Differential Equations Quiz: Systems Complex Eigenvalues

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Systems Complex EigenvaluesQuestion 1 of 4

A system $\mathbf{x}' = A\mathbf{x}$ has eigenvalues $\lambda = \sigma \pm i\omega$ where $\sigma < 0$ and $\omega > 0$ . If the distance from any trajectory to the origin decreases by a factor of $e$ while the trajectory rotates through angle $\theta$ , what is the relationship between $\sigma$ , $\omega$ , and $\theta$ ?

\sigma = -\frac{\omega}{\theta}

\sigma = -\frac{\omega \theta}{2\pi}

\sigma = -\frac{\omega}{2\pi} \theta

\sigma \theta = -\omega

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Differential Equations Quiz

Differential Equations Quiz: Systems Complex Eigenvalues

Practice Systems Complex Eigenvalues 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 Systems Complex Eigenvalues, 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

$\sigma = -\frac{\omega}{\theta}$ (correct answer)
$\sigma = -\frac{\omega \theta}{2\pi}$
$\sigma = -\frac{\omega}{2\pi} \theta$
$\sigma \theta = -\omega$

Explanation: The solution has form

\mathbf{x}(t) = e^{\sigma t}(\mathbf{u}\cos(\omega t) + \mathbf{v}\sin(\omega t))

. The distance from origin is proportional to

e^{\sigma t}

, and the angular displacement is

\omega t

. If the trajectory rotates through angle

\theta

, then

\omega t = \theta

, so

t = \frac{\theta}{\omega}

. During this time, the distance changes by factor

e^{\sigma t} = e^{\sigma \theta/\omega}

. For the distance to decrease by factor

e

(i.e., become

\frac{1}{e}

times the original), we need

e^{\sigma \theta/\omega} = \frac{1}{e} = e^{-1}

. Therefore

\frac{\sigma \theta}{\omega} = -1

, which gives

\sigma = -\frac{\omega}{\theta}

Question 2

Consider the system $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$ where $a + d = 0$ and $ad - bc = 25$ . If the solution satisfying $x(0) = 3, y(0) = 4$ has the property that $x^2(t) + y^2(t)$ is constant for all $t$ , which of the following must be true about the coefficient matrix?

$a = 0, d = 0$ , and $bc = -25$ with $b$ and $c$ having the same sign
$a = 0, d = 0$ , and $bc = -25$ with $b$ and $c$ having opposite signs (correct answer)
$a = -d \neq 0$ , and the eigenvalues are $\pm 5i$
$a = -d \neq 0$ , and the eigenvalues are $a \pm 5i$

Explanation: For

x^2(t) + y^2(t)

to be constant, we need

\frac{d}{dt}(x^2 + y^2) = 2x x' + 2y y' = 0

for all solutions. Substituting the system:

2x(ax + by) + 2y(cx + dy) = 2ax^2 + 2(b+c)xy + 2dy^2 = 0

. Since

a + d = 0

, we have

d = -a

, so

2ax^2 + 2(b+c)xy - 2ay^2 = 0

. For this to hold for all

x, y

, we need

a = 0

(so

d = 0

) and

b + c = 0

. With

a = d = 0

and

ad - bc = 25

, we get

-bc = 25

, so

bc = -25

. Since

b + c = 0

, we have

c = -b

, so

b(-b) = -b^2 = -25

, giving

b^2 = 25

. Thus

b = \pm 5

and

c = \mp 5

, meaning

b

and

c

have opposite signs.

Question 3

A system $\mathbf{x}' = A\mathbf{x}$ has complex eigenvalues $3 \pm 4i$ . The trajectory starting at $(1, 1)$ first returns to the positive $x$ -axis (where $y = 0$ and $x > 0$ ) at time $t^*$ . What is the $x$ -coordinate of this return point?

$e^{3\pi/2}$
$\sqrt{2} e^{3\pi/2}$ (correct answer)
$e^{3\pi/4}$
$\sqrt{2} e^{3\pi/4}$

Explanation: The eigenvalues

3 \pm 4i

give solutions of the form

\mathbf{x}(t) = e^{3t}(c_1\mathbf{u}\cos(4t) + c_2\mathbf{v}\sin(4t))

where

\mathbf{u}

and

\mathbf{v}

are the real and imaginary parts of an eigenvector. Starting at

(1,1)

, the trajectory spirals outward. The trajectory returns to the positive

x

-axis when it has completed one revolution, which occurs when

4t = 2\pi

, so

t^* = \frac{\pi}{2}

. At this time,

y = 0

and the

x

-coordinate is the initial distance from origin times

e^{3t^*}

. The initial distance is

\sqrt{1^2 + 1^2} = \sqrt{2}

. Therefore, the

x

-coordinate at return is

\sqrt{2} \cdot e^{3\pi/2}

Question 4

For the system $\mathbf{x}' = A\mathbf{x}$ where $A$ has complex eigenvalues $\lambda = -1 \pm 3i$ , two students propose different fundamental matrix solutions. Student 1 claims $\Phi_1(t) = e^{-t}\begin{pmatrix} \cos(3t) & \sin(3t) \\ 2\cos(3t) + \sin(3t) & 2\sin(3t) - \cos(3t) \end{pmatrix}$ . Student 2 claims $$\Phi_2(t) = e^{-t}\begin{pmatrix} \cos(3t) & -\sin(3t) \ 2\cos(3t) + \sin(3t) & 2\sin(3t) + \cos(3t) \end{pmatrix}

Both matrices are valid fundamental solutions since $\det(\Phi_1(0)) = \det(\Phi_2(0)) = -1 \neq 0$
Only $\Phi_1(t)$ can be valid since fundamental matrices must have $\det(\Phi(0)) > 0$
Neither can be determined valid without verifying $\Phi'(t) = A\Phi(t)$ and checking linear independence (correct answer)
Only $\Phi_2(t)$ can be valid since the off-diagonal terms must have opposite signs for complex eigenvalues

Explanation: A fundamental matrix

\Phi(t)

must satisfy two conditions: (1)

\Phi'(t) = A\Phi(t)

and (2)

\det(\Phi(t)) \neq 0

for all

t

. While both proposed matrices have

\det(\Phi(0)) = -1 \neq 0

, this alone doesn't guarantee they satisfy the differential equation. The specific form of the matrix

A

determines which solution is correct. We cannot conclude validity without explicitly checking that the derivative equals

A

times the matrix. The sign of the determinant and patterns in off-diagonal terms don't determine validity by themselves - only the fundamental differential equation relationship matters.