Differential Equations Quiz: Complex Roots And Oscillations
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Complex Roots And OscillationsQuestion 1 of 16

A physical system is modeled by a second-order linear homogeneous differential equation with constant coefficients. Its solution is observed to be y(t)=5e−2tcos⁡(3t−π/4)y(t) = 5e^{-2t}\cos(3t - \pi/4). Which of the following differential equations models this system?

y′′−4y′+13y=0y'' - 4y' + 13y = 0
y′′+4y′+13y=0y'' + 4y' + 13y = 0
y′′+2y′+10y=0y'' + 2y' + 10y = 0
y′′+4y′+5y=0y'' + 4y' + 5y = 0
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Differential Equations Quiz

Differential Equations Quiz: Complex Roots And Oscillations

Practice Complex Roots And Oscillations in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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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.

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Question 1

A physical system is modeled by a second-order linear homogeneous differential equation with constant coefficients. Its solution is observed to be y(t)=5e−2tcos⁡(3t−π/4)y(t) = 5e^{-2t}\cos(3t - \pi/4). Which of the following differential equations models this system?

  1. y′′−4y′+13y=0y'' - 4y' + 13y = 0
  2. y′′+4y′+13y=0y'' + 4y' + 13y = 0 (correct answer)
  3. y′′+2y′+10y=0y'' + 2y' + 10y = 0
  4. y′′+4y′+5y=0y'' + 4y' + 5y = 0
Explanation: The solution has the form y(t)=Aeλtcos⁡(ωt−ϕ)y(t) = Ae^{\lambda t}\cos(\omega t - \phi). By comparing this to the given solution, we can identify the decay rate λ=−2\lambda = -2 and the angular frequency ω=3\omega = 3. The characteristic roots of the differential equation must be complex conjugates r=λ±iω=−2±3ir = \lambda \pm i\omega = -2 \pm 3i. To find the characteristic equation, we can compute (r−(−2+3i))(r−(−2−3i))=0(r - (-2 + 3i))(r - (-2 - 3i)) = 0. This simplifies to ((r+2)−3i)((r+2)+3i)=(r+2)2−(3i)2=r2+4r+4−9i2=r2+4r+13=0((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=0y'' + 4y' + 13y = 0.

Question 2

The solution to the initial value problem y′′+2y′+5y=0y'' + 2y' + 5y = 0 with y(0)=2y(0) = 2 and y′(0)=βy'(0) = \beta is a pure damped cosine wave, meaning it has the form y(t)=Ae−tcos⁡(ωt)y(t) = Ae^{-t}\cos(\omega t). What is the value of β\beta?

  1. 22
  2. 00
  3. −1-1
  4. −2-2 (correct answer)
Explanation: First, find the general solution. The characteristic equation is r2+2r+5=0r^2 + 2r + 5 = 0. The roots are r=−2±4−202=−1±2ir = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i. The general solution is y(t)=e−t(C1cos⁡(2t)+C2sin⁡(2t))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 C2=0C_2=0. The solution is then y(t)=C1e−tcos⁡(2t)y(t) = C_1 e^{-t}\cos(2t). Using the first initial condition, y(0)=C1cos⁡(0)=C1=2y(0) = C_1\cos(0) = C_1 = 2. So, the specific solution is y(t)=2e−tcos⁡(2t)y(t) = 2e^{-t}\cos(2t). Now, we find its derivative: y′(t)=−2e−tcos⁡(2t)−4e−tsin⁡(2t)y'(t) = -2e^{-t}\cos(2t) - 4e^{-t}\sin(2t). The second initial condition is y′(0)=βy'(0)=\beta. Evaluating the derivative at t=0t=0: y′(0)=−2e0cos⁡(0)−4e0sin⁡(0)=−2(1)−4(0)=−2y'(0) = -2e^0\cos(0) - 4e^0\sin(0) = -2(1) - 4(0) = -2. Thus, β=−2\beta = -2.

