Differential Equations Quiz: Odes With Discontinuous Inputs
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Odes With Discontinuous InputsQuestion 1 of 20

Find the solution to the initial value problem y′′+4y=δ(t−π/2)y'' + 4y = \delta(t - \pi/2), with initial conditions y(0)=0y(0)=0 and y′(0)=0y'(0)=0.

y(t)=u(t−π/2)sin⁡(2(t−π/2))y(t) = u(t-\pi/2)\sin(2(t-\pi/2))
y(t)=12u(t−π/2)sin⁡(2t)y(t) = \frac{1}{2}u(t-\pi/2)\sin(2t)
y(t)=−12u(t−π/2)cos⁡(2t)y(t) = -\frac{1}{2}u(t-\pi/2)\cos(2t)
y(t)=−12u(t−π/2)sin⁡(2t)y(t) = -\frac{1}{2}u(t-\pi/2)\sin(2t)
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Differential Equations Quiz

Differential Equations Quiz: Odes With Discontinuous Inputs

Practice Odes With Discontinuous Inputs 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 Odes With Discontinuous Inputs, 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

Find the solution to the initial value problem y′′+4y=δ(t−π/2)y'' + 4y = \delta(t - \pi/2), with initial conditions y(0)=0y(0)=0 and y′(0)=0y'(0)=0.

  1. y(t)=u(t−π/2)sin⁡(2(t−π/2))y(t) = u(t-\pi/2)\sin(2(t-\pi/2))
  2. y(t)=12u(t−π/2)sin⁡(2t)y(t) = \frac{1}{2}u(t-\pi/2)\sin(2t)
  3. y(t)=−12u(t−π/2)cos⁡(2t)y(t) = -\frac{1}{2}u(t-\pi/2)\cos(2t)
  4. y(t)=−12u(t−π/2)sin⁡(2t)y(t) = -\frac{1}{2}u(t-\pi/2)\sin(2t) (correct answer)
Explanation: When you encounter a differential equation with a Dirac delta function (impulse) on the right side, you're dealing with a forced oscillation problem that requires finding the impulse response of the system. The homogeneous equation y′′+4y=0y'' + 4y = 0 has characteristic equation r2+4=0r^2 + 4 = 0, giving r=±2ir = \pm 2i. This means the natural frequency is ω=2\omega = 2, and the homogeneous solution involves sin⁡(2t)\sin(2t) and cos⁡(2t)\cos(2t). For an impulse δ(t−π/2)\delta(t - \pi/2) applied to this second-order system, the response is the impulse response function multiplied by the unit step function u(t−π/2)u(t - \pi/2). The impulse response for y′′+ω2y=δ(t−a)y'' + \omega^2 y = \delta(t - a) with zero initial conditions is 1ωsin⁡(ω(t−a))u(t−a)\frac{1}{\omega}\sin(\omega(t-a))u(t-a). With ω=2\omega = 2 and a=π/2a = \pi/2, this gives 12sin⁡(2(t−π/2))u(t−π/2)\frac{1}{2}\sin(2(t-\pi/2))u(t-\pi/2). Using the identity sin⁡(2(t−π/2))=sin⁡(2t−π)=−sin⁡(2t)\sin(2(t-\pi/2)) = \sin(2t - \pi) = -\sin(2t), the solution becomes −12sin⁡(2t)u(t−π/2)-\frac{1}{2}\sin(2t)u(t-\pi/2). Choice A incorrectly uses coefficient 1 instead of 12\frac{1}{2} and doesn't account for the phase shift. Choice B has the correct coefficient but wrong sign—it misses the negative sign from the trigonometric identity. Choice C incorrectly uses cosine instead of sine, suggesting confusion about which trigonometric function appears in the impulse response. Remember: for impulse problems, always check your trigonometric identities carefully when simplifying the phase-shifted terms, as sign errors are common pitfalls.

Question 2

The response of a mechanical system initially at rest to an external force is described by the displacement y(t)=e−(t−2)sin⁡(t−2)u2(t)y(t) = e^{-(t-2)}\sin(t-2)u_2(t). What is the instantaneous change in velocity, y′(2+)−y′(2−)y'(2^+) - y'(2^-), at time t=2t=2?

