Differential Equations Quiz: Stability From Phase Lines

What this quiz covers

This quiz focuses on Stability From Phase Lines, 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. $y=3$ becomes unstable and $y=-1$ becomes stable.
2. $y=3$ becomes semi-stable and $y=-1$ becomes semi-stable. (correct answer)
3. $y=3$ becomes stable and $y=-1$ becomes unstable.
4. $y=3$ becomes unstable and $y=-1$ becomes semi-stable.

Explanation: The equilibrium points for $y' = [f(y)]^2$ are the same as for $y' = f(y)$, since $f(y)=0$ if and only if $[f(y)]^2=0$.\nHowever, the dynamics between the equilibrium points change. Since $[f(y)]^2 \ge 0$ for all $y$, the derivative $y'$ is always non-negative. This means that on the new phase line, all arrows must point upward, except at the equilibrium points where the rate of change is zero.\nFor the equilibrium at $y=3$: The arrow just below it will point up, and the arrow just above it will also point up. This is the definition of a semi-stable equilibrium.\nFor the equilibrium at $y=-1$: The arrow just below it will point up, and the arrow just above it will also point up. This is also a semi-stable equilibrium.\nTherefore, both the originally unstable and stable equilibria become semi-stable under this transformation.

Question 2

1. Stable
2. Unstable
3. Semi-stable (correct answer)
4. The stability cannot be determined from the information given.

Explanation: Since $f'(c)=0$, the standard first-derivative test for stability is inconclusive. We must examine the behavior of $f(y)$ near $y=c$ using higher-order derivatives. The Taylor expansion of $f(y)$ around $y=c$ is $f(y) = f(c) + f'(c)(y-c) + \frac{f''(c)}{2!}(y-c)^2 + O((y-c)^3)$.\nGiven the conditions, this simplifies to $f(y) \approx \frac{f''(c)}{2}(y-c)^2$ for $y$ near $c$. Since we are given that $f''(c) > 0$, the term $(y-c)^2$ ensures that $f(y)$ is positive for $y$ on both sides of $c$ (for $y \neq c$).\nThis means that $y' = f(y)$ is positive for $y$ just above $c$ and also for $y$ just below $c$. On a phase line, the arrows on both sides of the equilibrium point $c$ would point upward (away from $c$). This is the definition of a semi-stable equilibrium. (If $f''(c)<0$, it would also be semi-stable, with both arrows pointing down.)

Question 3

1. The solution $y(t)$ is strictly increasing for all $t$ in its domain.
2. The solution $y(t)$ may become unbounded in finite time.
3. The graph of the solution $y(t)$ has a horizontal asymptote. (correct answer)
4. For any initial value $y_0$, the solution $y(t)$ with $y(0)=y_0$ satisfies $y(t) > y_0$ for $t>0$.

Explanation: The condition $f(y) > 0$ for all $y$ means that $y'(t)$ is always positive. This implies that any solution $y(t)$ must be strictly increasing (making A and D true). Since the equation is autonomous and has no equilibrium points (since $f(y)$ is never zero), a solution $y(t)$ cannot approach a finite limit $L$ as $t \to \infty$. If it did, $L$ would have to be an equilibrium point. A horizontal asymptote corresponds to a finite limit as $t \to \infty$. Therefore, the graph of $y(t)$ cannot have a horizontal asymptote, making statement C necessarily false. Statement B can be true; for example, the solution to $y' = 1+y^2$ is $y(t) = \tan(t+c)$, which becomes unbounded in finite time. This equation satisfies $f(y) \ge 1 > 0$.

Question 4

1. Zero
2. One (correct answer)
3. Two
4. Three

Explanation: The equation $y' = f(y) + \epsilon$ corresponds to a small upward vertical shift of the graph of $f(y)$ by $\epsilon$. We analyze what happens to the roots of $f(y)$, which are the equilibrium points.\n1. At the unstable equilibrium $y=0$: For $f(y)$ to have an unstable equilibrium, the graph of $f(y)$ must cross the y-axis at $y=0$ with a positive slope. Shifting this graph up by a small $\epsilon$ will move the root slightly to the left, but a root will still exist near $y=0$.\n2. At the semi-stable equilibrium $y=2$: The arrows point towards 2 from above (so $f(y)<0$ for $y>2$) and away from 2 from below (so $f(y)>0$ for $y<2$). This means the graph of $f(y)$ must touch the y-axis at $y=2$ from above and have a local maximum at $y=2$. For example, $f(y)$ could look like $-y(y-2)^2$. When we shift this graph up by $\epsilon$, the local maximum at $(2,0)$ becomes $(2, \epsilon)$. The function is now positive around $y=2$, so the root at $y=2$ vanishes.\nTherefore, the original two equilibrium points are reduced to one equilibrium point after the perturbation.

