Differential Equations Quiz: Parameter Sensitivity

What this quiz covers

This quiz focuses on Parameter Sensitivity, 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. The bifurcation occurs at $r = 0$, where a single stable equilibrium splits into two new stable equilibria.
2. The bifurcation occurs at $r = 0$, where a stable and an unstable equilibrium are created as $r$ increases. (correct answer)
3. The bifurcation occurs at $r = 1$, where a stable equilibrium becomes unstable and creates a new equilibrium.
4. No bifurcation occurs because the system's qualitative behavior is the same for all values of the parameter $r$.

Explanation: Equilibrium points are found by setting $dy/dt = 0$, which gives $y^2 = r$. For $r < 0$, there are no real equilibria. For $r = 0$, there is one equilibrium at $y=0$. For $r > 0$, there are two equilibria at $y = \pm\sqrt{r}$. The number of equilibria changes at $r=0$, so this is the bifurcation point. To determine stability, we find the derivative of $f(y, r) = r - y^2$ with respect to $y$, which is $f_y = -2y$. For $y^* = \sqrt{r}$, $f_y = -2\sqrt{r} < 0$, so this equilibrium is stable. For $y^* = -\sqrt{r}$, $f_y = 2\sqrt{r} > 0$, so this equilibrium is unstable. Thus, as $r$ increases through 0, a stable and an unstable equilibrium are created. This is a saddle-node bifurcation.

Question 2

1. The extinction equilibrium at $P=0$ changes from stable to unstable, while a stable, positive carrying capacity emerges. (correct answer)
2. The extinction equilibrium at $P=0$ is always unstable, and a stable positive equilibrium appears only when $r > 0$.
3. For $r<0$ the population collapses; for $r>0$ it grows to a positive limit. The bifurcation at $r=0$ only involves the creation of this limit.
4. A bifurcation occurs at $r = 1$, where the population behavior switches from guaranteed extinction to growth towards a stable equilibrium.

Explanation: The equilibria are found by solving $P(r-P)=0$, which yields $P^*=0$ and $P^*=r$. Let $f(P) = rP - P^2$. The derivative is $f'(P) = r - 2P$. For the equilibrium $P^*=0$, stability is determined by $f'(0) = r$. Thus, $P^*=0$ is stable for $r < 0$ and unstable for $r > 0$. For the equilibrium $P^*=r$, stability is determined by $f'(r) = r - 2r = -r$. Thus, $P^*=r$ is stable for $r > 0$ and unstable for $r < 0$. At the bifurcation point $r=0$, the two equilibria collide and exchange stability. As $r$ increases through 0, the stable extinction equilibrium becomes unstable, and the carrying capacity equilibrium $P^*=r$ becomes positive and stable. This is a transcritical bifurcation.

Question 3

1. For $\alpha < 0$, there is one stable equilibrium. For $\alpha > 0$, there is one unstable equilibrium and two stable equilibria. (correct answer)
2. For $\alpha < 0$, there is one unstable equilibrium. For $\alpha > 0$, there is one stable equilibrium and two unstable equilibria.
3. For all $\alpha$, $x=0$ is a stable equilibrium, and for $\alpha > 0$, two additional unstable equilibria appear.
4. A bifurcation occurs at $\alpha=1$, where a single stable equilibrium becomes unstable and splits into two new equilibria.

Explanation: Equilibria are given by $x(\alpha - x^2) = 0$, so $x^*=0$ is an equilibrium for all $\alpha$, and $x^* = \pm\sqrt{\alpha}$ are equilibria for $\alpha \ge 0$. Let $f(x) = \alpha x - x^3$, so $f'(x) = \alpha - 3x^2$. For $\alpha < 0$, the only real equilibrium is $x^*=0$. Stability is given by $f'(0) = \alpha < 0$, so it is stable. For $\alpha > 0$, there are three equilibria. For $x^*=0$, $f'(0) = \alpha > 0$, so it is unstable. For $x^* = \pm\sqrt{\alpha}$, $f'(\pm\sqrt{\alpha}) = \alpha - 3(\alpha) = -2\alpha < 0$, so both are stable. The bifurcation at $\alpha=0$ is a supercritical pitchfork bifurcation.