Question 3

Consider the differential equation y′′+γy′+9y=0y'' + \gamma y' + 9y = 0, where γ\gamma is a real parameter representing a damping coefficient. For which values of γ\gamma will the solution exhibit oscillations that decay over time?

  1. γ>6\gamma > 6
  2. −6<γ<6-6 < \gamma < 6
  3. 0<γ<60 < \gamma < 6 (correct answer)
  4. γ<0\gamma < 0
Explanation: For the solution to exhibit oscillations, the roots of the characteristic equation r2+γr+9=0r^2 + \gamma r + 9 = 0 must be complex. This occurs when the discriminant is negative: Δ=γ2−4(1)(9)<0\Delta = \gamma^2 - 4(1)(9) < 0, which simplifies to γ2<36\gamma^2 < 36, or −6<γ<6-6 < \gamma < 6. The roots are r=−γ±Δ2r = \frac{-\gamma \pm \sqrt{\Delta}}{2}. The real part of the roots is λ=−γ/2\lambda = -\gamma/2. For the oscillations to decay over time, the amplitude term eλte^{\lambda t} must approach zero as t→∞t \to \infty, which requires λ<0\lambda < 0. So, we need −γ/2<0-\gamma/2 < 0, which implies γ>0\gamma > 0. Combining both conditions, we need γ>0\gamma > 0 and −6<γ<6-6 < \gamma < 6. The intersection of these two intervals is 0<γ<60 < \gamma < 6.

Question 4

Two mass-spring systems, A and B, are described by the differential equations:

System A: y′′+2y′+10y=0y'' + 2y' + 10y = 0

System B: 2y′′+2y′+5y=02y'' + 2y' + 5y = 0

Let ωA\omega_A and ωB\omega_B be their respective angular frequencies of oscillation, and let the decay of their amplitudes be governed by factors eλAte^{\lambda_A t} and eλBte^{\lambda_B t}. Which of the following statements is true?

  1. ωA>ωB\omega_A > \omega_B and the amplitude of System A decays faster than System B. (correct answer)
  2. ωA>ωB\omega_A > \omega_B and the amplitude of System B decays faster than System A.
  3. ωA<ωB\omega_A < \omega_B and the amplitude of System A decays faster than System B.
  4. ωA<ωB\omega_A < \omega_B and the amplitude of System B decays faster than System A.
Explanation: For System A (y′′+2y′+10y=0y'' + 2y' + 10y = 0), the characteristic equation is r2+2r+10=0r^2 + 2r + 10 = 0. The roots are r=−2±4−402=−1±3ir = \frac{-2 \pm \sqrt{4-40}}{2} = -1 \pm 3i. Thus, λA=−1\lambda_A = -1 and ωA=3\omega_A = 3. For System B (2y′′+2y′+5y=02y'' + 2y' + 5y = 0), the characteristic equation is 2r2+2r+5=02r^2 + 2r + 5 = 0. The roots are r=−2±4−404=−12±32ir = \frac{-2 \pm \sqrt{4-40}}{4} = -\frac{1}{2} \pm \frac{3}{2}i. Thus, λB=−1/2\lambda_B = -1/2 and ωB=3/2=1.5\omega_B = 3/2 = 1.5. Comparing the frequencies, ωA=3>1.5=ωB\omega_A = 3 > 1.5 = \omega_B. Comparing the decay rates, the amplitude factors are e−te^{-t} for A and e−0.5te^{-0.5t} for B. Since the magnitude of λA\lambda_A is greater than the magnitude of λB\lambda_B (i.e., ∣−1∣>∣−0.5∣|-1| > |-0.5|), the amplitude of System A decays faster. Therefore, ωA>ωB\omega_A > \omega_B and System A's amplitude decays faster.

