Differential Equations Quiz: Phase Line Analysis

What this quiz covers

This quiz focuses on Phase Line Analysis, 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. 0 (correct answer)
2. 6
3. 9
4. 12

Explanation: Equilibrium points are the roots of the quadratic equation $y^2 - ay + 9 = 0$. A bifurcation (in this case, a saddle-node bifurcation) occurs when the number of real roots changes. This happens when the discriminant of the quadratic is equal to zero.\nThe discriminant $D$ is given by $D = b^2 - 4ac = (-a)^2 - 4(1)(9) = a^2 - 36$.\nSetting the discriminant to zero gives $a^2 - 36 = 0$, which solves to $a = 6$ and $a = -6$. These are the two values of $a$ for which a bifurcation occurs. The question asks for the sum of these values, which is $6 + (-6) = 0$.

Question 2

1. The equilibrium at $P=3$ is stable, representing the carrying capacity.
2. An initial population of $P(0)=4$ will decrease towards the equilibrium at $P=3$.
3. The equilibrium at $P=1$ is stable, representing an extinction threshold.
4. An initial population of $P(0)=2$ will increase and approach the equilibrium at $P=3$. (correct answer)

Explanation: When analyzing population dynamics through differential equations, you need to identify equilibrium points and determine their stability by examining the sign of the derivative around each point. The equilibrium points occur where $\frac{dP}{dt} = 0$, so $(P-3)^2(P-1) = 0$. This gives us equilibrium points at $P = 1$ and $P = 3$. To understand stability, examine the sign of $\frac{dP}{dt}$ in different regions:

• For $P < 1$: $(P-3)^2 > 0$ and $(P-1) < 0$, so $\frac{dP}{dt} < 0$ (population decreases)
• For $1 < P < 3$: $(P-3)^2 > 0$ and $(P-1) > 0$, so $\frac{dP}{dt} > 0$ (population increases)
• For $P > 3$: $(P-3)^2 > 0$ and $(P-1) > 0$, so $\frac{dP}{dt} > 0$ (population increases)

Question 3

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

Explanation: A saddle-node bifurcation occurs when two equilibrium points merge and disappear. This corresponds to a point where the graph of $f(y) = y^3 - 12y + \alpha$ is tangent to the $y$-axis. At such a point, we must have both $f(y)=0$ and $f'(y)=0$.\nFirst, find the values of $y$ where $f'(y)=0$. The derivative is $f'(y) = 3y^2 - 12$.\nSetting $f'(y)=0$ gives $3y^2 - 12 = 0$, so $3(y^2 - 4) = 0$. This yields $y=2$ and $y=-2$. These are the $y$-coordinates of the bifurcation points.\nThe question asks for the positive value of $y$ at which a bifurcation occurs, which is $y=2$. (The corresponding $\alpha$ values would be found by plugging these $y$ values back into $f(y)=0$: for $y=2$, $8-24+\alpha=0 \implies \alpha=16$; for $y=-2$, $-8+24+\alpha=0 \implies \alpha=-16$.)

