Differential Equations Quiz: Qualitative Behavior Analysis

What this quiz covers

This quiz focuses on Qualitative Behavior 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. $a < 0$ (correct answer)
2. $0 < a < 1$
3. $a > 1$
4. $a < 1$

Explanation: The stability and type of the equilibrium at the origin are determined by the eigenvalues of the matrix $A$. The characteristic equation is $\det(A - \lambda I) = 0$, which is $(-1-\lambda)(-1-\lambda) - a = 0$, or $\lambda^2 + 2\lambda + (1-a) = 0$. The eigenvalues are given by the quadratic formula: $\lambda = \frac{-2 \pm \sqrt{4 - 4(1-a)}}{2} = -1 \pm \sqrt{1 - (1-a)} = -1 \pm \sqrt{a}$. Trajectories spiral when the eigenvalues are complex conjugates. This occurs when the term under the square root is negative, i.e., $a < 0$. In this case, the eigenvalues are $\lambda = -1 \pm i\sqrt{|a|}$. Since the real part is $-1 < 0$, the trajectories are stable spirals; they spiral into the origin.

Question 2

1. There is a unique solution given by $y(t) = 0$ for all $t$.
2. There are exactly two distinct solutions for $t \ge 1$.
3. No solution exists because the differential equation's right-hand side is not differentiable at $y=0$.
4. There are infinitely many solutions that are defined for all real numbers $t$. (correct answer)

Explanation: The function $f(y) = y^{2/3}$ is continuous for all $y$, so the existence theorem guarantees at least one solution exists. However, its partial derivative with respect to $y$, $\frac{\partial f}{\partial y} = \frac{2}{3}y^{-1/3}$, is not continuous (in fact, it's unbounded) at $y=0$. Therefore, the uniqueness theorem does not apply to the initial condition $y(1)=0$. One obvious solution is the equilibrium solution $y_1(t) = 0$ for all $t$. We can find other solutions by separation of variables: $\int y^{-2/3} dy = \int dt \implies 3y^{1/3} = t+C \implies y(t) = (\frac{t+C}{3})^3$. Applying the initial condition $y(1)=0$ gives $0 = (\frac{1+C}{3})^3$, so $C=-1$. This yields the solution $y_2(t) = (\frac{t-1}{3})^3$. We can construct infinitely many solutions by 'splicing' the zero solution with shifted versions of the non-trivial solution. For any constant $c \ge 1$, the function $y_c(t) = \begin{cases} 0 & t \le c \\ (\frac{t-c}{3})^3 & t > c \end{cases}$ is continuous, differentiable, satisfies the DE for all $t$, and satisfies $y(1)=0$. Since there are infinitely many choices for $c$, there are infinitely many solutions.

Question 3

1. All solutions approach $y=1$.
2. All solutions approach the curve $y=e^{-t}$.
3. All solutions approach $y=0$. (correct answer)
4. Some solutions approach $y=0$, while others grow infinitely.

Explanation: The equation is $y' = e^{-t} - y$. This is a non-autonomous linear equation. We can analyze its long-term behavior by considering the forcing term $e^{-t}$ and the homogeneous part $y' = -y$. As $t \to \infty$, the forcing term $e^{-t}$ decays to 0. The equation becomes asymptotically similar to $y' = -y$. The solutions to $y'=-y$ are of the form $y_h(t) = Ce^{-t}$, which all decay to 0 as $t \to \infty$. Because both the forcing term and the homogeneous solutions decay to zero, we can infer that all solutions to the full equation must also approach 0. Alternatively, consider the curve $g(t) = e^{-t}$. If $y(t) > g(t)$, then $y' = g(t) - y(t) < 0$, so the solution decreases. If $y(t) < g(t)$, then $y' > 0$, so the solution increases. Solutions are thus 'funneled' towards the curve $g(t)=e^{-t}$. Since $\lim_{t \to \infty} e^{-t} = 0$, all solutions must asymptotically approach 0.