Question 5

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

Explanation: The transformation from $y' = f(y)$ to $y' = -f(y)$ multiplies the rate of change by -1, which reverses the direction of all arrows on the phase line.\n1. An equilibrium at $y=2$ is stable for $f(y)$. This means for $f(y)$, the arrow above $y=2$ points down, and the arrow below $y=2$ points up. For $-f(y)$, these arrows are reversed: the arrow above $y=2$ will point up, and the arrow below $y=2$ will point down. This configuration corresponds to an unstable equilibrium.\n2. An equilibrium at $y=0$ is semi-stable for $f(y)$. This means arrows on one side point toward $y=0$ and on the other side point away. For example, the arrow above points away (up) and the arrow below points away (up). When reversed for $-f(y)$, the arrow above will point toward $y=0$ (down) and the arrow below will point toward $y=0$ (down). This is still a semi-stable configuration. The same logic applies if the arrows were originally both pointing down. Therefore, the semi-stable equilibrium remains semi-stable.

Question 6

1. The number of stable equilibria changes from zero to one. (correct answer)
2. The number of stable equilibria changes from one to two.
3. The number of stable equilibria changes from zero to two.
4. The number of stable equilibria remains constant at one.

Explanation: We analyze the equilibrium points and their stability based on $\alpha$. Let $f(y, \alpha) = y^3 - \alpha y = y(y^2 - \alpha)$. The equilibria are solutions to $f(y, \alpha)=0$. We use the linearization theorem, checking the sign of $f_y = \frac{\partial f}{\partial y} = 3y^2 - \alpha$.\nCase 1: $\alpha < 0$. The only real solution to $y(y^2 - \alpha) = 0$ is $y=0$. At this point, $f_y(0, \alpha) = -\alpha > 0$. Since the derivative is positive, $y=0$ is an unstable equilibrium. There are zero stable equilibria.\nCase 2: $\alpha = 0$. The equation is $y'=y^3$. The only equilibrium is $y=0$. Since $f_y(0,0)=0$, the test is inconclusive. However, the phase line for $y'=y^3$ has arrows pointing up for $y>0$ and down for $y<0$, so $y=0$ is unstable. There are zero stable equilibria.\nCase 3: $\alpha > 0$. There are three equilibria: $y=0$ and $y=\pm\sqrt{\alpha}$.\n- At $y=0$: $f_y(0, \alpha) = -\alpha < 0$. Since the derivative is negative, $y=0$ is a stable equilibrium.\n- At $y=\pm\sqrt{\alpha}$: $f_y(\pm\sqrt{\alpha}, \alpha) = 3(\sqrt{\alpha})^2 - \alpha = 3\alpha - \alpha = 2\alpha > 0$. Since the derivative is positive, both $y=\pm\sqrt{\alpha}$ are unstable equilibria.\nThus, for $\alpha > 0$, there is exactly one stable equilibrium. As $\alpha$ increases from negative to positive, the number of stable equilibria changes from 0 to 1.

Question 7

1. The population first decreases, then increases towards a capacity of $4$.
2. The population dies out, approaching $0$.
3. The population remains constant at $3$.
4. The population thrives and approaches a carrying capacity of $4$. (correct answer)

Explanation: When you encounter a logistic equation with harvesting, you need to analyze the equilibrium points and their stability using a phase line. This tells you how populations behave over time from different starting points. First, find the equilibrium points by setting $P' = 0$: $P(5-P) - 4 = 0$, which gives $5P - P^2 - 4 = 0$, or $P^2 - 5P + 4 = 0$. Factoring yields $(P-1)(P-4) = 0$, so the equilibrium points are $P = 1$ and $P = 4$. To determine stability, check the sign of $P'$ in each interval. For $P < 1$: choose $P = 0$, then $P' = 0(5) - 4 = -4 < 0$. For $1 < P < 4$: choose $P = 2$, then $P' = 2(3) - 4 = 2 > 0$. For $P > 4$: choose $P = 5$, then $P' = 5(0) - 4 = -4 < 0$. This means $P = 1$ is unstable (populations move away from it) and $P = 4$ is stable (populations approach it). Since $P(0) = 3$ lies between these equilibrium points where $P' > 0$, the population increases toward the stable equilibrium at $P = 4$. Answer D is correct because the population thrives (increases) and approaches carrying capacity 4. Answer A is wrong because the population doesn't decrease first from $P = 3$. Answer B is incorrect because the population doesn't approach 0 from this initial condition. Answer C is wrong because $P = 3$ isn't an equilibrium point. Remember: always find equilibrium points first, then test intervals to build your phase line and predict long-term behavior.

Question 8

1. All solutions with initial conditions $y_0 > -1$ will eventually approach $y = 2$ as $t \to \infty$
2. Solutions with initial conditions $2 < y_0 < 4$ will approach $y = 4$ as $t \to \infty$
3. Solutions with initial conditions $y_0 > 4$ will grow without bound as $t \to \infty$
4. Solutions with initial conditions $-1 < y_0 < 2$ will approach $y = 2$ as $t \to \infty$ (correct answer)

Explanation: From the phase line description: $y = -1$ is unstable (arrows away), $y = 2$ is stable (arrows toward from both sides), and $y = 4$ is semistable (arrows toward from left, away from right). For solutions in $(-1, 2)$, arrows point toward $y = 2$, so these solutions approach 2. Choice A is wrong because solutions with $y_0 > 4$ move away from 4. Choice B is wrong because arrows in $(2, 4)$ point left toward 2, not toward 4. Choice C is wrong because we don't know the behavior as $y \to \infty$ without more information about $f(y)$ for large $y$.

