Differential Equations Quiz: Equilibrium Solutions And Stability

What this quiz covers

This quiz focuses on Equilibrium Solutions And Stability, 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. $f(y) = (y-2)^2(y-5)(y-8)^2$
2. $f(y) = (y-2)(y-5)^2(y-8)$ (correct answer)
3. $f(y) = (y-2)^2(y-5)^2(y-8)$
4. $f(y) = (y-2)(y-5)(y-8)^2$

Explanation: We need: $y = 2$ unstable (arrows away on both sides), $y = 5$ semi-stable (arrows toward from left, away from right), $y = 8$ stable (arrows toward from both sides). For $f(y) = (y-2)(y-5)^2(y-8)$: Near $y = 2$, the sign changes from negative to positive as $y$ increases through 2 (unstable). Near $y = 5$, the factor $(y-5)^2$ doesn't change sign, but the overall function is negative for $y < 5$ and positive for $y > 5$ due to the other factors, giving semi-stable behavior. Near $y = 8$, the sign changes from positive to negative (stable).

Question 2

1. 125 (correct answer)
2. 250
3. 500
4. 1000

Explanation: The rate of change of the population is $dP/dt = f(P) = 0.5P(1 - P/1000) - h$. The term $g(P) = 0.5P(1 - P/1000)$ represents the natural growth rate of the population. If the harvesting rate $h$ is greater than the maximum possible growth rate, then $dP/dt$ will always be negative, and the population will go to extinction. We need to find the maximum value of $g(P)$. This function is a downward-opening parabola with roots at $P=0$ and $P=1000$. The maximum value occurs at the vertex, which is halfway between the roots, at $P = 500$. The maximum growth rate is $g(500) = 0.5(500)(1 - 500/1000) = 250(1 - 0.5) = 250(0.5) = 125$. This value is the maximum sustainable yield. If the harvesting rate $h$ is greater than 125, then $dP/dt < 0$ for all $P>0$, and the population will be driven to extinction. Therefore, the critical value is $h_{crit} = 125$.

Question 3

1. $y_0 = 3$
2. $y_0 = -2$
3. $y_0 = 1$ (correct answer)
4. $y_0 = -3$

Explanation: When analyzing the long-term behavior of autonomous differential equations like $\frac{dy}{dt} = y^3 - 4y$, you need to identify equilibrium points and determine their stability. This tells you where solutions will go as time approaches infinity. First, find the equilibrium points by setting $y^3 - 4y = 0$, which gives $y(y^2 - 4) = 0$. The equilibrium points are $y = 0, y = 2,$ and $y = -2$. To determine stability, analyze the sign of $\frac{dy}{dt}$ in each region. Factor the right side: $y^3 - 4y = y(y-2)(y+2)$. Test the sign in each interval:

• For $y < -2$: negative × negative × negative = negative, so $\frac{dy}{dt} < 0$
• For $-2 < y < 0$: negative × negative × positive = positive, so $\frac{dy}{dt} > 0$
• For $0 < y < 2$: positive × negative × positive = negative, so $\frac{dy}{dt} < 0$
• For $y > 2$: positive × positive × positive = positive, so $\frac{dy}{dt} > 0$

Question 4

1. -12
2. 0
3. 4
4. 16 (correct answer)

Explanation: Bifurcations in this type of system occur when two or more equilibrium points coalesce. This corresponds to a point where the graph of $f(y) = y^3 - 12y + h$ is tangent to the y-axis. This means we need to find values of $h$ for which the system of equations $f(y)=0$ and $f'(y)=0$ has a solution. First, find the critical points of $f(y)$ by solving $f'(y) = 0$: $f'(y) = 3y^2 - 12 = 0$ $3(y^2 - 4) = 0$ $y = 2$ or $y = -2$. These are the $y$-values where bifurcations can occur. Now, substitute these values back into $f(y)=0$ to find the corresponding values of $h$. Case 1: $y=2$. $f(2) = (2)^3 - 12(2) + h = 0$ $8 - 24 + h = 0$ $-16 + h = 0 \implies h = 16$. Case 2: $y=-2$. $f(-2) = (-2)^3 - 12(-2) + h = 0$ $-8 + 24 + h = 0$ $16 + h = 0 \implies h = -16$. The bifurcations occur at $h=16$ and $h=-16$. Of the options provided, 16 is a bifurcation value.

