Differential Equations Quiz: Qualitative Behavior Analysis

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Qualitative Behavior AnalysisQuestion 1 of 20

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?

a < 0

0 < a < 1

a > 1

a < 1

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Differential Equations Quiz

Differential Equations Quiz: Qualitative Behavior Analysis

Practice Qualitative Behavior Analysis in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

$a < 0$ (correct answer)
$0 < a < 1$
$a > 1$
$a < 1$

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

Without solving the differential equation $\frac{dy}{dt} + y = e^{-t}$ , determine the behavior of all its solutions as $t \to \infty$ .

All solutions approach $y=1$ .
All solutions approach the curve $y=e^{-t}$ .
All solutions approach $y=0$ . (correct answer)
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 3

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?

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

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?

The origin is stable, but not necessarily asymptotically stable.
The origin is unstable.
The origin is asymptotically stable. (correct answer)
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

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 5

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., $r=3.9$ )?

There is one stable equilibrium and one unstable equilibrium.
There are two stable equilibria and one unstable equilibrium.
There is only one equilibrium, which is unstable.
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 6

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 $\alpha$ ?

$\alpha = 3$
$\alpha = 6$
$\alpha = 0$
$\alpha = 9$ (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 7

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?

The solution changes from increasing to decreasing, indicating a local maximum.
The solution changes from decreasing to increasing, indicating a local minimum. (correct answer)
The solution has an inflection point where it crosses the ellipse.
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 8

In which region of the $ty$ -plane are all non-trivial solution curves to the differential equation $\frac{dy}{dt} = t - 2y$ concave down?

The region defined by $y > \frac{1}{2}t$
The region defined by $y < \frac{1}{2}t$
The region defined by $y > \frac{1}{2}t - \frac{1}{4}$
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 9

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?

$H = 13$ (correct answer)
$H = 12$
$H = 25$
$H = 50$

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 10

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 $t > 0$ ?

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

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 11

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 $y > 2$ ?

Solutions have horizontal asymptotes approaching $y = 2$ as $x \to \pm\infty$ with monotonic approach
Solutions increase without bound as $x$ increases, with vertical asymptotes at finite $x$ -values where $y \to \infty$ (correct answer)
Solutions oscillate around $y = 2$ with decreasing amplitude as $x \to \infty$ due to the damping effect
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 12

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.

The origin is always a stable focus regardless of the value of $\mu$ due to the linear restoring force
The stability depends on $\mu$ : stable focus for small $\mu$ and unstable focus for large $\mu$ values
The origin is a center with purely periodic solutions in its neighborhood for any positive $\mu$ value
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 13

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?

When $h = r$ , the carrying capacity equilibrium collides with the extinction equilibrium and they exchange stability (correct answer)
When $h = \frac{r}{2}$ , a new equilibrium appears through a saddle-node bifurcation at the midpoint $y = \frac{K}{2}$
When $h = \frac{r}{4}$ , the system transitions from having two stable equilibria to having oscillatory solutions
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 14

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 $\mu = 0$ ?

A transcritical bifurcation where the origin changes stability and a new equilibrium branch appears
A pitchfork bifurcation creating two symmetric equilibria that branch from the origin when $\mu > 0$
A Hopf bifurcation where the origin changes from a stable to unstable spiral and a limit cycle is born (correct answer)
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 15

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?

The origin is a stable spiral with all trajectories eventually converging to it regardless of initial conditions
The origin is an unstable spiral, the unit circle is a stable limit cycle attracting all non-zero trajectories (correct answer)
Both the origin and unit circle are unstable, with trajectories escaping to infinity for most initial conditions
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 16

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 $0 < y_0 < 1$ ?

Solutions decrease monotonically and approach $y = 0$ asymptotically as $x \to \infty$ without reaching it (correct answer)
Solutions reach $y = 0$ in finite time, creating a boundary layer phenomenon at this value
Solutions initially decrease but then increase without bound after reaching a minimum positive value
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.

Question 17

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?

All equilibria are stable spirals, and every trajectory converges to one of the stable equilibrium points
The system has a unique stable limit cycle enclosing the origin due to the nonlinear restoring force
Equilibria alternate between stable spirals and saddle points, with separatrices dividing regions of attraction (correct answer)
All trajectories are bounded and eventually become periodic regardless of the damping parameter $\delta$

Explanation: 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.

Question 18

Consider the initial value problem $\frac{dy}{dt} = y^{2/3}$ , with $y(1) = 0$ . Which of the following statements about the solution(s) to this IVP is correct?

There is a unique solution given by $y(t) = 0$ for all $t$ .
There are exactly two distinct solutions for $t \ge 1$ .
No solution exists because the differential equation's right-hand side is not differentiable at $y=0$ .
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 19

Consider the autonomous differential equation $\frac{dy}{dt} = (y^2 - 1)(y - 3)^2$ . If a solution $y(t)$ satisfies the initial condition $y(0) = 1.1$ , what is the long-term behavior of $y(t)$ ?

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

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 20

For the autonomous differential equation $\frac{dy}{dt} = y(2-y)$ , for which values of $y$ are the solution curves $y(t)$ concave up?

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