Question 4

1. $\frac{dr}{dt} = r(1-r^2)$; trajectories spiral towards the origin, which is a stable equilibrium.
2. $\frac{dr}{dt} = r(1-r^2)$; trajectories spiral away from the origin towards a limit cycle at $r=1$. (correct answer)
3. $\frac{dr}{dt} = r(1-r^4)$; trajectories spiral away from the origin towards a limit cycle at $r=1$.
4. $\frac{dr}{dt} = r^2(1-r^2)$; trajectories move directly away from the origin along straight lines.

Explanation: When analyzing nonlinear systems with radial symmetry, converting to polar coordinates often reveals the radial behavior more clearly. This system has the form that suggests examining how the distance from the origin changes over time. To find $\frac{dr}{dt}$, use the chain rule: $\frac{dr}{dt} = \frac{d}{dt}\sqrt{x^2+y^2} = \frac{xx' + yy'}{\sqrt{x^2+y^2}} = \frac{xx' + yy'}{r}$ Substituting the given equations: $\frac{dr}{dt} = \frac{x[y + x(1-x^2-y^2)] + y[-x + y(1-x^2-y^2)]}{r}$ Expanding: $= \frac{xy + x^2(1-r^2) - yx + y^2(1-r^2)}{r} = \frac{(x^2+y^2)(1-r^2)}{r} = \frac{r^2(1-r^2)}{r} = r(1-r^2)$ This differential equation tells us that when $r < 1$, we have $\frac{dr}{dt} = r(1-r^2) > 0$, so trajectories move away from the origin. When $r > 1$, $\frac{dr}{dt} < 0$, so trajectories move inward. At $r = 1$, $\frac{dr}{dt} = 0$, indicating a limit cycle. Choice A incorrectly interprets the stability—trajectories don't spiral toward the origin. Choice C has the wrong differential equation ($r(1-r^4)$ instead of $r(1-r^2)$). Choice D also has an incorrect equation and wrongly suggests straight-line motion rather than spiraling. The answer is B: trajectories spiral away from the unstable origin toward the stable limit cycle at $r = 1$. Remember: when converting to polar coordinates, always check both the radial equation and the angular behavior to fully understand the system's dynamics.

Question 5

1. The origin is stable, but not necessarily asymptotically stable.
2. The origin is unstable.
3. The origin is asymptotically stable. (correct answer)
4. No conclusion can be drawn without solving the system.

Explanation: To use the Lyapunov function $V(x,y) = x^2 + 2y^2$, we first check if it is positive definite. For any $(x,y) \neq (0,0)$, $x^2 \ge 0$ and $y^2 \ge 0$, so $V(x,y) > 0$. At $(0,0)$, $V(0,0)=0$. Thus, $V$ is positive definite. Next, we compute the time derivative of $V$ along the system's trajectories: $\dot{V} = \frac{\partial V}{\partial x}x' + \frac{\partial V}{\partial y}y' = (2x)(-x - 2y^2) + (4y)(xy - y^3) = -2x^2 - 4xy^2 + 4xy^2 - 4y^4 = -2x^2 - 4y^4$. We must check if $\dot{V}$ is negative definite. For any $(x,y) \neq (0,0)$, either $x \neq 0$ or $y \neq 0$, so $-2x^2 - 4y^4 < 0$. At $(0,0)$, $\dot{V}=0$. Thus, $\dot{V}$ is negative definite. By Lyapunov's stability theorem, since there exists a positive definite function $V$ whose derivative $\dot{V}$ is negative definite in a neighborhood of the origin, the origin is an asymptotically stable equilibrium point.

Question 6

1. There is one stable equilibrium and one unstable equilibrium.
2. There are two stable equilibria and one unstable equilibrium.
3. There is only one equilibrium, which is unstable.
4. There is one stable equilibrium and two unstable equilibria. (correct answer)