Question 4

1. Solutions will converge to one of two new stable equilibria that appear for $\mu > 0$.
2. All non-zero solutions will diverge to $\pm\infty$. (correct answer)
3. All solutions will converge to the stable equilibrium at $x=0$.
4. Solutions will oscillate around $x=0$ with increasing amplitude.

Explanation: When you encounter a differential equation with a parameter like $\frac{dx}{dt} = \mu x + x^3$, you're dealing with a bifurcation problem where the system's qualitative behavior changes dramatically as the parameter crosses a critical value. To analyze this system, start by finding equilibria where $\frac{dx}{dt} = 0$: $\mu x + x^3 = 0$, which gives $x(\mu + x^2) = 0$. This yields $x = 0$ and $x = \pm\sqrt{-\mu}$ (when $\mu < 0$). For $\mu > 0$, only $x = 0$ exists as an equilibrium. To determine stability, examine the derivative: $\frac{d}{dx}(\mu x + x^3) = \mu + 3x^2$. At $x = 0$, this equals $\mu$. Since $\mu > 0$, the equilibrium at $x = 0$ is unstable. Now consider the behavior away from equilibrium. For any $x \neq 0$ when $\mu > 0$, both terms $\mu x$ and $x^3$ have the same sign as $x$. This means $\frac{dx}{dt} > 0$ when $x > 0$ and $\frac{dx}{dt} < 0$ when $x < 0$, causing solutions to diverge away from the unstable equilibrium toward $\pm\infty$. This confirms answer B. Option A is incorrect because no other equilibria exist for $\mu > 0$. Option C is wrong since $x = 0$ becomes unstable when $\mu > 0$. Option D is incorrect because this system doesn't produce oscillatory behavior—it's purely monotonic. Remember: in bifurcation problems, always check how equilibria and their stability change as you cross the bifurcation point.

Question 5

1. The system has two stable equilibria for $|r| > 2$ and no equilibria for $|r| < 2$.
2. A single bifurcation occurs at $r=0$, where the stability of the equilibrium changes.
3. Bifurcations occur at $r=\pm 2$, where a single equilibrium splits into two new stable equilibria.
4. Bifurcations occur at $r=\pm 2$, where a stable and an unstable equilibrium merge and annihilate. (correct answer)

Explanation: When analyzing differential equations with parameters, you're looking for bifurcation points where the qualitative behavior changes dramatically. For $\frac{dy}{dt} = y^2 - ry + 1$, start by finding equilibria where $\frac{dy}{dt} = 0$. Setting $y^2 - ry + 1 = 0$ and using the quadratic formula: $y = \frac{r \pm \sqrt{r^2 - 4}}{2}$. The key insight is that real equilibria only exist when the discriminant $r^2 - 4 \geq 0$, which means $|r| \geq 2$. For $|r| > 2$, you get two distinct equilibria. To determine stability, check the derivative: $\frac{d}{dy}(y^2 - ry + 1) = 2y - r$. The equilibrium $y_1 = \frac{r - \sqrt{r^2-4}}{2}$ gives $2y_1 - r = -\sqrt{r^2-4} < 0$ (stable), while $y_2 = \frac{r + \sqrt{r^2-4}}{2}$ gives $2y_2 - r = +\sqrt{r^2-4} > 0$ (unstable). At $r = \pm 2$, the discriminant equals zero, so the two equilibria collide and annihilate, leaving no equilibria for $|r| < 2$. This describes a saddle-node bifurcation. Answer D correctly identifies this scenario. Answer A incorrectly states both equilibria are stable. Answer B mislocates the bifurcation at $r = 0$ instead of $r = \pm 2$. Answer C wrongly claims a single equilibrium splits into two stable ones, when actually it's the reverse—two equilibria merge and disappear. Remember: in saddle-node bifurcations, equilibria are created or destroyed in stable-unstable pairs, and the critical parameter values occur where the discriminant of your equilibrium equation equals zero.

