Qualitative Behavior Analysis - Differential Equations

1. Consider the equation $\frac{dy}{dt} = ry\left(1 - \frac{y}{K}\right) - hy$ where $r, K, h > 0$ represent growth rate, carrying capacity, and harvesting rate respectively. For what relationship between the parameters does the system exhibit a transcritical bifurcation?
   Answer: When $h = r$, the carrying capacity equilibrium collides with the extinction equilibrium and they exchange stability 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.
2. Consider the differential equation $\frac{dy}{dt} = t^2 + 4y^2 - 4$. What can be said about a solution curve $y(t)$ as it crosses the ellipse $t^2 + 4y^2 = 4$ from the region inside the ellipse to the region outside?
   Answer: The solution changes from decreasing to increasing, indicating a local minimum. 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.
3. Analyze the system $\frac{dx}{dt} = y$, $\frac{dy}{dt} = -\sin(x) - \delta y$ where $\delta > 0$ represents damping. This models a damped pendulum. What can be concluded about the phase portrait structure?
   Answer: Equilibria alternate between stable spirals and saddle points, with separatrices dividing regions of attraction When analyzing nonlinear dynamical systems like the damped pendulum, you need to identify equilibria and classify their stability to understand the overall phase portrait structure. The equilibria occur where both derivatives equal zero: $\frac{dx}{dt} = y = 0$ and $\frac{dy}{dt} = -\sin(x) - \delta y = 0$. This gives us equilibrium points at $(n\pi, 0)$ for integer $n$, corresponding to the pendulum hanging down ($n$ even) or balanced upside-down ($n$ odd). To classify stability, you linearize around each equilibrium. The Jacobian matrix is $\begin{pmatrix} 0 & 1 \\ -\cos(x) & -\delta \end{pmatrix}$. At points like $(0,0)$ where $\cos(0) = 1$, the eigenvalues have negative real parts (since $\delta > 0$), making these stable spirals. At points like $(\pi,0)$ where $\cos(\pi) = -1$, you get one positive and one negative eigenvalue, creating saddle points. This alternating pattern continues for all equilibria. The separatrices (stable/unstable manifolds of the saddle points) divide the phase plane into basins of attraction for different stable equilibria. Option A is wrong because not all equilibria are stable—saddle points are unstable. Option B is incorrect because damped systems don't have limit cycles; energy dissipation prevents sustained oscillations. Option D fails because trajectories aren't periodic in the presence of damping—they spiral into equilibria. Remember: In nonlinear systems, always find equilibria first, then linearize to classify stability. The global behavior emerges from understanding how these local behaviors connect.
4. A system of differential equations is defined by $x' = y$ and $y' = \sin(x)$. Which of the following statements correctly describes the qualitative behavior of its phase portrait?
   Answer: The system has equilibria that are centers surrounded by periodic orbits, and other equilibria that are saddle points. This system is equivalent to the second-order equation $x'' = \sin(x)$, which describes the motion of a pendulum (with unconventional sign). It is a Hamiltonian system with Hamiltonian $H(x,y) = \frac{1}{2}y^2 + \cos(x)$, since $x' = \frac{\partial H}{\partial y} = y$ and $y' = -\frac{\partial H}{\partial x} = \sin(x)$. Equilibrium points occur when $x'=0$ and $y'=0$, which means $y=0$ and $\sin(x)=0$. This gives equilibria at $(n\pi, 0)$ for any integer $n$. The Jacobian matrix is $J(x,y) = \begin{pmatrix} 0 & 1 \\ \cos(x) & 0 \end{pmatrix}$. At equilibria where $n$ is even, $x=2k\pi$, $\cos(x)=1$, so $J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. The determinant is $-1$, so these are saddle points. At equilibria where $n$ is odd, $x=(2k+1)\pi$, $\cos(x)=-1$, so $J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. The determinant is $+1$ and the trace is $0$, so linearization predicts a center. For Hamiltonian systems, a linear center corresponds to a true nonlinear center. These centers are surrounded by closed, periodic orbits corresponding to the level curves of the Hamiltonian function.
5. At which of the following $y$-values could a non-constant solution to the differential equation $\frac{dy}{dt} = y^3 - 9y$ have an inflection point?
   Answer: $y = \sqrt{3}$ 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 $y = \sqrt{3}$.
6. For the autonomous differential equation $\frac{dy}{dt} = y(2-y)$, for which values of $y$ are the solution curves $y(t)$ concave up?
   Answer: For $y \in (0, 1) \cup (2, \infty)$ 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 $y \in (0, 1) \cup (2, \infty)$.
7. For the differential equation $\frac{dy}{dx} = y\ln|y|$ with $y \neq 0$, analyze the long-term behavior of solutions. What happens to solutions with initial conditions $y(0) = y_0$ where $0 < y_0 < 1$?
