Complex Roots & Oscillations - Differential Equations

1. Consider the differential equation $y'' + \beta y' + \gamma y = 0$ where $\beta^2 < 4\gamma$. If one solution is $y_1 = e^{-3t}\sin(2t)$, what is the value of $\beta + \gamma$?
   Answer: $19$ 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.
2. The motion of a damped oscillator satisfies $y'' + 2y' + 10y = 0$ with $y(0) = 0$ and $y'(0) = 6$. At what time $t > 0$ does the oscillator first return to its equilibrium position $y = 0$?
   Answer: $t = \frac{\pi}{3}$ 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.
3. A damped oscillating system is modeled by $y'' + 4y' + ky = 0$. It is observed that the time between successive maxima of the oscillation is $\pi/3$. What is the value of the parameter $k$?
   Answer: 40 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 $k=40$.
4. The solution to $y'' + 0.1y' + y = 0$ with initial conditions $y(0)=1, y'(0)=0$ represents a damped oscillation. Which of the following best describes the trajectory of the solution in the phase plane (the $y-y'$ plane) as $t$ increases from 0?
   Answer: A spiral moving inwards towards the origin in a clockwise direction. 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.
5. A physical system is modeled by a second-order linear homogeneous differential equation with constant coefficients. Its solution is observed to be $y(t) = 5e^{-2t}\cos(3t - \pi/4)$. Which of the following differential equations models this system?
   Answer: $y'' + 4y' + 13y = 0$ 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 $y'' + 4y' + 13y = 0$.
6. The motion of a damped oscillator is described by $y'' + 4y' + 20y = 0$. If $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions for this equation, what is the value of the Wronskian $W(y_1, y_2)(t)$ at $t=\ln(2)$?
   Answer: $1/4$ 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 $t=\ln(2)$: $W(\ln(2)) = 4e^{-4\ln(2)} = 4e^{\ln(2^{-4})} = 4 \cdot 2^{-4} = 4/16 = 1/4$.
7. Consider the family of differential equations $y'' + 2py' + (p^2 + \omega^2)y = 0$ where $p > 0$ and $\omega > 0$ are parameters. If the amplitude of oscillation decreases by a factor of $e^{-1}$ over exactly one complete period, what is the relationship between $p$ and $\omega$?
   Answer: $p = \frac{\omega}{2\pi}$ 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.
8. The solution to $y''+0.2y'+25.01y=0$ represents a weakly damped oscillation. Approximately how many full oscillations does the system complete before the amplitude of the oscillation drops to $1/e$ of its initial value?
   Answer: 8 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.
9. Consider two differential equations: (I) $y'' + 2y' + 5y = 0$ and (II) $y'' + 2y' + 2y = 0$. Both have solutions starting from the same initial conditions $y(0) = 1, y'(0) = 0$. At time $t = \frac{\pi}{4}$, which statement correctly compares their behaviors?
   Answer: Both solutions have the same sign, with equation (II) having larger magnitude due to slower oscillation 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.
10. The differential equation $y'' + ky' + 9y = 0$ has a solution of the form $y = e^{-2t}\cos(\sqrt{5}t)$. Which of the following statements about the general solution is correct?
    Answer: The general solution is $y = e^{-2t}(C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t))$ and $k = 4$ 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.
11. Two mass-spring systems, A and B, are described by the differential equations: System A: $y'' + 2y' + 10y = 0$ System B: $2y'' + 2y' + 5y = 0$ Let $\omega_A$ and $\omega_B$ be their respective angular frequencies of oscillation, and let the decay of their amplitudes be governed by factors $e^{\lambda_A t}$ and $e^{\lambda_B t}$. Which of the following statements is true?
    Answer: $\omega_A > \omega_B$ and the amplitude of System A decays faster than System B. 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.
12. An underdamped harmonic oscillator is described by $y'' + y' + \frac{5}{4}y = 0$, with initial conditions $y(0) = 4$ and $y'(0) = 2$. The solution can be written in the form $y(t) = Ae^{\lambda t} \cos(\omega t - \phi)$, where $A > 0$. What is the initial amplitude $A$?
    Answer: $4\sqrt{2}$ 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 $A = \sqrt{C_1^2 + C_2^2} = \sqrt{4^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$.
13. A second-order linear ODE has characteristic polynomial $r^2 + ar + b$ where $a$ and $b$ are real constants. If the general solution can be written as $y = e^{-3t}(C_1\cos(\omega t) + C_2\sin(\omega t))$ and the discriminant $a^2 - 4b = -64$, what is the natural frequency $\omega$?
    Answer: $\omega = 4$ 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.
14. The solution to the initial value problem $y'' + 2y' + 5y = 0$ with $y(0) = 2$ and $y'(0) = \beta$ is a pure damped cosine wave, meaning it has the form $y(t) = Ae^{-t}\cos(\omega t)$. What is the value of $\beta$?
    Answer: $-2$ 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, $\beta = -2$.
15. Consider the differential equation $y'' + 4y' + 13y = 0$. If the general solution can be written in the form $y = e^{\alpha t}(A\cos(\beta t) + B\sin(\beta t))$, what is the period of oscillation when $A = 1$ and $B = 0$?
    Answer: $\frac{2\pi}{3}$ 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).
16. Consider the differential equation $y'' + \gamma y' + 9y = 0$, where $\gamma$ is a real parameter representing a damping coefficient. For which values of $\gamma$ will the solution exhibit oscillations that decay over time?
    Answer: $0 < \gamma < 6$ 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 $0 < \gamma < 6$.

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