Explanation: Equilibrium points are found by solving $y' = y(y^2 - 4y + r) = 0$. One equilibrium is always $y_1=0$. The others are roots of $y^2 - 4y + r = 0$, which are $y = \frac{4 \pm \sqrt{16-4r}}{2} = 2 \pm \sqrt{4-r}$. When $r<4$, the term under the square root is positive, so there are two distinct real roots, $y_2 = 2 - \sqrt{4-r}$ and $y_3 = 2 + \sqrt{4-r}$. Thus, for $r<4$, there are three equilibria. To determine stability, let $f(y) = y^3 - 4y^2 + ry$. Then $f'(y) = 3y^2 - 8y + r$. For $r \approx 3.9 > 0$, $f'(0) = r > 0$, so $y_1=0$ is unstable. For the other two equilibria, we know they are the roots of $g(y) = 3y^2-8y+r$. The roots of $f'(y)=0$ are $y = \frac{8 \pm \sqrt{64-12r}}{6}$. Since $r<4$, these are real. Let $y_2$ and $y_3$ be the equilibria. $f'(y_2) = (y_2-y_3)(y_2-y_1)$ where $y_1, y_2, y_3$ are the roots of $f(y)$. A simpler approach is to sketch $f(y)$ for $r=3.9$. It's a cubic with roots at $0$, $2-\epsilon$, and $2+\epsilon$. As $y \to \infty$, $f(y) \to \infty$. The signs of $f(y)$ are: $(-)$ for $y<0$, $(+)$ for $0<y<2-\epsilon$, $(-)$ for $2-\epsilon<y<2+\epsilon$, and $(+)$ for $y>2+\epsilon$. This implies $y=0$ is unstable (slope changes from - to +), $y=2-\epsilon$ is stable (+ to -), and $y=2+\epsilon$ is unstable (- to +). Therefore, there is one stable and two unstable equilibria.

Question 7

1. $\lim_{t \to \infty} y(t) = 1$
2. $\lim_{t \to \infty} y(t) = 3$ (correct answer)
3. $\lim_{t \to \infty} y(t) = -1$
4. $\lim_{t \to \infty} y(t) = \infty$

Explanation: The equilibrium points of the differential equation are the roots of $f(y) = (y^2 - 1)(y - 3)^2 = 0$, which are $y = -1, 1, 3$. To determine the behavior of the solution, we analyze the sign of $\frac{dy}{dt}$ in the intervals between these equilibria. The initial condition is $y(0) = 1.1$, which is in the interval $(1, 3)$. Let's pick a test point in this interval, such as $y=2$. At $y=2$, $\frac{dy}{dt} = (2^2 - 1)(2 - 3)^2 = (3)(-1)^2 = 3 > 0$. Since $\frac{dy}{dt} > 0$ for $y \in (1, 3)$, the solution $y(t)$ is strictly increasing on this interval. Therefore, starting from $y(0) = 1.1$, the solution will increase and approach the next equilibrium point, which is $y = 3$. So, $\lim_{t \to \infty} y(t) = 3$.

Question 8

1. $\alpha = 3$
2. $\alpha = 6$
3. $\alpha = 0$
4. $\alpha = 9$ (correct answer)

Explanation: When analyzing bifurcations in differential equations, you're looking for parameter values where the qualitative behavior of the system changes dramatically—typically when equilibrium points appear, disappear, or change stability. To find equilibrium solutions, set $\frac{dy}{dt} = 0$, giving us $y^2 - 6y + \alpha = 0$. Using the quadratic formula: $y = \frac{6 \pm \sqrt{36 - 4\alpha}}{2} = 3 \pm \sqrt{9 - \alpha}$ The key insight is that equilibrium points exist only when the discriminant $9 - \alpha \geq 0$, or $\alpha \leq 9$. When $\alpha < 9$, we have two distinct equilibria. When $\alpha = 9$, the discriminant equals zero, giving us exactly one equilibrium point (a repeated root) at $y = 3$. When $\alpha > 9$, no real equilibria exist. This dramatic change from two equilibria to zero equilibria as $\alpha$ passes through 9 represents a saddle-node bifurcation, making $\alpha = 9$ our bifurcation value. Choice A ($\alpha = 3$) gives equilibria at $y = 3 \pm \sqrt{6}$—two distinct points with no special behavior. Choice B ($\alpha = 6$) yields equilibria at $y = 3 \pm \sqrt{3}$—again, just two regular equilibria. Choice C ($\alpha = 0$) produces equilibria at $y = 0$ and $y = 6$—two equilibria, nothing unusual. Remember: bifurcations often occur when discriminants equal zero or when equilibria collide and disappear. Always check what happens to the number and nature of equilibria as your parameter varies.