Question 5

The motion of a damped oscillator is described by y′′+4y′+20y=0y'' + 4y' + 20y = 0. If y1(t)y_1(t) and y2(t)y_2(t) form a fundamental set of solutions for this equation, what is the value of the Wronskian W(y1,y2)(t)W(y_1, y_2)(t) at t=ln⁡(2)t=\ln(2)?

  1. 6464
  2. 1/161/16
  3. 1/41/4 (correct answer)
  4. −1/8-1/8
Explanation: The Wronskian can be found using Abel's identity, W(t)=Ce−∫p(t)dtW(t) = C e^{-\int p(t) dt}, where p(t)=4p(t)=4. This gives W(t)=Ce−4tW(t) = C e^{-4t}. To find CC, we can compute the Wronskian for a specific fundamental set of solutions. The characteristic equation is r2+4r+20=0r^2+4r+20=0, with roots r=−4±16−802=−2±4ir = \frac{-4 \pm \sqrt{16-80}}{2} = -2 \pm 4i. The standard fundamental solutions are y1(t)=e−2tcos⁡(4t)y_1(t) = e^{-2t}\cos(4t) and y2(t)=e−2tsin⁡(4t)y_2(t) = e^{-2t}\sin(4t). The Wronskian W(y1,y2)(t)=y1y2′−y1′y2W(y_1, y_2)(t) = y_1y_2' - y_1'y_2. A known result for solutions of the form eλtcos⁡(ωt)e^{\lambda t}\cos(\omega t) and eλtsin⁡(ωt)e^{\lambda t}\sin(\omega t) is W(t)=ωe2λtW(t) = \omega e^{2\lambda t}. Here, λ=−2\lambda=-2 and ω=4\omega=4, so W(t)=4e2(−2)t=4e−4tW(t) = 4e^{2(-2)t} = 4e^{-4t}. We need to evaluate this at t=ln⁡(2)t=\ln(2): W(ln⁡(2))=4e−4ln⁡(2)=4eln⁡(2−4)=4⋅2−4=4/16=1/4W(\ln(2)) = 4e^{-4\ln(2)} = 4e^{\ln(2^{-4})} = 4 \cdot 2^{-4} = 4/16 = 1/4.

Question 6

The solution to y′′+0.2y′+25.01y=0y''+0.2y'+25.01y=0 represents a weakly damped oscillation. Approximately how many full oscillations does the system complete before the amplitude of the oscillation drops to 1/e1/e of its initial value?

  1. 2
  2. 8 (correct answer)
  3. 16
  4. 50
Explanation: The characteristic equation is r2+0.2r+25.01=0r^2 + 0.2r + 25.01 = 0. The roots are r=−0.2±0.04−4(25.01)2=−0.2±−1002=−0.1±5ir = \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)=A0e−0.1tcos⁡(5t−ϕ)y(t) = A_0 e^{-0.1t} \cos(5t - \phi). The amplitude is A(t)=A0e−0.1tA(t) = A_0 e^{-0.1t}. We want to find the time tt when the amplitude drops to 1/e1/e of its initial value, A0A_0. So, A(t)=A0/eA(t) = A_0/e. This means A0e−0.1t=A0e−1A_0 e^{-0.1t} = A_0 e^{-1}, which implies −0.1t=−1-0.1t = -1, or t=10t=10. Now we need to find the number of oscillations in this time. The angular frequency is ω=5\omega = 5 rad/s. The period of one oscillation is T=2π/ω=2π/5T = 2\pi/\omega = 2\pi/5 seconds. The number of oscillations in t=10t=10 seconds is N=t/T=10/(2π/5)=50/(2π)=25/πN = t/T = 10 / (2\pi/5) = 50/(2\pi) = 25/\pi. Using the approximation π≈3.14\pi \approx 3.14, we get N≈25/3.14≈7.96N \approx 25/3.14 \approx 7.96. This is approximately 8 full oscillations.