Question 4

1. $y = 1$
2. $y = 0$
3. $y = 2/3$ (correct answer)
4. $y = 1/2$

Explanation: When analyzing autonomous differential equations for inflection points, you need to find where the second derivative equals zero. An inflection point occurs where the concavity of the solution curve changes. Starting with $\frac{dy}{dt} = y^2 - y^3$, you find the second derivative by differentiating both sides with respect to $t$: $\frac{d^2y}{dt^2} = \frac{d}{dt}(y^2 - y^3) = (2y - 3y^2)\frac{dy}{dt}$ Substituting the original equation $\frac{dy}{dt} = y^2 - y^3$: $\frac{d^2y}{dt^2} = (2y - 3y^2)(y^2 - y^3) = y^2(2y - 3y^2)(1 - y)$ For an inflection point, set $\frac{d^2y}{dt^2} = 0$: $y^2(2y - 3y^2)(1 - y) = 0$ This gives $y = 0$, $2y - 3y^2 = 0$, or $y = 1$. From $2y - 3y^2 = 0$: $y(2 - 3y) = 0$, so $y = 0$ or $y = \frac{2}{3}$. However, you need non-constant solutions. At $y = 0$ and $y = 1$, the original equation gives $\frac{dy}{dt} = 0$, making these equilibrium points where solutions are constant. Only $y = \frac{2}{3}$ corresponds to a point where $\frac{dy}{dt} \neq 0$, allowing for genuine inflection points in non-constant solutions. Answer (A) $y = 1$ and (B) $y = 0$ are equilibrium points, not inflection points. Answer (D) $y = \frac{1}{2}$ doesn't satisfy the second derivative condition. Study tip: For autonomous equations, always check whether critical points of the second derivative correspond to equilibria of the original equation—true inflection points occur only where the solution is actually changing.

Question 5

1. -4
2. 0
3. 4 (correct answer)
4. It cannot be determined from the information given.

Explanation: The information about the stability of the equilibrium points allows us to determine the sign of $f(y)$ in the intervals between them.\n- Since $y=2$ is an unstable equilibrium (a source), solutions starting near $y=2$ move away from it. This means for $y$ slightly greater than 2, $dy/dt = f(y)$ must be positive.\n- Since $y=8$ is a stable equilibrium (a sink), solutions starting near $y=8$ move towards it. This means for $y$ slightly less than 8, $dy/dt = f(y)$ must be positive.\nBoth conditions imply that for the entire interval $(2, 8)$, we must have $f(y) > 0$. The value $y=5$ is in this interval. Therefore, $f(5)$ must be positive. Of the choices given, only 4 is positive.

Question 6

1. $y \in (0, 2)$
2. $y \in (1, \infty)$
3. $y \in (-\infty, 0) \cup (1, 2)$
4. $y \in (0, 1) \cup (2, \infty)$ (correct answer)

Explanation: Concavity of solution curves $y(t)$ is determined by the sign of the second derivative, $y''(t)$. We use the chain rule: $y'' = \frac{d}{dt}(y') = \frac{d(f(y))}{dy} \frac{dy}{dt} = f'(y)f(y)$.\nGiven $f(y) = y(2-y) = 2y-y^2$. The first derivative is $f'(y) = 2-2y$.\nSo, $y'' = (2-2y)(2y-y^2) = 2(1-y)y(2-y)$.\nWe want to find where $y''>0$. The roots of $y''$ are at $y=0, y=1,$ and $y=2$. We test the intervals between these roots:\n- For $y>2$: $y''>0$ since $(1-y)<0$, $y>0$, and $(2-y)<0$. The product of three negatives is positive. Concave up.\n- For $1<y<2$: $y''<0$ since $(1-y)<0$, $y>0$, and $(2-y)>0$. The product is negative. Concave down.\n- For $0<y<1$: $y''>0$ since $(1-y)>0$, $y>0$, and $(2-y)>0$. The product is positive. Concave up.\n- For $y<0$: $y''<0$ since $(1-y)>0$, $y<0$, and $(2-y)>0$. The product is negative. Concave down.\nThus, the solution curves are concave up for $y \in (0, 1) \cup (2, \infty)$.

Question 7

1. 0
2. 200 (correct answer)
3. 600
4. 1000

Explanation: First, we find the equilibrium points of the population model by setting $dP/dt = 0$. The factors give the equilibria:\n- $P=0$ (Extinction)\n- $1 - P/1000 = 0 \implies P = 1000$ (Carrying capacity)\n- $P/200 - 1 = 0 \implies P = 200$ (Allee threshold)\nThe equilibria are $P=0, 200, 1000$. To understand the dynamics, we create a phase line by checking the sign of $dP/dt$ in the intervals:\n- For $P > 1000$: $dP/dt = (+)(-)(+) < 0$ (decreasing). The population declines towards 1000.\n- For $200 < P < 1000$: $dP/dt = (+)(+)(+) > 0$ (increasing). The population grows towards 1000.\n- For $0 < P < 200$: $dP/dt = (+)(+)(-) < 0$ (decreasing). The population declines towards 0 (extinction).\nFrom this analysis, $P=1000$ is a stable equilibrium, and $P=200$ is an unstable equilibrium. This unstable equilibrium acts as the critical threshold. If the initial population is below 200, it will go extinct. If it is above 200, it will grow towards the carrying capacity of 1000. Therefore, the threshold is 200.