Question 9

1. The solution changes from increasing to decreasing, indicating a local maximum.
2. The solution changes from decreasing to increasing, indicating a local minimum. (correct answer)
3. The solution has an inflection point where it crosses the ellipse.
4. The solution maintains a constant slope as it crosses the ellipse.

Explanation: The sign of $\frac{dy}{dt}$ determines whether a solution is increasing or decreasing. A local extremum occurs where $\frac{dy}{dt} = 0$. From the equation, $\frac{dy}{dt} = 0$ when $t^2 + 4y^2 - 4 = 0$, which is exactly the equation of the ellipse.

• Inside the ellipse, $t^2 + 4y^2 < 4$, which means $t^2 + 4y^2 - 4 < 0$. So, $\frac{dy}{dt} < 0$ and the solution is decreasing.
• Outside the ellipse, $t^2 + 4y^2 > 4$, which means $t^2 + 4y^2 - 4 > 0$. So, $\frac{dy}{dt} > 0$ and the solution is increasing. When a solution curve crosses the ellipse from the inside to the outside, its derivative $\frac{dy}{dt}$ changes from negative to positive. A function that changes from decreasing to increasing has a local minimum.

Question 10

1. The region defined by $y > \frac{1}{2}t$
2. The region defined by $y < \frac{1}{2}t$
3. The region defined by $y > \frac{1}{2}t - \frac{1}{4}$
4. The region defined by $y < \frac{1}{2}t - \frac{1}{4}$ (correct answer)

Explanation: Concavity is determined by the sign of the second derivative, $\frac{d^2y}{dt^2}$. Let $f(t,y) = t - 2y$. We calculate the second derivative using the multivariable chain rule: $\frac{d^2y}{dt^2} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial y} \frac{dy}{dt}$. The partial derivatives are $\frac{\partial f}{\partial t} = 1$ and $\frac{\partial f}{\partial y} = -2$. Substituting these and the expression for $\frac{dy}{dt}$ gives: $\frac{d^2y}{dt^2} = 1 + (-2)(t - 2y) = 1 - 2t + 4y$. The solution curves are concave down where $\frac{d^2y}{dt^2} < 0$. $1 - 2t + 4y < 0 \implies 4y < 2t - 1 \implies y < \frac{1}{2}t - \frac{1}{4}$.

Question 11

1. $H = 13$ (correct answer)
2. $H = 12$
3. $H = 25$
4. $H = 50$

Explanation: When you encounter a logistic equation with harvesting, you're analyzing the balance between natural population growth and external removal. The key insight is finding the critical harvesting rate where equilibrium points disappear, guaranteeing extinction regardless of initial conditions. Start by finding equilibrium points where $\frac{dP}{dt} = 0$. Setting the equation equal to zero: $0.5P(1 - P/100) - H = 0$, which rearranges to $0.5P - 0.005P^2 = H$. The right side represents the natural growth rate at population P, while H is the constant harvesting rate. For extinction to be guaranteed, the harvesting rate must exceed the maximum possible growth rate. To find this maximum, take the derivative: $\frac{d}{dP}(0.5P - 0.005P^2) = 0.5 - 0.01P = 0$. This gives $P = 50$ thousand as the population where growth is maximized. The maximum growth rate is $0.5(50) - 0.005(50)^2 = 25 - 12.5 = 12.5$ thousand per year. Therefore, any harvesting rate above 12.5 guarantees extinction. The smallest integer value satisfying this is 13. Choice A ($H = 13$) is correct because it's the smallest integer above the critical threshold. Choice B ($H = 12$) fails because it's below 12.5, allowing two equilibrium points to exist. Choice C ($H = 25$) equals the maximum growth rate at zero harvesting, but that's not the relevant threshold here. Choice D ($H = 50$) would certainly cause extinction but isn't the smallest such integer. Remember: for harvesting problems, find where the harvest rate equals the maximum natural growth rate to identify the extinction threshold.

