Differential Equations Quiz: Complex Roots And Oscillations

What this quiz covers

This quiz focuses on Complex Roots And Oscillations, 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. 13
2. 36
3. 40 (correct answer)
4. 52

Explanation: The time between successive maxima is the quasi-period, $T_d$. We are given $T_d = \pi/3$. The quasi-period is related to the angular frequency $\omega$ by $T_d = 2\pi/\omega$. So, $\pi/3 = 2\pi/\omega$, which implies $\omega = 6$. The angular frequency $\omega$ is the imaginary part of the complex roots of the characteristic equation $r^2+4r+k=0$. The roots are $r = \frac{-4 \pm \sqrt{16-4k}}{2} = -2 \pm \sqrt{4-k}$. For oscillations, the term under the square root must be negative. We can write the roots as $r = -2 \pm i\sqrt{k-4}$. The imaginary part is $\omega = \sqrt{k-4}$. We set this equal to 6: $\sqrt{k-4} = 6$. Squaring both sides gives $k-4=36$, so $k=40$.

Question 2

1. $y'' - 4y' + 13y = 0$
2. $y'' + 4y' + 13y = 0$ (correct answer)
3. $y'' + 2y' + 10y = 0$
4. $y'' + 4y' + 5y = 0$

Explanation: The solution has the form $y(t) = Ae^{\lambda t}\cos(\omega t - \phi)$. By comparing this to the given solution, we can identify the decay rate $\lambda = -2$ and the angular frequency $\omega = 3$. The characteristic roots of the differential equation must be complex conjugates $r = \lambda \pm i\omega = -2 \pm 3i$. To find the characteristic equation, we can compute $(r - (-2 + 3i))(r - (-2 - 3i)) = 0$. This simplifies to $((r+2) - 3i)((r+2) + 3i) = (r+2)^2 - (3i)^2 = r^2 + 4r + 4 - 9i^2 = r^2 + 4r + 13 = 0$. This characteristic equation corresponds to the differential equation $y'' + 4y' + 13y = 0$.

Question 3

1. $2$
2. $0$
3. $-1$
4. $-2$ (correct answer)

Explanation: First, find the general solution. The characteristic equation is $r^2 + 2r + 5 = 0$. The roots are $r = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i$. The general solution is $y(t) = e^{-t}(C_1 \cos(2t) + C_2 \sin(2t))$. For the solution to be a 'pure' damped cosine wave of the form given, the sine term must be absent, so $C_2=0$. The solution is then $y(t) = C_1 e^{-t}\cos(2t)$. Using the first initial condition, $y(0) = C_1\cos(0) = C_1 = 2$. So, the specific solution is $y(t) = 2e^{-t}\cos(2t)$. Now, we find its derivative: $y'(t) = -2e^{-t}\cos(2t) - 4e^{-t}\sin(2t)$. The second initial condition is $y'(0)=\beta$. Evaluating the derivative at $t=0$: $y'(0) = -2e^0\cos(0) - 4e^0\sin(0) = -2(1) - 4(0) = -2$. Thus, $\beta = -2$.

Question 4

1. $\gamma > 6$
2. $-6 < \gamma < 6$
3. $0 < \gamma < 6$ (correct answer)
4. $\gamma < 0$

Explanation: For the solution to exhibit oscillations, the roots of the characteristic equation $r^2 + \gamma r + 9 = 0$ must be complex. This occurs when the discriminant is negative: $\Delta = \gamma^2 - 4(1)(9) < 0$, which simplifies to $\gamma^2 < 36$, or $-6 < \gamma < 6$. The roots are $r = \frac{-\gamma \pm \sqrt{\Delta}}{2}$. The real part of the roots is $\lambda = -\gamma/2$. For the oscillations to decay over time, the amplitude term $e^{\lambda t}$ must approach zero as $t \to \infty$, which requires $\lambda < 0$. So, we need $-\gamma/2 < 0$, which implies $\gamma > 0$. Combining both conditions, we need $\gamma > 0$ and $-6 < \gamma < 6$. The intersection of these two intervals is $0 < \gamma < 6$.

Question 5

1. $\omega_A > \omega_B$ and the amplitude of System A decays faster than System B. (correct answer)
2. $\omega_A > \omega_B$ and the amplitude of System B decays faster than System A.
3. $\omega_A < \omega_B$ and the amplitude of System A decays faster than System B.
4. $\omega_A < \omega_B$ and the amplitude of System B decays faster than System A.