  1. 1 (correct answer)
  2. 0
  3. e−2(cos⁡(2)−sin⁡(2))e^{-2}(\cos(2) - \sin(2))
  4. -1
Explanation: The velocity is y′(t)y'(t). For t<2t<2, the system is at rest, so y(t)=0y(t)=0 and the velocity y′(t)=0y'(t)=0. Thus, y′(2−)=0y'(2^-) = 0. For t>2t>2, u2(t)=1u_2(t)=1, so y(t)=e−(t−2)sin⁡(t−2)y(t) = e^{-(t-2)}\sin(t-2). We find the velocity by differentiating with respect to tt using the product rule: y′(t)=−e−(t−2)sin⁡(t−2)+e−(t−2)cos⁡(t−2)y'(t) = -e^{-(t-2)}\sin(t-2) + e^{-(t-2)}\cos(t-2). To find the velocity just after the impulse, we evaluate this expression at t=2+t=2^+. y′(2+)=−e−(2−2)sin⁡(2−2)+e−(2−2)cos⁡(2−2)=−e0sin⁡(0)+e0cos⁡(0)=0+1=1y'(2^+) = -e^{-(2-2)}\sin(2-2) + e^{-(2-2)}\cos(2-2) = -e^0\sin(0) + e^0\cos(0) = 0 + 1 = 1. The instantaneous change in velocity is y′(2+)−y′(2−)=1−0=1y'(2^+) - y'(2^-) = 1 - 0 = 1. This jump in the first derivative is characteristic of the response of a second-order system to a Dirac delta impulse.

Question 3

An RLC circuit with inductance L=1L=1 H, resistance R=2R=2 Ω\Omega, and capacitance C=1C=1 F is initially inert (zero charge and current). At time t=πt=\pi seconds, a constant voltage of 1 V is applied. What is the charge q(t)q(t) on the capacitor for t>πt > \pi?

  1. q(t)=1+(1+t−π)e−(t−π)q(t) = 1 + (1+t-\pi)e^{-(t-\pi)}
  2. q(t)=1−(1+t)e−tq(t) = 1 - (1+t)e^{-t}
  3. q(t)=(1−e−(t−π)−(t−π)e−(t−π))u0(t)q(t) = (1 - e^{-(t-\pi)} - (t-\pi)e^{-(t-\pi)})u_0(t)
  4. q(t)=1−(1+t−π)e−(t−π)q(t) = 1 - (1+t-\pi)e^{-(t-\pi)} (correct answer)
Explanation: When analyzing RLC circuits with delayed voltage inputs, you need to set up and solve a second-order differential equation with shifted initial conditions. For this RLC circuit, Kirchhoff's voltage law gives us Ld2qdt2+Rdqdt+qC=V(t)L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = V(t). With the given values (L=1L=1, R=2R=2, C=1C=1) and voltage applied at t=πt=\pi, this becomes d2qdt2+2dqdt+q=1\frac{d^2q}{dt^2} + 2\frac{dq}{dt} + q = 1 for t>πt > \pi. The solution has the form q(t)=qh(t)+qp(t)q(t) = q_h(t) + q_p(t). The characteristic equation r2+2r+1=0r^2 + 2r + 1 = 0 gives (r+1)2=0(r+1)^2 = 0, so r=−1r = -1 (repeated root). This yields qh(t)=(A+Bt)e−tq_h(t) = (A + Bt)e^{-t}. The particular solution for the constant forcing function is qp=1q_p = 1. Since the voltage starts at t=πt = \pi with zero initial charge and current, we use shifted conditions: q(π)=0q(\pi) = 0 and q′(π)=0q'(\pi) = 0. Substituting u=t−πu = t - \pi transforms this to a problem starting at u=0u = 0, giving us q(u)=1−(1+u)e−uq(u) = 1 - (1 + u)e^{-u}. Converting back: q(t)=1−(1+t−π)e−(t−π)q(t) = 1 - (1 + t - \pi)e^{-(t-\pi)}. Answer choice D is correct. Choice A has the wrong sign on the exponential term. Choice B ignores the time shift entirely, treating the voltage as applied from t=0t = 0. Choice C includes an unnecessary unit step function and incorrect form. Study tip: Always identify when voltage sources are delayed and shift your time variable accordingly. The initial conditions must be applied at the moment the source turns on, not at t=0t = 0.