Question 6

1. As $\lambda$ approaches 0, the single stable equilibrium becomes unstable, indicating a bifurcation at $\lambda=0$.
2. The system undergoes a transcritical bifurcation at $\lambda = 1$, where stability is exchanged with another equilibrium.
3. The equilibrium $Q = Q_{env}$ is stable for $0 < \lambda < 1$ but becomes unstable for $\lambda > 1$.
4. For any $\lambda > 0$, the system has one stable equilibrium, so no bifurcation occurs in this parameter range. (correct answer)

Explanation: When analyzing how a differential equation's behavior changes with parameters, you're looking for bifurcations—points where the qualitative structure (like stability or number of equilibria) changes dramatically. For the equation $\frac{dQ}{dt} = -\lambda(Q - Q_{env})$, start by finding equilibria where $\frac{dQ}{dt} = 0$. This gives us $-\lambda(Q - Q_{env}) = 0$, which means $Q = Q_{env}$ regardless of $\lambda$ value (as long as $\lambda \neq 0$). To determine stability, examine the derivative: $\frac{d}{dQ}\left[-\lambda(Q - Q_{env})\right] = -\lambda$. Since $\lambda > 0$, this derivative is always negative, meaning the equilibrium $Q = Q_{env}$ is always stable for any positive $\lambda$. The system maintains the same qualitative structure—one stable equilibrium—across all positive $\lambda$ values. Choice A is wrong because the equilibrium remains stable as $\lambda \to 0^+$; there's no stability change. Choice B incorrectly identifies a transcritical bifurcation at $\lambda = 1$—no such bifurcation exists since stability doesn't change there. Choice C falsely claims the equilibrium becomes unstable for $\lambda > 1$, but we showed it remains stable ($-\lambda < 0$) for all $\lambda > 0$. Choice D correctly recognizes that the system maintains one stable equilibrium throughout the parameter range $\lambda > 0$, so no bifurcation occurs. Study tip: When checking for bifurcations, always examine both the number of equilibria and their stability as you vary the parameter. Real bifurcations involve qualitative changes in the phase portrait's structure.

Question 7

1. Solutions grow to $+\infty$, and the rate of growth is uniformly large and independent of the small value of $\epsilon$.
2. Solutions converge to a new stable equilibrium near $y=0$, and the convergence is faster for smaller $\epsilon$.
3. Solutions starting near $y=0$ pass through a region of slow movement before eventually growing to $+\infty$. (correct answer)
4. Solutions oscillate with a period proportional to $1/\epsilon$ before escaping to $+\infty$.

Explanation: When analyzing differential equations near bifurcation points, you need to understand how small parameter changes create dramatic behavioral shifts. A saddle-node bifurcation at $r = 0$ means equilibrium points appear or disappear as $r$ crosses zero. For $dy/dt = r + y^2$ with small positive $r = \epsilon$, let's find the equilibrium by setting $dy/dt = 0$: $\epsilon + y^2 = 0$, giving $y = \pm\sqrt{-\epsilon}$. Since $\epsilon > 0$, there are no real equilibrium points - solutions must grow without bound. However, the key insight is understanding how they grow. Near $y = 0$, we have $dy/dt = \epsilon + y^2 \approx \epsilon$ (since $y^2$ is very small). This means $dy/dt$ is approximately the tiny value $\epsilon$, creating very slow movement - almost like the solution "remembers" where the equilibrium used to be at $r = 0$. As $y$ increases, $y^2$ eventually dominates and growth accelerates rapidly. Choice A is wrong because growth isn't uniformly large - it starts very slowly. Choice B incorrectly assumes equilibrium points exist for $r > 0$. Choice D describes oscillatory behavior, but this first-order equation cannot produce oscillations. Choice C correctly captures this "ghost" of the destroyed equilibrium: solutions experience slow passage through the region where equilibrium points existed at $r = 0$, then accelerate to infinity once they escape this bottleneck. Study tip: Near bifurcation points, always check for "ghosts" of destroyed equilibria that create temporary slow dynamics before the ultimate long-term behavior takes over.