   Answer: Solutions decrease monotonically and approach $y = 0$ asymptotically as $x \to \infty$ without reaching it 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.
8. For the system $\frac{dx}{dt} = \mu x - y - x(x^2 + y^2)$, $\frac{dy}{dt} = x + \mu y - y(x^2 + y^2)$ where $\mu$ is a parameter, analyze the bifurcation that occurs as $\mu$ varies. What type of bifurcation occurs at $\mu = 0$?
   Answer: A Hopf bifurcation where the origin changes from a stable to unstable spiral and a limit cycle is born 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.
9. For the system $\frac{dx}{dt} = -y + x(x^2 + y^2 - 1)$, $\frac{dy}{dt} = x + y(x^2 + y^2 - 1)$, analyze the behavior near the origin and the unit circle. Which statement best describes the global phase portrait?
   Answer: The origin is an unstable spiral, the unit circle is a stable limit cycle attracting all non-zero trajectories 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.
10. Consider the linear system $\frac{d\vec{x}}{dt} = A\vec{x}$ where $A = \begin{pmatrix} -1 & a \\ 1 & -1 \end{pmatrix}$. The qualitative behavior of the phase portrait near the origin depends on the parameter $a$. For which range of $a$ values will trajectories spiral into the origin?
    Answer: $a < 0$ 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.
11. Consider the equation $\frac{d^2y}{dt^2} + \mu(y^2 - 1)\frac{dy}{dt} + y = 0$ where $\mu > 0$. Using phase plane analysis with $x = y$ and $v = \frac{dy}{dt}$, determine the nature of the equilibrium at the origin.
    Answer: The origin is an unstable focus, and the system admits a stable limit cycle surrounding the origin 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.
12. Consider the differential equation $\frac{dy}{dt} = y^3 - 4y^2 + ry$, where $r$ is a real parameter. A bifurcation occurs at $r=4$. Which statement accurately describes the number and stability of equilibrium points when $r$ is slightly less than 4 (e.g., $r=3.9$)?
    Answer: There is one stable equilibrium and two unstable equilibria. 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.
13. Let $y(t)$ be the solution to the initial value problem $\frac{dy}{dt} = t + \cos(y)$ with $y(0) = \pi/2$. Without solving the equation, which of the following provides a valid lower bound for $y(t)$ for all $t > 0$?
    Answer: $y(t) \ge \frac{1}{2}t^2 - t + \pi/2$ 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.
14. Consider the system $x' = y + x(1 - x^2 - y^2)$ and $y' = -x + y(1 - x^2 - y^2)$. To analyze the behavior near the origin, the system is converted to polar coordinates $(r, \theta)$, where $r^2 = x^2+y^2$. What is the resulting differential equation for $r$, and what does it imply about trajectories near the origin?
    Answer: $\frac{dr}{dt} = r(1-r^2)$; trajectories spiral away from the origin towards a limit cycle at $r=1$. 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.
15. Consider the one-parameter family of differential equations $\frac{dy}{dt} = y^2 - 6y + \alpha$. A bifurcation occurs when a change in the parameter $\alpha$ alters the number or stability of the equilibrium solutions. What is the bifurcation value for $\alpha$?
    Answer: $\alpha = 9$ 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.
16. Consider the differential equation $\frac{dy}{dx} = \frac{y^2 - 4}{x^2 + 1}$. Analyze the phase portrait behavior. Which statement correctly describes the solution curves in the region where $y > 2$?
    Answer: Solutions increase without bound as $x$ increases, with vertical asymptotes at finite $x$-values where $y \to \infty$ 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 $x$.
17. A fish population is modeled by the logistic equation with harvesting: $\frac{dP}{dt} = 0.5P(1 - P/100) - H$, where $P$ is the population in thousands and $H$ is the constant harvesting rate in thousands per year. What is the smallest integer harvesting rate $H$ that guarantees the population will eventually go extinct, regardless of the initial population size?
    Answer: $H = 13$ 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.
18. Consider the system $x' = -x - 2y^2$, $y' = xy - y^3$, which has an equilibrium at the origin $(0,0)$. Let $V(x,y) = x^2 + 2y^2$ be a candidate Lyapunov function. What can be concluded about the stability of the origin?
    Answer: The origin is asymptotically stable. 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.
19. Without solving the differential equation $\frac{dy}{dt} + y = e^{-t}$, determine the behavior of all its solutions as $t \to \infty$.
    Answer: All solutions approach $y=0$. 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.
20. In which region of the $ty$-plane are all non-trivial solution curves to the differential equation $\frac{dy}{dt} = t - 2y$ concave down?
    Answer: The region defined by $y < \frac{1}{2}t - \frac{1}{4}$ 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 $\frac{d^2y}{dt^2} < 0$. $1 - 2t + 4y < 0 \implies 4y < 2t - 1 \implies y < \frac{1}{2}t - \frac{1}{4}$.

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