Phase Line Analysis - Differential Equations

How many semi-stable equilibrium points does the equation $dy/dt = (y^2-1)(e^y-1)$ have?
Answer: 0 First, find the equilibrium points by setting $dy/dt=0$ . This occurs when $(y^2-1)=0$ or $(e^y-1)=0$ . - $y^2-1 = (y-1)(y+1) = 0$ gives $y=1$ and $y=-1$ . - $e^y-1=0$ gives $e^y=1$ , which means $y=0$ . The equilibrium points are $y=-1, y=0,$ and $y=1$ . An 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). Let's check the signs of $f(y)=(y^2-1)(e^y-1)$ in the intervals: - $y>1$ : $(+)(+)=+$ . Increasing. - $0<y<1$ : $(-)(+)=-$ . Decreasing. (Sign change at $y=1$ , so it's unstable). - $-1<y<0$ : $(-)(-)=+$ . Increasing. (Sign change at $y=0$ , so it's unstable). - $y<-1$ : $(+)(-)=-$ . Decreasing. (Sign change at $y=-1$ , so it's stable). Based 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.
For $\frac{dy}{dt} = f(y)$ where $f(y)$ is continuous, suppose the phase line shows equilibria at $y = -2, 0, 1, 3$ with stability pattern: stable, unstable, stable, unstable (respectively). If $f(0.5) = 2$ , what can be concluded about $f(-1)$ ?
Answer: $f(-1) < 0$ because solutions between stable and unstable equilibria flow toward the stable one 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.
For the autonomous system $dy/dt = y^2 - y^3$ , non-constant solutions $y(t)$ have an inflection point when $y$ is equal to which value?
Answer: $y = 2/3$ 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.
Consider the autonomous differential equation $\frac{dy}{dt} = y(y-2)(y-4)^2$ . If a solution curve starts at $y(0) = 1.5$ , what is the long-term behavior of $y(t)$ as $t \to \infty$ ?
Answer: $y(t) \to 0$ because it is the nearest stable equilibrium point 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 $y=0$ .
Consider the family of differential equations $dy/dt = y^2 - ay + 9$ , where $a$ is a real parameter. A bifurcation occurs when a qualitative change in the number of equilibrium points happens. What is the sum of the values of $a$ for which a bifurcation occurs?
Answer: 0 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. The discriminant $D$ is given by $D = b^2 - 4ac = (-a)^2 - 4(1)(9) = a^2 - 36$ . Setting 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 $6 + (-6) = 0$ .
For the differential equation $dy/dt = y^3 - 12y + \alpha$ , bifurcations occur at two distinct values of the parameter $\alpha$ . What is the positive value of $y$ at which one of these bifurcations occurs?
Answer: 2 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 $f'(y)=0$ . First, find the values of $y$ where $f'(y)=0$ . The derivative is $f'(y) = 3y^2 - 12$ . Setting $f'(y)=0$ 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. The 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 $y=-2$ , $-8+24+\alpha=0 \implies \alpha=-16$ .)
A solution $y(t)$ to the differential equation $dy/dt = y^2(9-y^2)$ has the initial condition $y(0) = -2$ . What is the value of $\lim_{t \to \infty} y(t)$ ?
Answer: 0 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. - If $y>3$ (e.g., $y=4$ ), $dy/dt = 16(9-16) < 0$ (decreasing). - If $0<y<3$ (e.g., $y=1$ ), $dy/dt = 1(9-1) > 0$ (increasing). - If $-3<y<0$ (e.g., $y=-1$ ), $dy/dt = 1(9-1) > 0$ (increasing). - If $y<-3$ (e.g., $y=-4$ ), $dy/dt = 16(9-16) < 0$ (decreasing). The 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, $\lim_{t \to \infty} y(t) = 0$ .
Consider the phase line for $\frac{dy}{dt} = y^3 - 4y$ . If two solutions start at $y_1(0) = 1.9$ and $y_2(0) = 2.1$ , what can be concluded about their long-term behaviors?
Answer: $y_1(t) \to 0$ while $y_2(t) \to +\infty$ , demonstrating sensitive dependence on initial conditions near the unstable equilibrium 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.
Consider the autonomous system $\frac{dy}{dt} = (y-1)^3(y+2)^2$ . If the solution curve passes through the point $(0, 0.5)$ , which statement best describes the solution's behavior as $t$ increases?
Answer: The solution approaches $y = -2$ asymptotically from above since $\frac{dy}{dt} < 0$ for $y \in (-2, 1)$ 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 $\frac{dy}{dt} > 0$ .
For the equation $\frac{dy}{dt} = y^2(y^2 - 4)$ , determine the number of equilibrium points and classify the stability of the equilibrium at $y = 2$ .