Question 7

The solution to y′′+0.1y′+y=0y'' + 0.1y' + y = 0 with initial conditions y(0)=1,y′(0)=0y(0)=1, y'(0)=0 represents a damped oscillation. Which of the following best describes the trajectory of the solution in the phase plane (the y−y′y-y' plane) as tt increases from 0?

  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 r2+0.1r+1=0r^2+0.1r+1=0. The discriminant is (0.1)2−4(1)=−3.99<0(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<0-0.1/2 = -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′′)(y', y'') at the initial point. At t=0t=0, the position is (y,y′)=(1,0)(y, y') = (1, 0). From the differential equation, y′′=−0.1y′−yy'' = -0.1y' - y. At the initial point, y′′=−0.1(0)−1=−1y'' = -0.1(0) - 1 = -1. The velocity vector at (1,0)(1,0) is (y′,y′′)=(0,−1)(y', y'') = (0, -1), which points straight down in the phase plane. A trajectory starting on the positive y-axis at (1,0)(1,0) and immediately moving into the fourth quadrant (where y>0,y′<0y>0, y'<0) must be spiraling in a clockwise direction.

Question 8

An underdamped harmonic oscillator is described by y′′+y′+54y=0y'' + y' + \frac{5}{4}y = 0, with initial conditions y(0)=4y(0) = 4 and y′(0)=2y'(0) = 2. The solution can be written in the form y(t)=Aeλtcos⁡(ωt−ϕ)y(t) = Ae^{\lambda t} \cos(\omega t - \phi), where A>0A > 0. What is the initial amplitude AA?

  1. 424\sqrt{2} (correct answer)
  2. 252\sqrt{5}
  3. 44
  4. 88
Explanation: First, find the general solution. The characteristic equation is r2+r+5/4=0r^2 + r + 5/4 = 0. The roots are r=−1±1−52=−12±ir = \frac{-1 \pm \sqrt{1-5}}{2} = -\frac{1}{2} \pm i. The general solution is y(t)=e−t/2(C1cos⁡(t)+C2sin⁡(t))y(t) = e^{-t/2}(C_1 \cos(t) + C_2 \sin(t)). We apply the initial conditions to find C1C_1 and C2C_2. From y(0)=4y(0)=4, we get 4=e0(C1cos⁡(0)+C2sin⁡(0))4 = e^0(C_1\cos(0) + C_2\sin(0)), which gives C1=4C_1=4. Next, we find the derivative: y′(t)=−12e−t/2(C1cos⁡(t)+C2sin⁡(t))+e−t/2(−C1sin⁡(t)+C2cos⁡(t))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)=2y'(0)=2, we get 2=−12(C1)+C22 = -\frac{1}{2}(C_1) + C_2. Substituting C1=4C_1=4, we have 2=−12(4)+C22 = -\frac{1}{2}(4) + C_2, which gives C2=4C_2 = 4. The solution is y(t)=e−t/2(4cos⁡(t)+4sin⁡(t))y(t) = e^{-t/2}(4\cos(t) + 4\sin(t)). The term in the parentheses can be written as Acos⁡(t−ϕ)A\cos(t - \phi), where the amplitude is A=C12+C22=42+42=16+16=32=42A = \sqrt{C_1^2 + C_2^2} = \sqrt{4^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}.

Question 9

Consider the family of differential equations y′′+2py′+(p2+ω2)y=0y'' + 2py' + (p^2 + \omega^2)y = 0 where p>0p > 0 and ω>0\omega > 0 are parameters. If the amplitude of oscillation decreases by a factor of e−1e^{-1} over exactly one complete period, what is the relationship between pp and ω\omega?