Question 12

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

Explanation: The concavity of a solution curve $y(t)$ is determined by the sign of its second derivative, $\frac{d^2y}{dt^2}$. For an autonomous equation $\frac{dy}{dt} = f(y)$, we use the chain rule: $\frac{d^2y}{dt^2} = \frac{d}{dt}(f(y)) = f'(y) \frac{dy}{dt} = f'(y)f(y)$. Here, $f(y) = y(2-y) = 2y - y^2$. The first derivative is $f'(y) = 2 - 2y$. Thus, $\frac{d^2y}{dt^2} = (2 - 2y)(2y - y^2) = 2(1-y)y(2-y)$. The solution is concave up where $\frac{d^2y}{dt^2} > 0$. The roots of this expression are $y=0, 1, 2$. We test the intervals:

• For $y>2$: $2(-)(+)(-) > 0$ (Concave up).
• For $1<y<2$: $2(-)(+)(+) < 0$ (Concave down).
• For $0<y<1$: $2(+)(+)(+) > 0$ (Concave up).
• For $y<0$: $2(+)(-)(+) < 0$ (Concave down). Therefore, the solution curves are concave up for $y \in (0, 1) \cup (2, \infty)$.

Question 13

1. $y(t) \ge \frac{1}{2}t^2 - t + \pi/2$ (correct answer)
2. $y(t) \ge \frac{1}{2}t^2 + \pi/2$
3. $y(t) \le \frac{1}{2}t^2 + t + \pi/2$
4. $y(t) \ge t - 1$

Explanation: When you encounter a differential equation problem asking for bounds without solving, you're being tested on comparison techniques and understanding how differential inequalities work. The key insight is that if you have two functions with the same initial condition, and one has a consistently smaller derivative than the other, then that function will remain below the other for all future time. Here, we have $\frac{dy}{dt} = t + \cos(y)$ with $y(0) = \pi/2$. Since $\cos(y) \geq -1$ for all $y$, we know that $\frac{dy}{dt} = t + \cos(y) \geq t - 1$. Now consider the auxiliary function $w(t) = \frac{1}{2}t^2 - t + \pi/2$ with $w(0) = \pi/2$. Its derivative is $\frac{dw}{dt} = t - 1$. Since $\frac{dy}{dt} \geq \frac{dw}{dt}$ and both functions start at the same point, we have $y(t) \geq w(t)$ for all $t > 0$. Choice A is correct because it represents this valid lower bound. Choice B fails because it uses $\cos(y) \geq 0$, which isn't always true. Choice C gives an upper bound (note the $\leq$ symbol), not a lower bound. Choice D represents $y(t) \geq t - 1$, which would only be valid if $y$ were linear, but our function curves upward faster due to the $\cos(y)$ term. Strategy tip: For differential inequality problems, always look for the extreme values of variable terms (like $\cos(y)$ ranging from -1 to 1) to construct comparison equations you can integrate directly.

Question 14

1. $y = 0$
2. $y = 3$
3. $y = \sqrt{3}$ (correct answer)
4. $y = \sqrt{6}$

Explanation: An inflection point on a solution curve $y(t)$ occurs where $\frac{d^2y}{dt^2} = 0$ and the concavity changes. For a non-constant solution, we must have $\frac{dy}{dt} \neq 0$. For an autonomous equation $\frac{dy}{dt} = f(y)$, we have $\frac{d^2y}{dt^2} = f'(y)f(y)$. For an inflection point, we need $f'(y)f(y) = 0$ and $f(y) \neq 0$. This implies that we must have $f'(y) = 0$ and $f(y) \neq 0$. Here, $f(y) = y^3 - 9y$. The equilibrium points where $f(y) = 0$ are $y(y^2-9)=0$, so $y=0, 3, -3$. At these points, $\frac{dy}{dt}=0$, so they cannot be inflection points for a non-constant solution. Next, we find $f'(y) = 3y^2 - 9$. Setting $f'(y) = 0$ gives $3y^2 - 9 = 0$, so $y^2 = 3$, which means $y = \pm\sqrt{3}$. At $y = \sqrt{3}$, $f(\sqrt{3}) = (\sqrt{3})^3 - 9\sqrt{3} = 3\sqrt{3} - 9\sqrt{3} = -6\sqrt{3} \neq 0$. Since $f'(\sqrt{3})=0$ and $f(\sqrt{3})\neq 0$, an inflection point can occur at $y = \sqrt{3}$.