Question 8

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

Explanation: First, find the equilibrium points by setting $dy/dt=0$. This occurs when $(y^2-1)=0$ or $(e^y-1)=0$.\n- $y^2-1 = (y-1)(y+1) = 0$ gives $y=1$ and $y=-1$.\n- $e^y-1=0$ gives $e^y=1$, which means $y=0$.\nThe equilibrium points are $y=-1, y=0,$ and $y=1$.\nAn equilibrium point $y_c$ is semi-stable if the function $f(y)$ has a root of even multiplicity at $y_c$, meaning $f(y)$ does not change sign across $y_c$. All the roots here are simple roots (multiplicity 1).\nLet's check the signs of $f(y)=(y^2-1)(e^y-1)$ in the intervals:\n- $y>1$: $(+)(+)=+$. Increasing.\n- $0<y<1$: $(-)(+)=-$. Decreasing. (Sign change at $y=1$, so it's unstable).\n- $-1<y<0$: $(-)(-)=+$. Increasing. (Sign change at $y=0$, so it's unstable).\n- $y<-1$: $(+)(-)=-$. Decreasing. (Sign change at $y=-1$, so it's stable).\nBased on this analysis, $y=1$ is unstable, $y=0$ is unstable, and $y=-1$ is stable. There are no points where the direction of the arrows is the same on both sides, so there are no semi-stable equilibrium points.

Question 9

1. -3
2. 0 (correct answer)
3. 3
4. $\infty$

Explanation: First, find the equilibrium points by setting $dy/dt = 0$. This gives $y^2(9-y^2) = 0$, so the equilibrium points are $y=-3, y=0, y=3$. Next, we construct a phase line by testing the sign of $dy/dt$ in the intervals between the equilibria.\n- If $y>3$ (e.g., $y=4$), $dy/dt = 16(9-16) < 0$ (decreasing).\n- If $0<y<3$ (e.g., $y=1$), $dy/dt = 1(9-1) > 0$ (increasing).\n- If $-3<y<0$ (e.g., $y=-1$), $dy/dt = 1(9-1) > 0$ (increasing).\n- If $y<-3$ (e.g., $y=-4$), $dy/dt = 16(9-16) < 0$ (decreasing).\nThe phase line shows that $y=3$ is stable, $y=0$ is semi-stable, and $y=-3$ is unstable. The initial condition is $y(0)=-2$, which is in the interval $(-3, 0)$. In this interval, $dy/dt > 0$, so the solution $y(t)$ is increasing. The solution will approach the next equilibrium point above its starting value, which is $y=0$. Therefore, $\lim_{t \to \infty} y(t) = 0$.

Question 10

1. Three equilibrium points; $y = 2$ is unstable since $f'(2) > 0$ where $f(y) = y^2(y^2 - 4)$ (correct answer)
2. Three equilibrium points; $y = 2$ is semi-stable since it has multiplicity 2 in the factorization
3. Five equilibrium points; $y = 2$ is stable since solutions approach it from the left
4. Three equilibrium points; $y = 2$ is semi-stable since $f'(2) = 0$ but $f''(2) \neq 0$