Question 5

1. -1
2. 0 (correct answer)
3. 1
4. 2

Explanation: First, find the equilibrium solutions by setting $dy/dt = 0$. This occurs when $e^y - 1 = 0$ or $y^2 - 1 = 0$. $e^y - 1 = 0 \implies e^y = 1 \implies y = 0$. $y^2 - 1 = 0 \implies (y-1)(y+1) = 0 \implies y = 1$ or $y = -1$. The equilibrium solutions are $y = -1, 0, 1$. Next, determine the stability of each. Let $f(y) = (e^y - 1)(y - 1)(y + 1)$. We analyze the sign of $f(y)$ in the intervals between the equilibria.

• Interval $y > 1$ (e.g., $y=2$): $f(2) = (e^2-1)(1)(3) > 0$. (Arrow up)
• Interval $0 < y < 1$ (e.g., $y=0.5$): $f(0.5) = (e^{0.5}-1)(-0.5)(1.5) < 0$. (Arrow down)
• Interval $-1 < y < 0$ (e.g., $y=-0.5$): $f(-0.5) = (e^{-0.5}-1)(-1.5)(0.5) > 0$. (Arrow up)
• Interval $y < -1$ (e.g., $y=-2$): $f(-2) = (e^{-2}-1)(-3)(-1) < 0$. (Arrow down)

• At $y=1$: solutions below move away (down), solutions above move away (up). Unstable.
• At $y=0$: solutions below move towards (up), solutions above move towards (down). Stable.
• At $y=-1$: solutions below move away (down), solutions above move away (up). Unstable.

Question 6

1. $f'(c) \ge 0$.
2. $f(y) < 0$ for $y > c$ and $f(y) > 0$ for $y < c$ in a neighborhood of $c$.
3. $f'(c) = 0$ and $f''(c) > 0$.
4. $f'(c) > 0$. (correct answer)

Explanation: When analyzing the stability of equilibrium solutions for autonomous differential equations, you need to understand how solutions behave near the equilibrium point. For $dy/dt = f(y)$, an equilibrium $y = c$ occurs when $f(c) = 0$, meaning the rate of change is zero. The key insight is that $f'(c)$ determines the local behavior. If $f'(c) > 0$, then near the equilibrium, $f(y) \approx f'(c)(y-c)$. This means when $y > c$, we have $f(y) > 0$, so $dy/dt > 0$ and $y$ increases further from $c$. When $y < c$, we have $f(y) < 0$, so $dy/dt < 0$ and $y$ decreases further from $c$. In both cases, solutions move away from the equilibrium, making it unstable. Looking at the wrong answers: Choice A ($f'(c) \geq 0$) is insufficient because $f'(c) = 0$ could lead to either stable or unstable behavior depending on higher-order terms. Choice B describes the opposite scenario where solutions converge to $c$, making it stable, not unstable. Choice C ($f'(c) = 0$ and $f''(c) > 0$) requires second-derivative analysis and isn't sufficient on its own—you'd need to examine whether $f$ changes sign appropriately. Remember this pattern: for autonomous equations $dy/dt = f(y)$, a positive derivative at an equilibrium ($f'(c) > 0$) creates a "repelling" effect that pushes solutions away, guaranteeing instability. This is the linearization test for stability.

Question 7

1. -4
2. 0 (correct answer)
3. 4
4. 8

Explanation: An autonomous differential equation has equilibrium solutions where $dy/dt = 0$. For the quadratic equation $y^2 - ay + 4 = 0$, there is a single solution for $y$ when the discriminant is zero. The discriminant is $b^2 - 4ac = (-a)^2 - 4(1)(4) = a^2 - 16$. Setting the discriminant to zero gives $a^2 - 16 = 0$, which yields $a=4$ and $a=-4$. Case 1: $a=4$. The equation becomes $dy/dt = y^2 - 4y + 4 = (y-2)^2$. The single equilibrium solution is $y=2$. Case 2: $a=-4$. The equation becomes $dy/dt = y^2 + 4y + 4 = (y+2)^2$. The single equilibrium solution is $y=-2$. The two equilibrium solutions corresponding to the two values of $a$ are $2$ and $-2$. Their sum is $2 + (-2) = 0$.