  1. p=ω2πp = \frac{\omega}{2\pi} (correct answer)
  2. p=ωπp = \frac{\omega}{\pi}
  3. ω=p2π\omega = \frac{p}{2\pi}
  4. ω=pπ\omega = \frac{p}{\pi}
Explanation: The characteristic equation gives r=−p±iωr = -p \pm i\omega, so the solution is y=e−pt(Acos⁡(ωt)+Bsin⁡(ωt))y = e^{-pt}(A\cos(\omega t) + B\sin(\omega t)). The amplitude envelope is e−pte^{-pt}. One complete period is T=2πωT = \frac{2\pi}{\omega}. After one period, the amplitude becomes e−pT=e−p⋅2πωe^{-pT} = e^{-p \cdot \frac{2\pi}{\omega}}. For this to equal e−1e^{-1}: −p⋅2πω=−1-p \cdot \frac{2\pi}{\omega} = -1, giving p=ω2πp = \frac{\omega}{2\pi}. Choice B omits the factor of 2. Choices C and D invert the relationship incorrectly.

Question 10

A second-order linear ODE has characteristic polynomial r2+ar+br^2 + ar + b where aa and bb are real constants. If the general solution can be written as y=e−3t(C1cos⁡(ωt)+C2sin⁡(ωt))y = e^{-3t}(C_1\cos(\omega t) + C_2\sin(\omega t)) and the discriminant a2−4b=−64a^2 - 4b = -64, what is the natural frequency ω\omega?

  1. ω=6\omega = 6
  2. ω=8\omega = 8
  3. ω=4\omega = 4 (correct answer)
  4. ω=10\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(C1cos⁡(ωt)+C2sin⁡(ωt))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±iωr = -3 \pm i\omega. For a characteristic polynomial r2+ar+br^2 + ar + b, these roots satisfy the quadratic formula: r=−a±a2−4b2r = \frac{-a \pm \sqrt{a^2 - 4b}}{2}. Since the real part is −3-3, you know −a2=−3\frac{-a}{2} = -3, which gives a=6a = 6. The imaginary part comes from −(a2−4b)2=−(−64)2=642=82=4\frac{\sqrt{-(a^2 - 4b)}}{2} = \frac{\sqrt{-(-64)}}{2} = \frac{\sqrt{64}}{2} = \frac{8}{2} = 4. Therefore, ω=4\omega = 4. Looking at the wrong answers: (A) ω=6\omega = 6 confuses the natural frequency with the real part of the characteristic root. (B) ω=8\omega = 8 takes 64\sqrt{64} directly without dividing by 2 in the quadratic formula. (D) ω=10\omega = \sqrt{10} likely results from algebraic errors when manipulating the discriminant relationship. Study tip: Remember that for complex roots α±iβ\alpha \pm i\beta, the solution is eαt(C1cos⁡(βt)+C2sin⁡(βt))e^{\alpha t}(C_1\cos(\beta t) + C_2\sin(\beta t)). The coefficient of tt inside the trigonometric functions is always the imaginary part of the characteristic root, which equals ∣discriminant∣2\frac{\sqrt{|discriminant|}}{2} when the discriminant is negative.

Question 11

Consider two differential equations: (I) y′′+2y′+5y=0y'' + 2y' + 5y = 0 and (II) y′′+2y′+2y=0y'' + 2y' + 2y = 0. Both have solutions starting from the same initial conditions y(0)=1,y′(0)=0y(0) = 1, y'(0) = 0. At time t=π4t = \frac{\pi}{4}, which statement correctly compares their behaviors?

  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±2ir = -1 \pm 2i, so y1=e−t(cos⁡(2t)+12sin⁡(2t))y_1 = e^{-t}(\cos(2t) + \frac{1}{2}\sin(2t)). For (II): r=−1±ir = -1 \pm i, so y2=e−t(cos⁡(t)+sin⁡(t))y_2 = e^{-t}(\cos(t) + \sin(t)). At t=π4t = \frac{\pi}{4}: y1=e−π/4(0+12)=e−π/42y_1 = e^{-\pi/4}(0 + \frac{1}{2}) = \frac{e^{-\pi/4}}{2} and y2=e−π/4(22+22)=e−π/42y_2 = e^{-\pi/4}(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) = e^{-\pi/4}\sqrt{2}. Both are positive, and 2>12\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 12

Consider the differential equation y′′+4y′+13y=0y'' + 4y' + 13y = 0. If the general solution can be written in the form y=eαt(Acos⁡(βt)+Bsin⁡(βt))y = e^{\alpha t}(A\cos(\beta t) + B\sin(\beta t)), what is the period of oscillation when A=1A = 1 and B=0B = 0?