Explanation: Setting $\frac{dy}{dt} = y^2(y^2 - 4) = y^2(y-2)(y+2) = 0$ gives equilibria at $y = 0$ (with multiplicity 2), $y = 2$, and $y = -2$, for a total of 3 distinct equilibrium points. To classify $y = 2$, compute $f'(y) = \frac{d}{dy}[y^2(y^2-4)] = 2y(y^2-4) + y^2(2y) = 2y(y^2-4) + 2y^3 = 2y(2y^2-4) = 4y(y^2-2)$. At $y = 2$: $f'(2) = 4(2)(4-2) = 4(2)(2) = 16 > 0$, indicating $y = 2$ is unstable. Choice B is wrong because multiplicity in factorization doesn't automatically imply semi-stability; we need $f'(y_0) = 0$. Choice C is wrong about the number of equilibria. Choice D incorrectly states $f'(2) = 0$.

Question 11

1. The solution approaches $y = 1$ monotonically since it is the only stable equilibrium point
2. The solution decreases toward $y = -2$ since it is a stable equilibrium with even multiplicity
3. The solution approaches $y = -2$ asymptotically from above since $\frac{dy}{dt} < 0$ for $y \in (-2, 1)$ (correct answer)
4. The solution increases toward $y = 1$ since $\frac{dy}{dt} > 0$ for $y \in (-2, 1)$ and $y = 1$ attracts nearby solutions

Explanation: The equilibria are $y = 1$ (multiplicity 3) and $y = -2$ (multiplicity 2). For $y = 0.5 \in (-2, 1)$, we evaluate $\frac{dy}{dt} = (0.5-1)^3(0.5+2)^2 = (-0.5)^3(2.5)^2 = -0.125 \times 6.25 < 0$. Since $\frac{dy}{dt} < 0$, the solution decreases. For stability analysis: since $(y-1)^3$ has odd multiplicity, $y = 1$ allows flow across it (semi-stable or unstable depending on direction). Since $(y+2)^2$ has even multiplicity, $y = -2$ is semi-stable. Checking the sign throughout $(-2, 1)$: $(y-1)^3 < 0$ and $(y+2)^2 > 0$, so $\frac{dy}{dt} < 0$ for all $y \in (-2, 1)$. Therefore, starting at $y(0) = 0.5$, the solution decreases toward $y = -2$. Choice A incorrectly identifies $y=1$ as stable. Choice B gives wrong reasoning about even multiplicity implying stability. Choice D incorrectly states $\frac{dy}{dt} > 0$.

Question 12

1. Both solutions converge to the same equilibrium point $y = 2$ since their initial conditions are very close
2. $y_1(t) \to 0$ while $y_2(t) \to +\infty$, demonstrating sensitive dependence on initial conditions near the unstable equilibrium (correct answer)
3. Both solutions diverge to $+\infty$ since they start above the stable equilibrium at $y = 0$
4. $y_1(t) \to -2$ while $y_2(t) \to +2$, showing symmetric behavior about the unstable equilibrium point

Explanation: First find equilibria: $y^3 - 4y = y(y^2 - 4) = y(y-2)(y+2) = 0$, giving $y = -2, 0, 2$. Analyzing stability by checking signs: For $y \in (-2,0)$: $\frac{dy}{dt} > 0$ (moves toward 0). For $y \in (0,2)$: $\frac{dy}{dt} < 0$ (moves toward 0). For $y > 2$: $\frac{dy}{dt} > 0$ (moves toward $+\infty$). So $y = 0$ is stable, $y = \pm 2$ are unstable. Since $y_1(0) = 1.9 < 2$, the solution moves toward the stable equilibrium $y = 0$. Since $y_2(0) = 2.1 > 2$, the solution moves away from the unstable equilibrium $y = 2$ toward $+\infty$. This illustrates how small changes in initial conditions near an unstable equilibrium lead to vastly different long-term behaviors.

