hand; redo the logistic and cooling worked examples.
Below 5: Differential equations are the capstone of Part III. Re-read the chapter, paying special attention to the worked examples in §19.3, §19.4, and §19.8, then retake this quiz.