Explanation: For System A ($y'' + 2y' + 10y = 0$), the characteristic equation is $r^2 + 2r + 10 = 0$. The roots are $r = \frac{-2 \pm \sqrt{4-40}}{2} = -1 \pm 3i$. Thus, $\lambda_A = -1$ and $\omega_A = 3$. For System B ($2y'' + 2y' + 5y = 0$), the characteristic equation is $2r^2 + 2r + 5 = 0$. The roots are $r = \frac{-2 \pm \sqrt{4-40}}{4} = -\frac{1}{2} \pm \frac{3}{2}i$. Thus, $\lambda_B = -1/2$ and $\omega_B = 3/2 = 1.5$. Comparing the frequencies, $\omega_A = 3 > 1.5 = \omega_B$. Comparing the decay rates, the amplitude factors are $e^{-t}$ for A and $e^{-0.5t}$ for B. Since the magnitude of $\lambda_A$ is greater than the magnitude of $\lambda_B$ (i.e., $|-1| > |-0.5|$), the amplitude of System A decays faster. Therefore, $\omega_A > \omega_B$ and System A's amplitude decays faster.

Question 6

1. $64$
2. $1/16$
3. $1/4$ (correct answer)
4. $-1/8$

Explanation: The Wronskian can be found using Abel's identity, $W(t) = C e^{-\int p(t) dt}$, where $p(t)=4$. This gives $W(t) = C e^{-4t}$. To find $C$, we can compute the Wronskian for a specific fundamental set of solutions. The characteristic equation is $r^2+4r+20=0$, with roots $r = \frac{-4 \pm \sqrt{16-80}}{2} = -2 \pm 4i$. The standard fundamental solutions are $y_1(t) = e^{-2t}\cos(4t)$ and $y_2(t) = e^{-2t}\sin(4t)$. The Wronskian $W(y_1, y_2)(t) = y_1y_2' - y_1'y_2$. A known result for solutions of the form $e^{\lambda t}\cos(\omega t)$ and $e^{\lambda t}\sin(\omega t)$ is $W(t) = \omega e^{2\lambda t}$. Here, $\lambda=-2$ and $\omega=4$, so $W(t) = 4e^{2(-2)t} = 4e^{-4t}$. We need to evaluate this at $t=\ln(2)$: $W(\ln(2)) = 4e^{-4\ln(2)} = 4e^{\ln(2^{-4})} = 4 \cdot 2^{-4} = 4/16 = 1/4$.

Question 7

1. 2
2. 8 (correct answer)
3. 16
4. 50

Explanation: The characteristic equation is $r^2 + 0.2r + 25.01 = 0$. The roots are $r = \frac{-0.2 \pm \sqrt{0.04 - 4(25.01)}}{2} = \frac{-0.2 \pm \sqrt{-100}}{2} = -0.1 \pm 5i$. The solution has the form $y(t) = A_0 e^{-0.1t} \cos(5t - \phi)$. The amplitude is $A(t) = A_0 e^{-0.1t}$. We want to find the time $t$ when the amplitude drops to $1/e$ of its initial value, $A_0$. So, $A(t) = A_0/e$. This means $A_0 e^{-0.1t} = A_0 e^{-1}$, which implies $-0.1t = -1$, or $t=10$. Now we need to find the number of oscillations in this time. The angular frequency is $\omega = 5$ rad/s. The period of one oscillation is $T = 2\pi/\omega = 2\pi/5$ seconds. The number of oscillations in $t=10$ seconds is $N = t/T = 10 / (2\pi/5) = 50/(2\pi) = 25/\pi$. Using the approximation $\pi \approx 3.14$, we get $N \approx 25/3.14 \approx 7.96$. This is approximately 8 full oscillations.

Question 8

1. A spiral moving inwards towards the origin in a clockwise direction. (correct answer)
2. A spiral moving inwards towards the origin in a counter-clockwise direction.
3. A closed ellipse centered at the origin.
4. A trajectory that approaches the origin along a straight line without spiraling.

Explanation: The characteristic equation is $r^2+0.1r+1=0$. The discriminant is $(0.1)^2 - 4(1) = -3.99 < 0$, so the roots are complex. The real part of the roots is $-0.1/2 = -0.05 < 0$. This corresponds to a damped oscillation. In the phase plane, a damped oscillation is represented by a spiral trajectory moving inwards toward the equilibrium point at the origin. To determine the direction of the spiral, we can check the velocity vector $(y', y'')$ at the initial point. At $t=0$, the position is $(y, y') = (1, 0)$. From the differential equation, $y'' = -0.1y' - y$. At the initial point, $y'' = -0.1(0) - 1 = -1$. The velocity vector at $(1,0)$ is $(y', y'') = (0, -1)$, which points straight down in the phase plane. A trajectory starting on the positive y-axis at $(1,0)$ and immediately moving into the fourth quadrant (where $y>0, y'<0$) must be spiraling in a clockwise direction.