Question 13

1. $f(-1) > 0$ because solutions must flow away from the unstable equilibrium at $y = 0$
2. $f(-1)$ cannot be determined without additional information about the specific form of $f(y)$
3. $f(-1) = 0$ because $y = -1$ might be an additional equilibrium point not initially identified
4. $f(-1) < 0$ because solutions between stable and unstable equilibria flow toward the stable one (correct answer)

Explanation: When you encounter autonomous differential equations with phase line analysis, you need to understand how solution behavior connects to the sign of $f(y)$ between equilibria. The key insight is that solutions flow toward stable equilibria and away from unstable ones. Given the stability pattern (stable at $y = -2$, unstable at $y = 0$, stable at $y = 1$, unstable at $y = 3$), you can determine the direction of flow between equilibria. Since $f(0.5) = 2 > 0$, solutions move upward between the unstable equilibrium at $y = 0$ and stable equilibrium at $y = 1$, which makes sense—solutions flow away from unstable points toward stable ones. For $y = -1$, which lies between the stable equilibrium at $y = -2$ and unstable equilibrium at $y = 0$, solutions must flow toward the stable point at $y = -2$. This requires $\frac{dy}{dt} = f(y) < 0$, meaning solutions move downward. Therefore, $f(-1) < 0$. Choice A incorrectly focuses only on flowing away from $y = 0$ without considering the stable equilibrium at $y = -2$ that attracts nearby solutions. Choice B is wrong because the stability pattern and continuity of $f(y)$ provide sufficient information to determine the sign. Choice C misunderstands the problem—we're not looking for additional equilibria, and $f(-1) = 0$ would contradict the given equilibrium locations. Study tip: Always trace solution flow from the stability pattern. Between any two equilibria, solutions flow toward the stable one and away from the unstable one, which directly determines whether $f(y)$ is positive or negative in that interval.

Question 14

1. $P = 1$ thousand, since populations below this value cannot reach the carrying capacity
2. $P = 4$ thousand, since populations above the carrying capacity will decline to extinction
3. $P = 3$ thousand, since this represents the inflection point of the logistic growth model
4. $P = 2$ thousand, since this is the unstable equilibrium separating extinction from growth (correct answer)

Explanation: When analyzing population models with differential equations, phase line analysis helps you understand long-term behavior by examining equilibrium points and the direction of change between them. First, find the equilibrium points by setting $\frac{dP}{dt} = 0$: $P(P-2)(4-P) = 0$, giving us $P = 0$, $P = 2$, and $P = 4$. Next, determine the sign of $\frac{dP}{dt}$ in each interval by testing values:

• For $P < 0$: negative × negative × positive = positive (population increases)
• For $0 < P < 2$: positive × negative × positive = negative (population decreases)
• For $2 < P < 4$: positive × positive × positive = positive (population increases)
• For $P > 4$: positive × positive × negative = negative (population decreases)

Question 15

1. $y(t) \to 0$ because it is the nearest stable equilibrium point (correct answer)
2. $y(t) \to 2$ because it is the only stable equilibrium point below the initial condition
3. $y(t) \to 4$ because it is a semi-stable equilibrium point that attracts from below
4. The solution oscillates between $y = 0$ and $y = 2$ indefinitely without approaching any equilibrium

Explanation: First, find equilibrium points by setting $\frac{dy}{dt} = 0$: $y(y-2)(y-4)^2 = 0$, giving $y = 0, 2, 4$. To determine stability, analyze the sign of $f'(y) = \frac{d}{dy}[y(y-2)(y-4)^2]$ at each equilibrium. $f'(y) = (y-2)(y-4)^2 + y(y-4)^2 + 2y(y-2)(y-4) = (y-4)[(y-2)(y-4) + y(y-4) + 2y(y-2)] = (y-4)(3y^2-12y+8)$. At $y=0$: $f'(0) = (-4)(8) = -32 < 0$, so $y=0$ is stable. At $y=2$: $f'(2) = (-2)(-4) = 8 > 0$, so $y=2$ is unstable. At $y=4$: $f'(4) = 0$, requiring higher-order analysis, but since $(y-4)^2$ appears in the original equation, $y=4$ is semi-stable. From the phase line analysis, starting at $y=1.5 < 2$, the solution moves toward the stable equilibrium $y=0$.