Question 8

1. The derivative $\partial f / \partial x$ evaluated at an equilibrium point $(x^*, r_c)$ is equal to zero.
2. The numerical value of at least one equilibrium point $x^*$ changes as the parameter $r$ is varied through $r_c$.
3. The qualitative structure of the phase line, such as the number of equilibria or their stability, changes as $r$ passes through $r_c$. (correct answer)
4. The function $f(x, r)$ is not differentiable with respect to the parameter $r$ at the point $r_c$.

Explanation: When analyzing autonomous differential equations with parameters, you're looking at how the system's behavior changes as you vary the parameter. A bifurcation occurs when small changes in a parameter cause dramatic qualitative changes in the system's dynamics. The correct answer is C because bifurcations are fundamentally about qualitative structural changes in the phase portrait. At a bifurcation point $r = r_c$, the phase line undergoes a topological change - equilibrium points may appear, disappear, or change stability. For example, in a saddle-node bifurcation, two equilibria collide and annihilate each other, while in a transcritical bifurcation, equilibria exchange stability. Option A describes a necessary condition for certain bifurcations but isn't the defining characteristic. The linearization $\partial f/\partial x = 0$ at equilibrium indicates marginal stability, which often occurs at bifurcation points, but bifurcations can happen without this condition being met. Option B confuses quantitative changes with qualitative ones. Equilibrium positions changing continuously with the parameter is normal behavior - what matters is when the fundamental structure changes discontinuously. Option D is incorrect because bifurcations typically occur at points where $f$ is smooth in both variables. Non-differentiability would indicate a more pathological situation, not a standard bifurcation. Remember: bifurcations are about qualitative changes in dynamics, not just numerical shifts. When studying bifurcation theory, focus on how phase portraits transform rather than getting caught up in specific mathematical conditions.

Question 9

1. The population collapses to extinction, regardless of the initial population size. (correct answer)
2. The population stabilizes at a new, lower carrying capacity that is close to zero.
3. The system gains a new unstable equilibrium, leading to unpredictable population dynamics.
4. The population begins to oscillate around a positive value before eventually collapsing.

Explanation: Equilibria are found by solving $N(1-N) - H = 0$, or $-N^2 + N - H = 0$. Using the quadratic formula, $N^* = \frac{-1 \pm \sqrt{1 - 4(-1)(-H)}}{-2} = \frac{1 \pm \sqrt{1 - 4H}}{2}$. Real-valued equilibria exist only if the discriminant is non-negative, i.e., $1 - 4H \ge 0$, which means $H \le 1/4$. The critical harvesting rate is therefore $H_c = 1/4$. At this point, the two equilibria merge. If $H$ is increased just beyond $1/4$, the discriminant $1-4H$ becomes negative, and there are no real equilibria. The function $f(N) = -N^2 + N - H$ is a downward-opening parabola that is now entirely below the horizontal axis for $H>1/4$. This means $dN/dt$ is always negative for any $N \ge 0$. Consequently, the population will decline and collapse to extinction, regardless of the starting population.

Question 10

1. The laser turns on for $G_0 > K$. At $G_0=K$, a transcritical bifurcation occurs where the zero-intensity state loses stability to a positive state. (correct answer)
2. The laser turns on for $G_0 > K$. At $G_0=K$, a saddle-node bifurcation occurs where a stable and an unstable positive intensity state are created.
3. The laser turns on for $G_0 > 1$. At $G_0=1$, a pitchfork bifurcation occurs where the zero state splits into two stable positive states.
4. The laser is on for any $G_0 > 0$. The parameter $G_0$ only affects the brightness, not the on/off state of the laser.