  1. π3\frac{\pi}{3}
  2. 2π3\frac{2\pi}{3} (correct answer)
  3. π2\frac{\pi}{2}
  4. 2π9\frac{2\pi}{9}
Explanation: The characteristic equation is r2+4r+13=0r^2 + 4r + 13 = 0. Using the quadratic formula: r=−4±16−522=−4±−362=−2±3ir = \frac{-4 \pm \sqrt{16-52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i. So α=−2\alpha = -2 and β=3\beta = 3. The period of oscillation is T=2πβ=2π3T = \frac{2\pi}{\beta} = \frac{2\pi}{3}. Choice A uses β=6\beta = 6 (doubling error). Choice C uses β=4\beta = 4 (confusing with coefficient of y′y'). Choice D uses β=9\beta = 9 (squaring error).

Question 13

The motion of a damped oscillator satisfies y′′+2y′+10y=0y'' + 2y' + 10y = 0 with y(0)=0y(0) = 0 and y′(0)=6y'(0) = 6. At what time t>0t > 0 does the oscillator first return to its equilibrium position y=0y = 0?

  1. t=π6t = \frac{\pi}{6}
  2. t=π2t = \frac{\pi}{2}
  3. t=π3t = \frac{\pi}{3} (correct answer)
  4. t=2π3t = \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=0y'' + 2y' + 10y = 0 is r2+2r+10=0r^2 + 2r + 10 = 0. Using the quadratic formula: r=−2±4−402=−2±−362=−1±3ir = \frac{-2 \pm \sqrt{4-40}}{2} = \frac{-2 \pm \sqrt{-36}}{2} = -1 \pm 3i. Since we have complex roots r=−1±3ir = -1 \pm 3i, this represents underdamped motion. For complex roots α±βi\alpha \pm \beta i, the general solution is y(t)=eαt(c1cos⁡(βt)+c2sin⁡(βt))y(t) = e^{\alpha t}(c_1 \cos(\beta t) + c_2 \sin(\beta t)). Here, α=−1\alpha = -1 and β=3\beta = 3, so y(t)=e−t(c1cos⁡(3t)+c2sin⁡(3t))y(t) = e^{-t}(c_1 \cos(3t) + c_2 \sin(3t)). Applying initial conditions: y(0)=0y(0) = 0 gives us c1=0c_1 = 0. Then y′(t)=e−t(3c2cos⁡(3t)−c2sin⁡(3t))y'(t) = e^{-t}(3c_2 \cos(3t) - c_2 \sin(3t)), and y′(0)=6y'(0) = 6 gives us 3c2=63c_2 = 6, so c2=2c_2 = 2. Therefore, y(t)=2e−tsin⁡(3t)y(t) = 2e^{-t}\sin(3t). The oscillator returns to equilibrium when y(t)=0y(t) = 0. Since e−t≠0e^{-t} \neq 0, we need sin⁡(3t)=0\sin(3t) = 0, which occurs when 3t=nπ3t = n\pi for integer nn. The first positive solution is t=π3t = \frac{\pi}{3}. Answer choice A (π6\frac{\pi}{6}) would correspond to sin⁡(π2)=1≠0\sin(\frac{\pi}{2}) = 1 \neq 0. Choice B (π2\frac{\pi}{2}) gives sin⁡(3π2)=−1≠0\sin(\frac{3\pi}{2}) = -1 \neq 0. Choice D (2π3\frac{2\pi}{3}) gives sin⁡(2π)=0\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.