Question 8

1. One stable and two semi-stable equilibria. (correct answer)
2. One stable and two unstable equilibria.
3. Three stable equilibria.
4. One stable, one unstable, and one semi-stable equilibrium.

Explanation: Equilibrium solutions occur when $dy/dt = 0$. This requires $y^2-4=0$ or $\sin(\frac{\pi y}{2})=0$. The first gives $y=\pm 2$. The second gives $\frac{\pi y}{2} = n\pi$ for integer $n$, so $y=2n$. In the interval $[-3, 3]$, this gives $y=-2, 0, 2$. The distinct equilibria are $y=-2, 0, 2$. Let $f(y) = (y^2 - 4) \sin(\frac{\pi y}{2})$. We use the first derivative test for stability where possible: $f'(y) = 2y\sin(\frac{\pi y}{2}) + (y^2-4)\frac{\pi}{2}\cos(\frac{\pi y}{2})$. At $y=0$: $f'(0) = 0 + (-4)(\frac{\pi}{2})\cos(0) = -2\pi < 0$. So $y=0$ is stable. At $y=2$: $f'(2) = 4\sin(\pi) + (0)... = 0$. The test is inconclusive. At $y=-2$: $f'(-2) = -4\sin(-\pi) + (0)... = 0$. The test is inconclusive. For $y=2$ and $y=-2$, we must analyze the sign of $f(y)$ nearby. Near $y=2$: For $y$ slightly greater than 2, $y^2-4>0$ and $\sin(\frac{\pi y}{2})<0$, so $f(y)<0$. For $y$ slightly less than 2, $y^2-4<0$ and $\sin(\frac{\pi y}{2})>0$, so $f(y)<0$. Since $f(y)$ is negative on both sides, $y=2$ is semi-stable. Near $y=-2$: For $y$ slightly greater than -2, $y^2-4<0$ and $\sin(\frac{\pi y}{2})<0$, so $f(y)>0$. For $y$ slightly less than -2, $y^2-4>0$ and $\sin(\frac{\pi y}{2})>0$, so $f(y)>0$. Since $f(y)$ is positive on both sides, $y=-2$ is semi-stable. Thus, there is one stable equilibrium ($y=0$) and two semi-stable equilibria ($y=\pm 2$).

Question 9

1. $y=2$ must be semi-stable.
2. $y=2$ must be unstable. (correct answer)
3. $y=2$ must be asymptotically stable.
4. The stability of $y=2$ cannot be determined from the information given.

Explanation: We can construct a phase line to analyze the stability. The equilibria are at $y=0, 2, 5$. Since $y=0$ is unstable, solutions near it move away from 0. Since there are no equilibria between 0 and 2, and solutions starting just above 0 must go somewhere, they must move toward the next equilibrium at $y=2$. This means $f(y) > 0$ for $y \in (0,2)$. Similarly, since $y=5$ is stable, solutions near it move toward 5. Since there are no equilibria between 2 and 5, solutions starting just above 2 must move toward 5. This means $f(y) > 0$ for $y \in (2,5)$. Now consider $y=2$: solutions just below 2 move toward 2 (from the left), while solutions just above 2 move away from 2 (toward 5). Therefore, $y=2$ must be unstable.

Question 10

1. $y=\alpha$ is semi-stable.
2. $y=\alpha$ is stable.
3. $y=\alpha$ is unstable. (correct answer)
4. The stability depends on the sign of $\alpha$.