Question 15

1. Solutions have horizontal asymptotes approaching $y = 2$ as $x \to \pm\infty$ with monotonic approach
2. Solutions increase without bound as $x$ increases, with vertical asymptotes at finite $x$-values where $y \to \infty$ (correct answer)
3. Solutions oscillate around $y = 2$ with decreasing amplitude as $x \to \infty$ due to the damping effect
4. Solutions decrease monotonically toward $y = 2$ but never cross this horizontal line in finite $x$-intervals

Explanation: For $y > 2$, we have $y^2 - 4 > 0$ and $x^2 + 1 > 0$, so $\frac{dy}{dx} > 0$. This means $y$ is always increasing when $y > 2$. Since there's no mechanism to bound the growth (the right side grows as $y^2$ while the denominator remains bounded), solutions will increase without bound. The equation can be separated: $\frac{dy}{y^2-4} = \frac{dx}{x^2+1}$, and the left side integral diverges as $y \to \infty$, confirming blow-up in finite $x$.

Question 16

1. The origin is always a stable focus regardless of the value of $\mu$ due to the linear restoring force
2. The stability depends on $\mu$: stable focus for small $\mu$ and unstable focus for large $\mu$ values
3. The origin is a center with purely periodic solutions in its neighborhood for any positive $\mu$ value
4. The origin is an unstable focus, and the system admits a stable limit cycle surrounding the origin (correct answer)

Explanation: When analyzing the stability of nonlinear differential equations, you need to examine both the linearized behavior near equilibria and the global dynamics. This Van der Pol-type oscillator requires phase plane analysis to understand how the nonlinear damping term $\mu(y^2 - 1)\frac{dy}{dt}$ affects system behavior. Converting to the phase plane system: $\frac{dx}{dt} = v$ and $\frac{dv}{dt} = -\mu(x^2 - 1)v - x$. At the origin $(0,0)$, linearizing gives $\frac{d}{dt}\begin{pmatrix}x\\v\end{pmatrix} = \begin{pmatrix}0 & 1\\\-1 & \mu\end{pmatrix}\begin{pmatrix}x\\v\end{pmatrix}$. The characteristic equation $\lambda^2 - \mu\lambda + 1 = 0$ has roots $\lambda = \frac{\mu \pm \sqrt{\mu^2 - 4}}{2}$. For $\mu > 0$, these have positive real parts, making the origin an unstable focus. The key insight is the nonlinear damping: when $|y| < 1$, we have $y^2 - 1 < 0$, so the damping term opposes motion (energy increases). When $|y| > 1$, the damping becomes positive (energy decreases). This creates a stable limit cycle where trajectories spiral outward from the unstable origin and inward from large amplitudes. Option A ignores the destabilizing linear term $\mu v$. Option B incorrectly suggests the stability changes with $\mu$ - the origin remains unstable for any $\mu > 0$. Option C misses that the damping term breaks the conservative structure needed for a center. Study tip: For Van der Pol oscillators, always check the sign of the damping coefficient in the linearization - positive damping at the origin typically indicates an unstable equilibrium with possible limit cycles.

Question 17

1. When $h = r$, the carrying capacity equilibrium collides with the extinction equilibrium and they exchange stability (correct answer)
2. When $h = \frac{r}{2}$, a new equilibrium appears through a saddle-node bifurcation at the midpoint $y = \frac{K}{2}$
3. When $h = \frac{r}{4}$, the system transitions from having two stable equilibria to having oscillatory solutions
4. When $h = rK$, both equilibria become unstable and solutions exhibit unbounded growth or extinction

Explanation: The equation becomes $\frac{dy}{dt} = y(r - \frac{ry}{K} - h) = y(r - h - \frac{ry}{K})$. Equilibria: $y = 0$ and $y = K(1 - \frac{h}{r})$. The second equilibrium exists and is positive when $h < r$. At $h = r$, the second equilibrium becomes $y = 0$, coinciding with the trivial equilibrium. For $h < r$: $y = 0$ is unstable, $y = K(1-\frac{h}{r})$ is stable. For $h > r$: only $y = 0$ exists and is stable. At $h = r$, the equilibria collide and exchange stability - this is the definition of a transcritical bifurcation.