Answer: Three equilibrium points; $y = 2$ is unstable since $f'(2) > 0$ where $f(y) = y^2(y^2 - 4)$ 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 $f'(2) = 0$ .
An autonomous system $dy/dt=f(y)$ has exactly two equilibrium points: an unstable equilibrium at $y=2$ and a stable equilibrium at $y=8$ . Which of the following is a possible value for $f(5)$ ?
Answer: 4 The information about the stability of the equilibrium points allows us to determine the sign of $f(y)$ in the intervals between them. - 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. - 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. Both 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.
A fish population $P(t)$ in a lake is modeled by the logistic equation with an Allee effect: $dP/dt = 0.2 P(1 - P/1000)(P/200 - 1)$ . There is a critical population threshold below which the population will decline to extinction. What is this threshold value?
Answer: 200 First, we find the equilibrium points of the population model by setting $dP/dt = 0$ . The factors give the equilibria: - $P=0$ (Extinction) - $1 - P/1000 = 0 \implies P = 1000$ (Carrying capacity) - $P/200 - 1 = 0 \implies P = 200$ (Allee threshold) The 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: - For $P > 1000$ : $dP/dt = (+)(-)(+) < 0$ (decreasing). The population declines towards 1000. - For $200 < P < 1000$ : $dP/dt = (+)(+)(+) > 0$ (increasing). The population grows towards 1000. - For $0 < P < 200$ : $dP/dt = (+)(+)(-) < 0$ (decreasing). The population declines towards 0 (extinction). From 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.
A population model is governed by $\frac{dP}{dt} = P(P-2)(4-P)$ where $P$ represents population in thousands. Based on phase line analysis, which initial population value represents the threshold below which the population will become extinct?
Answer: $P = 2$ thousand, since this is the unstable equilibrium separating extinction from growth 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) This reveals that $P = 0$ and $P = 4$ are stable equilibria (arrows point toward them), while $P = 2$ is unstable (arrows point away). The critical insight: populations starting below $P = 2$ will decline toward extinction ( $P = 0$ ), while those starting above $P = 2$ will grow toward the carrying capacity ( $P = 4$ ). Choice A incorrectly identifies $P = 1$ as significant when it's just a value in the declining region. Choice B misunderstands that $P = 4$ is the carrying capacity, not an extinction threshold. Choice C incorrectly references inflection points, which aren't relevant for determining stability. Choice D correctly identifies $P = 2$ as the unstable equilibrium that separates extinction from survival. Remember: in phase line analysis, unstable equilibria often represent critical thresholds that determine a system's ultimate fate.
The population $P(t)$ of a species is modeled by the differential equation $dP/dt = (P-3)^2(P-1)$ . Which statement accurately describes the long-term behavior of the population?
Answer: An initial population of $P(0)=2$ will increase and approach the equilibrium at $P=3$ . 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) With initial population $P(0) = 2$ , we're in the middle region where the population increases toward $P = 3$ , making answer D correct. Answer A is wrong because $P = 3$ is unstable—populations above 3 continue growing indefinitely. Answer B incorrectly suggests a population starting at 4 would decrease, but our analysis shows it increases. Answer C mischaracterizes $P = 1$ as stable when it's actually unstable—populations slightly above 1 grow toward 3, while those below 1 decline toward extinction. Remember: For autonomous differential equations, sketch the direction field by testing the sign of the derivative in each region between equilibrium points. This reveals the flow pattern and stability behavior.
For the logistic equation $dy/dt = y(2-y)$ , the solution curves $y(t)$ are concave up on which of the following intervals for $y$ ?
Answer: $y \in (0, 1) \cup (2, \infty)$ Concavity of solution curves $y(t)$ is determined by the sign of the second derivative, $y''(t)$ . We use the chain rule: $y'' = \frac{d}{dt}(y') = \frac{d(f(y))}{dy} \frac{dy}{dt} = f'(y)f(y)$ . Given $f(y) = y(2-y) = 2y-y^2$ . The first derivative is $f'(y) = 2-2y$ . So, $y'' = (2-2y)(2y-y^2) = 2(1-y)y(2-y)$ . We 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: - For $y>2$ : $y''>0$ since $(1-y)<0$ , $y>0$ , and $(2-y)<0$ . The product of three negatives is positive. Concave up. - For $1<y<2$ : $y''<0$ since $(1-y)<0$ , $y>0$ , and $(2-y)>0$ . The product is negative. Concave down. - For $0<y<1$ : $y''>0$ since $(1-y)>0$ , $y>0$ , and $(2-y)>0$ . The product is positive. Concave up. - For $y<0$ : $y''<0$ since $(1-y)>0$ , $y<0$ , and $(2-y)>0$ . The product is negative. Concave down. Thus, the solution curves are concave up for $y \in (0, 1) \cup (2, \infty)$ .

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