Question 9

1. $4\sqrt{2}$ (correct answer)
2. $2\sqrt{5}$
3. $4$
4. $8$

Explanation: First, find the general solution. The characteristic equation is $r^2 + r + 5/4 = 0$. The roots are $r = \frac{-1 \pm \sqrt{1-5}}{2} = -\frac{1}{2} \pm i$. The general solution is $y(t) = e^{-t/2}(C_1 \cos(t) + C_2 \sin(t))$. We apply the initial conditions to find $C_1$ and $C_2$. From $y(0)=4$, we get $4 = e^0(C_1\cos(0) + C_2\sin(0))$, which gives $C_1=4$. Next, we find the derivative: $y'(t) = -\frac{1}{2}e^{-t/2}(C_1\cos(t) + C_2\sin(t)) + e^{-t/2}(-C_1\sin(t) + C_2\cos(t))$. From $y'(0)=2$, we get $2 = -\frac{1}{2}(C_1) + C_2$. Substituting $C_1=4$, we have $2 = -\frac{1}{2}(4) + C_2$, which gives $C_2 = 4$. The solution is $y(t) = e^{-t/2}(4\cos(t) + 4\sin(t))$. The term in the parentheses can be written as $A\cos(t - \phi)$, where the amplitude is $A = \sqrt{C_1^2 + C_2^2} = \sqrt{4^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$.

Question 10

1. $p = \frac{\omega}{2\pi}$ (correct answer)
2. $p = \frac{\omega}{\pi}$
3. $\omega = \frac{p}{2\pi}$
4. $\omega = \frac{p}{\pi}$

Explanation: The characteristic equation gives $r = -p \pm i\omega$, so the solution is $y = e^{-pt}(A\cos(\omega t) + B\sin(\omega t))$. The amplitude envelope is $e^{-pt}$. One complete period is $T = \frac{2\pi}{\omega}$. After one period, the amplitude becomes $e^{-pT} = e^{-p \cdot \frac{2\pi}{\omega}}$. For this to equal $e^{-1}$: $-p \cdot \frac{2\pi}{\omega} = -1$, giving $p = \frac{\omega}{2\pi}$. Choice B omits the factor of 2. Choices C and D invert the relationship incorrectly.

Question 11

1. $\omega = 6$
2. $\omega = 8$
3. $\omega = 4$ (correct answer)
4. $\omega = \sqrt{10}$

Explanation: When you encounter a second-order linear ODE with complex characteristic roots, the solution form reveals crucial information about the system's behavior. The given solution $y = e^{-3t}(C_1\cos(\omega t) + C_2\sin(\omega t))$ tells you the characteristic roots are complex conjugates of the form $r = -3 \pm i\omega$. For a characteristic polynomial $r^2 + ar + b$, these roots satisfy the quadratic formula: $r = \frac{-a \pm \sqrt{a^2 - 4b}}{2}$. Since the real part is $-3$, you know $\frac{-a}{2} = -3$, which gives $a = 6$. The imaginary part comes from $\frac{\sqrt{-(a^2 - 4b)}}{2} = \frac{\sqrt{-(-64)}}{2} = \frac{\sqrt{64}}{2} = \frac{8}{2} = 4$. Therefore, $\omega = 4$. Looking at the wrong answers: (A) $\omega = 6$ confuses the natural frequency with the real part of the characteristic root. (B) $\omega = 8$ takes $\sqrt{64}$ directly without dividing by 2 in the quadratic formula. (D) $\omega = \sqrt{10}$ likely results from algebraic errors when manipulating the discriminant relationship. Study tip: Remember that for complex roots $\alpha \pm i\beta$, the solution is $e^{\alpha t}(C_1\cos(\beta t) + C_2\sin(\beta t))$. The coefficient of $t$ inside the trigonometric functions is always the imaginary part of the characteristic root, which equals $\frac{\sqrt{|discriminant|}}{2}$ when the discriminant is negative.

Question 12

1. $9$
2. $10$
3. $13$
4. $19$ (correct answer)

Explanation: From $y_1 = e^{-3t}\sin(2t)$, we identify the characteristic roots as $r = -3 \pm 2i$. The characteristic equation is $(r+3)^2 + 4 = r^2 + 6r + 13 = 0$. Comparing with $r^2 + \beta r + \gamma = 0$, we get $\beta = 6$ and $\gamma = 13$. Therefore $\beta + \gamma = 6 + 13 = 19$. We can verify: $\beta^2 = 36 < 4\gamma = 52$, confirming complex roots. Choice A gives $\gamma$ only. Choice B uses $\beta + 4$ (confusing imaginary part with $\gamma$). Choice C gives $\gamma$ only.