Question 18

1. A transcritical bifurcation where the origin changes stability and a new equilibrium branch appears
2. A pitchfork bifurcation creating two symmetric equilibria that branch from the origin when $\mu > 0$
3. A Hopf bifurcation where the origin changes from a stable to unstable spiral and a limit cycle is born (correct answer)
4. A saddle-node bifurcation where two equilibria collide and annihilate each other at the critical parameter value

Explanation: When analyzing nonlinear systems with parameters, you need to identify how equilibria and their stability change as the parameter varies. This system has cubic nonlinearities that suggest potential limit cycle behavior. First, find the equilibria by setting both derivatives to zero. The origin $(0,0)$ is always an equilibrium. To analyze stability, linearize around the origin by examining the Jacobian matrix of the linear terms: $J = \begin{pmatrix} \mu & -1 \\ 1 & \mu \end{pmatrix}$. The eigenvalues are $\lambda = \mu \pm i$, giving complex eigenvalues with real part $\mu$. For $\mu < 0$, the real part is negative, making the origin a stable spiral. For $\mu > 0$, the real part becomes positive, making it an unstable spiral. At $\mu = 0$, we have purely imaginary eigenvalues, which is the hallmark condition for a Hopf bifurcation. The cubic terms $-x(x^2 + y^2)$ and $-y(x^2 + y^2)$ act as nonlinear damping that becomes significant away from the origin. When $\mu > 0$ makes the origin unstable, trajectories spiral outward until this nonlinear damping balances the linear instability, creating a stable limit cycle. Option A is wrong because transcritical bifurcations involve equilibrium branches crossing, not limit cycles. Option B is incorrect since pitchfork bifurcations create additional equilibrium points, not periodic orbits. Option D describes saddle-node behavior with equilibrium collision, which doesn't occur here. Study tip: Hopf bifurcations occur when complex eigenvalues cross the imaginary axis (real part changes sign) combined with appropriate nonlinear terms that can sustain periodic motion.

Question 19

1. The origin is a stable spiral with all trajectories eventually converging to it regardless of initial conditions
2. The origin is an unstable spiral, the unit circle is a stable limit cycle attracting all non-zero trajectories (correct answer)
3. Both the origin and unit circle are unstable, with trajectories escaping to infinity for most initial conditions
4. The system has a stable limit cycle at radius $r = 2$ with the unit circle being an unstable periodic orbit

Explanation: Converting to polar coordinates: $r' = r(r^2 - 1)$ and $\theta' = 1$. For $r' = r(r^2 - 1)$: when $0 < r < 1$, $r' < 0$ (trajectories move inward); when $r > 1$, $r' > 0$ (trajectories move outward); when $r = 1$, $r' = 0$ (unit circle is invariant). The origin ($r = 0$) is unstable since nearby trajectories move away. The unit circle ($r = 1$) is stable since trajectories approach it from both inside and outside. $\theta' = 1 > 0$ means clockwise rotation, confirming spiral behavior.

Question 20

1. Solutions decrease monotonically and approach $y = 0$ asymptotically as $x \to \infty$ without reaching it (correct answer)
2. Solutions reach $y = 0$ in finite time, creating a boundary layer phenomenon at this value
3. Solutions initially decrease but then increase without bound after reaching a minimum positive value
4. Solutions oscillate with decreasing amplitude around $y = \frac{1}{e}$ due to the logarithmic term

Explanation: For $0 < y < 1$, we have $\ln|y| = \ln(y) < 0$ since $y < 1$. Therefore $\frac{dy}{dx} = y\ln(y) < 0$, so $y$ decreases. Separating variables: $\frac{dy}{y\ln(y)} = dx$. Let $u = \ln(y)$, so $du = \frac{dy}{y}$ and $\frac{du}{u} = dx$. Integrating: $\ln|u| = x + C$, so $\ln|\ln(y)| = x + C$. This gives $|\ln(y)| = Ae^x$. Since $\ln(y) < 0$ for our case, $\ln(y) = -Ae^x$ where $A > 0$. Thus $y = e^{-Ae^x}$. As $x \to \infty$, $y \to 0$ but never reaches it in finite time.