Explanation: When you encounter a differential equation like $\frac{dy}{dt} = (y - \alpha)^3$, you need to analyze the stability of equilibrium points by examining how solutions behave near those points. First, find the equilibrium by setting $\frac{dy}{dt} = 0$, which gives us $(y - \alpha)^3 = 0$, so $y = \alpha$ is indeed an equilibrium solution. To determine stability, examine what happens when $y$ is slightly perturbed from $\alpha$. If $y > \alpha$, then $(y - \alpha) > 0$, making $(y - \alpha)^3 > 0$, so $\frac{dy}{dt} > 0$. This means $y$ increases further away from $\alpha$. If $y < \alpha$, then $(y - \alpha) < 0$, making $(y - \alpha)^3 < 0$, so $\frac{dy}{dt} < 0$. This means $y$ decreases further away from $\alpha$. In both cases, small perturbations grow larger over time, confirming that $y = \alpha$ is unstable. Option A is incorrect because semi-stability occurs when solutions approach the equilibrium from one side but diverge from the other—that's not the case here. Option B is wrong since stable equilibria attract nearby solutions, but we've shown solutions move away from $y = \alpha$. Option D is incorrect because the sign of $\alpha$ doesn't affect the local behavior near the equilibrium—only the relative position of $y$ to $\alpha$ matters. Study tip: For polynomial right-hand sides, always check the sign of the derivative on both sides of an equilibrium point. If it has the same sign as the perturbation direction, the equilibrium is unstable.

Question 11

1. $y = 0$ is unstable, $y = 1$ is stable, and $y = 3$ is semi-stable (correct answer)
2. $y = 0$ is semi-stable, $y = 1$ is unstable, and $y = 3$ is stable
3. $y = 0$ is unstable, $y = 1$ is stable, and $y = 3$ is unstable
4. $y = 0$ is stable, $y = 1$ is unstable, and $y = 3$ is semi-stable

Explanation: To determine stability, analyze the sign of $f(y) = y^2(y-1)(y-3)^2$ around each equilibrium. For $y = 0$: $f(y) > 0$ for $y$ slightly positive and $f(y) = 0$ for $y$ slightly negative, so solutions move away from 0 (unstable). For $y = 1$: $f(y) < 0$ for $y < 1$ and $f(y) > 0$ for $y > 1$, so solutions approach 1 from both sides (stable). For $y = 3$: the factor $(y-3)^2$ creates a double root, making $f(y) > 0$ on both sides of $y = 3$, so solutions approach from below but move away from above (semi-stable).

Question 12

1. $a > 0$, and the cubic polynomial has three distinct real roots with negative derivative at the middle root (correct answer)
2. $a < 0$, and the cubic polynomial has three distinct real roots with positive derivative at the middle root
3. $a > 0$, and the cubic polynomial has three distinct real roots with positive derivative at the middle root
4. $a < 0$, and the cubic polynomial has three distinct real roots with negative derivative at the middle root

Explanation: For stability analysis, we examine $f'(y)$ at equilibria where $f(y) = ay^3 + by^2 + cy + d$. A equilibrium $y_0$ is stable if $f'(y_0) < 0$. Since we have three distinct roots with the middle one stable, we need $f'(y_{middle}) < 0$. Also, $f'(y) = 3ay^2 + 2by + c$. For a cubic with three real roots and $a > 0$, the function decreases then increases then decreases as we move from left to right through the roots, making the middle root stable ($f'(y_{middle}) < 0$) and the outer roots unstable. If $a < 0$, the behavior would be reversed.

Question 13

1. There are exactly two equilibrium solutions, both of which are stable
2. There are exactly two equilibrium solutions, both of which are unstable
3. There are no real equilibrium solutions, indicating population extinction is inevitable (correct answer)
4. There is exactly one equilibrium solution, which is semi-stable at $N = \frac{K}{2}$

Explanation: The equilibrium condition $rN(1-\frac{N}{K}) = H$ can be rewritten as $rN - \frac{rN^2}{K} = H$, or $\frac{rN^2}{K} - rN + H = 0$. This is a quadratic in $N$ with discriminant $\Delta = r^2 - 4 \cdot \frac{r}{K} \cdot H = r^2 - \frac{4rH}{K} = r^2(1 - \frac{4H}{rK})$. When $H > \frac{rK}{4}$, we have $\frac{4H}{rK} > 1$, so $\Delta < 0$. This means no real solutions exist for the equilibrium condition, indicating that harvesting exceeds the maximum sustainable yield and the population will inevitably go extinct.