Explanation: Equilibria are found by setting $dI/dt = 0$. This gives $I=0$ or $\frac{G_0}{1 + I} - K = 0$. The second condition yields $G_0 = K(1+I)$, or $I^* = \frac{G_0}{K} - 1$. This positive equilibrium exists only if $\frac{G_0}{K} - 1 > 0$, which means $G_0 > K$. Let $f(I) = (\frac{G_0}{1 + I} - K)I$. The derivative is $f'(I) = (\frac{G_0}{1+I} - K) - \frac{G_0 I}{(1+I)^2}$. Stability of $I^*=0$: $f'(0) = G_0 - K$. So, $I=0$ is stable if $G_0 < K$ and unstable if $G_0 > K$. Stability of $I^*=\frac{G_0}{K}-1$: At this equilibrium, $\frac{G_0}{1+I^*} = K$. So, $f'(I^*) = 0 - \frac{G_0 I^*}{(1+I^*)^2} = -\frac{G_0 (G_0/K - 1)}{(G_0/K)^2}$. Since $G_0, K > 0$ and $G_0/K > 1$ for this equilibrium to be positive, $f'(I^*)$ is negative. Thus, the positive equilibrium is stable whenever it exists. At the threshold $G_0=K$, the equilibria $I=0$ and $I=G_0/K-1$ collide and exchange stability. This is a transcritical bifurcation.

Question 11

1. The critical value is $\gamma = \omega_0$ where the system transitions from underdamped to overdamped (correct answer)
2. The critical value is $\gamma = \omega_0/2$ where the system transitions from underdamped to critically damped
3. The critical value is $\gamma = 2\omega_0$ where the discriminant of the characteristic equation becomes zero
4. The critical value is $\gamma = \omega_0/4$ where the natural frequency equals the damping frequency

Explanation: The characteristic equation is $r^2 + 2\gamma r + \omega_0^2 = 0$ with roots $r = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}$. For oscillatory behavior (underdamped), we need complex roots, which occurs when $\gamma^2 - \omega_0^2 < 0$, or $\gamma < \omega_0$. The transition occurs when $\gamma^2 - \omega_0^2 = 0$, giving $\gamma = \omega_0$ (critical damping). For $\gamma > \omega_0$, the system is overdamped (non-oscillatory). Choice B incorrectly identifies $\omega_0/2$ as critical. Choice C gives $2\omega_0$ which would be well into the overdamped regime. Choice D uses $\omega_0/4$ and mentions an incorrect concept about natural vs damping frequency.

Question 12

1. $H > \frac{rK}{2}$ ensures extinction because the maximum sustainable yield is exceeded
2. $H > rK$ ensures extinction because harvesting exceeds the maximum possible growth rate
3. $H > \frac{rK}{4}$ ensures extinction because this corresponds to the maximum of the logistic growth function (correct answer)
4. $H > \frac{r}{K}$ ensures extinction because the per capita growth rate becomes negative for all populations

Explanation: The equation becomes $\frac{dP}{dt} = -\frac{r}{K}P^2 + rP - H$. For equilibria to exist, we need $-\frac{r}{K}P^2 + rP - H = 0$, or equivalently $\frac{r}{K}P^2 - rP + H = 0$. The discriminant is $\Delta = r^2 - 4 \cdot \frac{r}{K} \cdot H = r^2(1 - \frac{4H}{rK})$. For real equilibria to exist, we need $\Delta \geq 0$, which requires $H \leq \frac{rK}{4}$. When $H > \frac{rK}{4}$, there are no equilibria and $\frac{dP}{dt} < 0$ for all $P > 0$, so the population always decreases to extinction. The value $\frac{rK}{4}$ corresponds to the maximum of the function $g(P) = rP(1-P/K) = rP - \frac{r}{K}P^2$, which occurs at $P = K/2$ and equals $\frac{rK}{4}$. Choice A gives $\frac{rK}{2}$ which exceeds the threshold. Choice B gives $rK$ which is much too large. Choice D gives $\frac{r}{K}$ which has wrong dimensions and wrong magnitude.