Question 13

1. Both solutions have the same sign, with equation (I) having larger magnitude due to slower oscillation
2. Both solutions have the same sign, with equation (II) having larger magnitude due to slower oscillation (correct answer)
3. The solutions have opposite signs, with equation (I) having completed exactly one-half cycle
4. The solutions have the same sign and approximately equal magnitudes despite different frequencies

Explanation: For (I): $r = -1 \pm 2i$, so $y_1 = e^{-t}(\cos(2t) + \frac{1}{2}\sin(2t))$. For (II): $r = -1 \pm i$, so $y_2 = e^{-t}(\cos(t) + \sin(t))$. At $t = \frac{\pi}{4}$: $y_1 = e^{-\pi/4}(0 + \frac{1}{2}) = \frac{e^{-\pi/4}}{2}$ and $y_2 = e^{-\pi/4}(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) = e^{-\pi/4}\sqrt{2}$. Both are positive, and $\sqrt{2} > \frac{1}{2}$, so (II) has larger magnitude. Equation (II) oscillates slower (frequency 1 vs 2). Choice A reverses the magnitude comparison. Choice C incorrectly predicts opposite signs. Choice D incorrectly claims equal magnitudes.

Question 14

1. $\frac{\pi}{3}$
2. $\frac{2\pi}{3}$ (correct answer)
3. $\frac{\pi}{2}$
4. $\frac{2\pi}{9}$

Explanation: The characteristic equation is $r^2 + 4r + 13 = 0$. Using the quadratic formula: $r = \frac{-4 \pm \sqrt{16-52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i$. So $\alpha = -2$ and $\beta = 3$. The period of oscillation is $T = \frac{2\pi}{\beta} = \frac{2\pi}{3}$. Choice A uses $\beta = 6$ (doubling error). Choice C uses $\beta = 4$ (confusing with coefficient of $y'$). Choice D uses $\beta = 9$ (squaring error).

Question 15

1. The general solution is $y = e^{-2t}(C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t))$ and $k = 4$ (correct answer)
2. The general solution is $y = e^{-2t}(C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t))$ and $k = -4$
3. The general solution is $y = e^{2t}(C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t))$ and $k = 4$
4. The general solution requires $k^2 - 36 < 0$ and the given particular solution is impossible

Explanation: From the given solution $y = e^{-2t}\cos(\sqrt{5}t)$, we identify $\alpha = -2$ and $\beta = \sqrt{5}$. The characteristic roots are $r = -2 \pm i\sqrt{5}$. From $r^2 + kr + 9 = 0$, we have $r = \frac{-k \pm \sqrt{k^2-36}}{2}$. Comparing: $\frac{-k}{2} = -2$ gives $k = 4$, and $\frac{\sqrt{36-16}}{2} = \frac{\sqrt{20}}{2} = \sqrt{5}$ confirms this. The general solution includes both cosine and sine terms. Choice B has wrong sign for $k$. Choice C has wrong sign in exponential. Choice D incorrectly analyzes the discriminant condition.

Question 16

1. $t = \frac{\pi}{6}$
2. $t = \frac{\pi}{2}$
3. $t = \frac{\pi}{3}$ (correct answer)
4. $t = \frac{2\pi}{3}$

Explanation: When you encounter a second-order linear differential equation with constant coefficients like this damped oscillator problem, you need to find the characteristic equation and determine what type of damping occurs. The characteristic equation for $y'' + 2y' + 10y = 0$ is $r^2 + 2r + 10 = 0$. Using the quadratic formula: $r = \frac{-2 \pm \sqrt{4-40}}{2} = \frac{-2 \pm \sqrt{-36}}{2} = -1 \pm 3i$. Since we have complex roots $r = -1 \pm 3i$, this represents underdamped motion. For complex roots $\alpha \pm \beta i$, the general solution is $y(t) = e^{\alpha t}(c_1 \cos(\beta t) + c_2 \sin(\beta t))$. Here, $\alpha = -1$ and $\beta = 3$, so $y(t) = e^{-t}(c_1 \cos(3t) + c_2 \sin(3t))$. Applying initial conditions: $y(0) = 0$ gives us $c_1 = 0$. Then $y'(t) = e^{-t}(3c_2 \cos(3t) - c_2 \sin(3t))$, and $y'(0) = 6$ gives us $3c_2 = 6$, so $c_2 = 2$. Therefore, $y(t) = 2e^{-t}\sin(3t)$. The oscillator returns to equilibrium when $y(t) = 0$. Since $e^{-t} \neq 0$, we need $\sin(3t) = 0$, which occurs when $3t = n\pi$ for integer $n$. The first positive solution is $t = \frac{\pi}{3}$. Answer choice A ($\frac{\pi}{6}$) would correspond to $\sin(\frac{\pi}{2}) = 1 \neq 0$. Choice B ($\frac{\pi}{2}$) gives $\sin(\frac{3\pi}{2}) = -1 \neq 0$. Choice D ($\frac{2\pi}{3}$) gives $\sin(2\pi) = 0$, but this is the second zero, not the first. Remember: for underdamped oscillators, the zeros occur at regular intervals determined by the imaginary part of the characteristic roots.