Question 9

1. $n$ must be odd and $n \geq 3$ to ensure the semistable behavior occurs
2. $n$ must be even and $n \geq 2$ to prevent sign changes across the equilibrium (correct answer)
3. $n$ can be any positive integer since $g(y)$ determines the stability behavior independently
4. $n$ must equal 2 specifically to create the required one-sided approach pattern

Explanation: For semistable behavior at $y = 1$, solutions must approach from one side and move away from the other side, which requires $f(y)$ to maintain the same sign on both sides of $y = 1$ (no sign change). Since $f(y) = (y-1)^n g(y)$ and $g(y)$ doesn't change sign near $y = 1$, the factor $(y-1)^n$ must not change sign either. This occurs when $n$ is even. If $n$ were odd, $(y-1)^n$ would change sign across $y = 1$, making it either stable or unstable, not semistable. We need $n \geq 2$ for an equilibrium to exist.

Question 10

1. $y = -2$ could be either stable or semistable depending on the multiplicities of the roots
2. $y = -2$ must be unstable since solutions cannot have two consecutive stable points
3. $y = -2$ must be stable since the alternating pattern requires it to attract solutions (correct answer)
4. $y = -2$ must be semistable since it precedes the unstable-stable-unstable pattern

Explanation: When analyzing autonomous differential equations like $\frac{dy}{dt} = f(y)$, understanding stability patterns of equilibrium points is crucial. The key insight is that for typical autonomous equations, stability must alternate between consecutive equilibrium points. Given the information, you know that solutions near $y = 0$ move toward $y = 1$, meaning $y = 0$ is unstable and $y = 1$ is stable. You also know that solutions near $y = 1$ move toward $y = 1$, confirming $y = 1$ is stable. Additionally, solutions near $y = 3$ move away from $y = 3$, so $y = 3$ is unstable. This gives you the pattern: $y = 1$ (stable) and $y = 3$ (unstable). Since $y = 0$ is between $y = -2$ and $y = 1$, and $y = 0$ is unstable (solutions move toward $y = 1$), the alternating stability pattern requires $y = -2$ to be stable. Looking at the wrong answers: A) incorrectly suggests that multiplicities determine stability when we already have enough information from the phase line behavior. B) makes a false claim about consecutive stable points - the alternating pattern actually supports having $y = -2$ stable followed by $y = 0$ unstable. D) incorrectly identifies $y = -2$ as semistable and mischaracterizes the overall pattern. The correct answer is C: $y = -2$ must be stable because the alternating stability pattern in autonomous equations requires it. Study tip: For phase line problems, always map out the stability pattern systematically - stability typically alternates between consecutive equilibrium points in simple autonomous equations.

Question 11

1. Both equations have identical stability patterns since they share the same equilibrium points
2. The stability of each equilibrium point is exactly reversed between the two equations (correct answer)
3. Equation (I) has all stable points while equation (II) has all unstable points
4. The equations have the same stable points but different semistable points

Explanation: For equation (I), $f_1(y) = y(1-y)(2-y)$: $y = 0$ unstable, $y = 1$ stable, $y = 2$ unstable. For equation (II), $f_2(y) = -y(1-y)(2-y) = -f_1(y)$: since the right-hand side is negated, the direction of all arrows on the phase line reverses. This means $y = 0$ becomes stable, $y = 1$ becomes unstable, and $y = 2$ becomes stable. The stability of each equilibrium point is exactly opposite between the two equations. Choice A is wrong because equilibrium locations don't determine stability patterns. Choice C is wrong because neither equation has uniform stability.

Question 12

1. This scenario is impossible since alternating stability requires exactly one stable point among three equilibria (correct answer)
2. The parameter $a$ must equal $b/2$ to create the symmetric stability pattern required
3. The parameters satisfy $a > 0$ and $b > a$, which naturally produces two stable equilibria
4. This scenario requires modification of the differential equation since the given form cannot produce two stable points

Explanation: For $f(y) = y(y-a)(y-b)$ with $0 < a < b$, the equilibria are at $y = 0, a, b$. Analyzing signs: $f(y) < 0$ for $y \in (-\infty, 0) \cup (a, b)$ and $f(y) > 0$ for $y \in (0, a) \cup (b, \infty)$. This gives: $y = 0$ unstable (arrows away), $y = a$ stable (arrows toward), $y = b$ unstable (arrows away). The alternating pattern of stability for consecutive equilibria is a fundamental property of autonomous equations with simple roots - stability must alternate. Having two stable points among three consecutive equilibria violates this principle.