Question 13

1. The system transitions from a stable spiral to a stable limit cycle, with increasing nonlinearity in the oscillations
2. The system transitions from a center to an unstable focus surrounded by a stable limit cycle of increasing amplitude (correct answer)
3. The system transitions from simple harmonic motion to chaotic oscillations as the damping term becomes dominant
4. The system transitions from an unstable focus to a stable node as the nonlinear damping stabilizes the origin

Explanation: When $\epsilon = 0$, the equation becomes $\frac{d^2x}{dt^2} + x = 0$, which is a harmonic oscillator with a center at the origin (periodic orbits). For small $\epsilon > 0$, the term $-\epsilon(1-x^2)\frac{dx}{dt}$ acts as negative damping when $|x| < 1$ (since $1-x^2 > 0$) and positive damping when $|x| > 1$. This destabilizes the origin (making it an unstable focus) while creating a stable limit cycle. As $\epsilon$ increases, the limit cycle amplitude grows and becomes more nonlinear. Choice A incorrectly states the origin becomes a stable spiral. Choice C is wrong because the system doesn't become chaotic - it has a stable limit cycle. Choice D incorrectly suggests the origin becomes stable and node-like.

Question 14

1. The equation is separable for all $\alpha$, and the parameter affects only the rate of convergence along trajectories, not their geometric shape
2. The equation is separable for all $\alpha$, but the critical value $\alpha = 1$ marks a transition in the stability of trajectories through the origin
3. The equation is separable only when $\alpha = 1$, which corresponds to a homogeneous equation with special symmetry properties
4. The equation is separable for all $\alpha$, and increasing $\alpha$ causes solution trajectories to become more curved in the $xy$-plane (correct answer)

Explanation: When you encounter a differential equation with a parameter and are asked about substitution effects, start by applying the given substitution systematically to see what conditions make the equation separable. Let's apply the substitution $v = \frac{y}{x}$, so $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting into the original equation: $v + x\frac{dv}{dx} = \frac{(vx)^2 - \alpha x^2}{x \cdot vx} = \frac{x^2(v^2 - \alpha)}{x^2v} = \frac{v^2 - \alpha}{v}$ Rearranging: $x\frac{dv}{dx} = \frac{v^2 - \alpha}{v} - v = \frac{v^2 - \alpha - v^2}{v} = \frac{-\alpha}{v}$ This gives us $\frac{dv}{dx} = \frac{-\alpha}{vx}$, which separates as $v\,dv = \frac{-\alpha}{x}\,dx$. This separation works for any value of $\alpha \neq 0$, confirming the equation is separable for all $\alpha$. The parameter $\alpha$ appears as a coefficient affecting the rate of change, and larger $|\alpha|$ values create steeper slopes in the $v$-$x$ space, which translates to more curved trajectories in the original $xy$-plane. Choice A incorrectly claims $\alpha$ only affects convergence rates, not geometric shape. Choice B wrongly identifies $\alpha = 1$ as special for stability properties. Choice C is completely wrong—the equation isn't separable only when $\alpha = 1$. Study tip: When analyzing parametric differential equations, always complete the substitution algebra first before making conclusions about separability or solution behavior. The parameter's role often becomes clear only after the full transformation.

Question 15

1. At $\mu = 0$, a stable limit cycle of radius $\sqrt{\mu}$ appears for $\mu > 0$ (supercritical Hopf bifurcation) (correct answer)
2. At $\mu = 1$, the origin changes stability and a limit cycle of radius 1 appears for all $\mu > 1$
3. At $\mu = 0$, an unstable limit cycle of radius $\sqrt{|\mu|}$ exists for $\mu < 0$ (subcritical Hopf bifurcation)
4. At $\mu = 0$, the origin changes from a stable to unstable focus, but no limit cycle appears for any value of $\mu$