Question 14

A damped oscillating system is modeled by y′′+4y′+ky=0y'' + 4y' + ky = 0. It is observed that the time between successive maxima of the oscillation is π/3\pi/3. What is the value of the parameter kk?

  1. 13
  2. 36
  3. 40 (correct answer)
  4. 52
Explanation: The time between successive maxima is the quasi-period, TdT_d. We are given Td=π/3T_d = \pi/3. The quasi-period is related to the angular frequency ω\omega by Td=2π/ωT_d = 2\pi/\omega. So, π/3=2π/ω\pi/3 = 2\pi/\omega, which implies ω=6\omega = 6. The angular frequency ω\omega is the imaginary part of the complex roots of the characteristic equation r2+4r+k=0r^2+4r+k=0. The roots are r=−4±16−4k2=−2±4−kr = \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±ik−4r = -2 \pm i\sqrt{k-4}. The imaginary part is ω=k−4\omega = \sqrt{k-4}. We set this equal to 6: k−4=6\sqrt{k-4} = 6. Squaring both sides gives k−4=36k-4=36, so k=40k=40.

Question 15

Consider the differential equation y′′+βy′+γy=0y'' + \beta y' + \gamma y = 0 where β2<4γ\beta^2 < 4\gamma. If one solution is y1=e−3tsin⁡(2t)y_1 = e^{-3t}\sin(2t), what is the value of β+γ\beta + \gamma?

  1. 99
  2. 1010
  3. 1313
  4. 1919 (correct answer)
Explanation: From y1=e−3tsin⁡(2t)y_1 = e^{-3t}\sin(2t), we identify the characteristic roots as r=−3±2ir = -3 \pm 2i. The characteristic equation is (r+3)2+4=r2+6r+13=0(r+3)^2 + 4 = r^2 + 6r + 13 = 0. Comparing with r2+βr+γ=0r^2 + \beta r + \gamma = 0, we get β=6\beta = 6 and γ=13\gamma = 13. Therefore β+γ=6+13=19\beta + \gamma = 6 + 13 = 19. We can verify: β2=36<4γ=52\beta^2 = 36 < 4\gamma = 52, confirming complex roots. Choice A gives γ\gamma only. Choice B uses β+4\beta + 4 (confusing imaginary part with γ\gamma). Choice C gives γ\gamma only.

Question 16

The differential equation y′′+ky′+9y=0y'' + ky' + 9y = 0 has a solution of the form y=e−2tcos⁡(5t)y = e^{-2t}\cos(\sqrt{5}t). Which of the following statements about the general solution is correct?

  1. The general solution is y=e−2t(C1cos⁡(5t)+C2sin⁡(5t))y = e^{-2t}(C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t)) and k=4k = 4 (correct answer)
  2. The general solution is y=e−2t(C1cos⁡(5t)+C2sin⁡(5t))y = e^{-2t}(C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t)) and k=−4k = -4
  3. The general solution is y=e2t(C1cos⁡(5t)+C2sin⁡(5t))y = e^{2t}(C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t)) and k=4k = 4
  4. The general solution requires k2−36<0k^2 - 36 < 0 and the given particular solution is impossible
Explanation: From the given solution y=e−2tcos⁡(5t)y = e^{-2t}\cos(\sqrt{5}t), we identify α=−2\alpha = -2 and β=5\beta = \sqrt{5}. The characteristic roots are r=−2±i5r = -2 \pm i\sqrt{5}. From r2+kr+9=0r^2 + kr + 9 = 0, we have r=−k±k2−362r = \frac{-k \pm \sqrt{k^2-36}}{2}. Comparing: −k2=−2\frac{-k}{2} = -2 gives k=4k = 4, and 36−162=202=5\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 kk. Choice C has wrong sign in exponential. Choice D incorrectly analyzes the discriminant condition.