Explanation: Converting to polar coordinates with $x = r\cos\theta$, $y = r\sin\theta$: $\frac{dr}{dt} = r(\mu - r^2)$ and $\frac{d\theta}{dt} = 1$. The origin is an equilibrium with $r = 0$. For $\mu < 0$, $\frac{dr}{dt} = r(\mu - r^2) < 0$ for small $r > 0$, so the origin is stable. For $\mu > 0$, $\frac{dr}{dt} = r(\mu - r^2) > 0$ for small $r$, so the origin becomes unstable. Additionally, there's an equilibrium at $r = \sqrt{\mu}$ when $\mu > 0$, corresponding to a limit cycle of radius $\sqrt{\mu}$. Since $\frac{dr}{dt} = r(\mu - r^2) < 0$ for $r > \sqrt{\mu}$, this limit cycle is stable. This is a supercritical Hopf bifurcation occurring at $\mu = 0$. Choice B incorrectly identifies $\mu = 1$ as the bifurcation point. Choice C describes subcritical behavior but gets the parameter range wrong. Choice D misses the limit cycle entirely.

Question 16

1. Both predator and prey populations at equilibrium increase proportionally to the change in $a$
2. The prey population at equilibrium remains constant while the predator population increases (correct answer)
3. The predator population at equilibrium remains constant while the prey population increases
4. The equilibrium point shifts but the relationship depends on the relative magnitudes of all parameters

Explanation: The nontrivial equilibrium occurs when both derivatives equal zero: $ax - bxy = 0$ and $-cy + dxy = 0$. From the first equation: $x(a - by) = 0$, so $y = a/b$. From the second equation: $y(dx - c) = 0$, so $x = c/d$. Therefore, the equilibrium is $(c/d, a/b)$. When $a$ increases, the prey equilibrium $x = c/d$ is unchanged (doesn't depend on $a$), while the predator equilibrium $y = a/b$ increases proportionally to $a$. Choice A is wrong because prey doesn't change. Choice C reverses the correct relationship. Choice D is incorrect because the equilibrium values have explicit dependence that doesn't require knowing relative magnitudes.

Question 17

1. Equilibrium concentration increases as $\sqrt{k_1/k_2}$, and approach rate decreases proportionally to $k_2$
2. Equilibrium concentration increases as $\sqrt{k_1/k_2}$, and approach rate increases proportionally to $\sqrt{k_1 k_2}$
3. Equilibrium concentration increases as $k_1/k_2$, and approach rate decreases proportionally to $k_2$
4. Equilibrium concentration increases as $\sqrt{k_1/k_2}$, and approach rate decreases proportionally to $\sqrt{k_1 k_2}$ (correct answer)

Explanation: When you encounter a differential equation modeling chemical reactions, you need to analyze both equilibrium behavior and stability properties. This equation represents a balance between constant production ($k_1$) and quadratic consumption ($k_2 x^2$). To find equilibrium, set $\frac{dx}{dt} = 0$: $k_1 - k_2 x^2 = 0$, which gives $x_{eq} = \sqrt{\frac{k_1}{k_2}}$. When $k_2$ decreases while $k_1$ stays fixed, the equilibrium concentration increases proportionally to $\sqrt{k_1/k_2}$. For the approach rate, you need to linearize around equilibrium. Taking the derivative: $\frac{d}{dx}\left(\frac{dx}{dt}\right) = -2k_2 x$. At equilibrium $x_{eq} = \sqrt{k_1/k_2}$, this becomes $-2k_2\sqrt{k_1/k_2} = -2\sqrt{k_1 k_2}$. The magnitude of this derivative determines how quickly the system returns to equilibrium, so the approach rate is proportional to $\sqrt{k_1 k_2}$. When $k_2$ decreases, this rate decreases. Answer A incorrectly states the approach rate depends only on $k_2$, missing the coupling with $k_1$. Answer B has the wrong sign—it claims the approach rate increases when $k_2$ decreases. Answer C gives the wrong scaling for equilibrium concentration (linear instead of square root) and wrong scaling for approach rate. Answer D correctly captures both relationships: equilibrium scales as $\sqrt{k_1/k_2}$ and approach rate decreases proportionally to $\sqrt{k_1 k_2}$. Remember: for nonlinear differential equations, always linearize around equilibrium points to determine stability and approach rates.