Question 4

The Laplace transform of the solution to an initial value problem is given by Y(s)=se−2ss2−1Y(s) = \frac{s e^{-2s}}{s^2-1}. What is the solution y(t)y(t)?

  1. y(t)=12(et−2−e−(t+2))u2(t)y(t) = \frac{1}{2}(e^{t-2} - e^{-(t+2)}) u_2(t)
  2. y(t)=cosh⁡(t)u2(t)y(t) = \cosh(t) u_2(t)
  3. y(t)=sinh⁡(t−2)u2(t)y(t) = \sinh(t-2) u_2(t)
  4. y(t)=cosh⁡(t−2)u2(t)y(t) = \cosh(t-2) u_2(t) (correct answer)
Explanation: When you encounter a Laplace transform with an exponential factor like e−2se^{-2s}, this signals a time shift (translation) in the original function. The key insight is recognizing how to handle both the exponential factor and the rational function separately. To find y(t)y(t), you need to use the time-shifting property: if L−1{F(s)}=f(t)\mathcal{L}^{-1}\{F(s)\} = f(t), then L−1{e−asF(s)}=f(t−a)ua(t)\mathcal{L}^{-1}\{e^{-as}F(s)\} = f(t-a)u_a(t), where ua(t)u_a(t) is the unit step function. First, identify F(s)=ss2−1F(s) = \frac{s}{s^2-1}. The inverse Laplace transform of this is cosh⁡(t)\cosh(t), since L−1{ss2−1}=cosh⁡(t)\mathcal{L}^{-1}\{\frac{s}{s^2-1}\} = \cosh(t). With the factor e−2se^{-2s}, this becomes cosh⁡(t−2)u2(t)\cosh(t-2)u_2(t) by the time-shifting property. Looking at the wrong answers: Choice A incorrectly expands the hyperbolic cosine using exponentials but makes sign and algebra errors in the process. Choice B gives cosh⁡(t)u2(t)\cosh(t)u_2(t), which ignores the time shift—this would be correct if there were no e−2se^{-2s} factor. Choice C gives sinh⁡(t−2)u2(t)\sinh(t-2)u_2(t), which would correspond to L−1{e−2ss2−1}\mathcal{L}^{-1}\{\frac{e^{-2s}}{s^2-1}\}, but our numerator has ss, not 1. Study tip: Always decompose problems with e−ase^{-as} factors into two steps: find the inverse transform of the rational part first, then apply the time-shifting property. Remember that L−1{ss2−a2}=cosh⁡(at)\mathcal{L}^{-1}\{\frac{s}{s^2-a^2}\} = \cosh(at) and L−1{1s2−a2}=sinh⁡(at)a\mathcal{L}^{-1}\{\frac{1}{s^2-a^2}\} = \frac{\sinh(at)}{a}.

Question 5

The impulse response of a linear time-invariant system is h(t)=e−tsin⁡(t)h(t) = e^{-t}\sin(t). This is the solution to ay′′+by′+cy=δ(t)ay''+by'+cy=\delta(t) with zero initial conditions. What is the system's step response, i.e., the solution for an input of u0(t)u_0(t)?

  1. y(t)=e−t(cos⁡(t)−sin⁡(t))y(t) = e^{-t}(\cos(t) - \sin(t))
  2. y(t)=12(1−e−t(cos⁡(t)+sin⁡(t)))y(t) = \frac{1}{2}(1 - e^{-t}(\cos(t) + \sin(t))) (correct answer)
  3. y(t)=1−e−tcos⁡(t)y(t) = 1 - e^{-t}\cos(t)
  4. y(t)=e−tsin⁡(t)u0(t)y(t) = e^{-t}\sin(t)u_0(t)
Explanation: When you encounter problems connecting impulse and step responses, remember that these are fundamental relationships in linear systems theory. The step response is the integral of the impulse response. Since the impulse response is h(t)=e−tsin⁡(t)h(t) = e^{-t}\sin(t), the step response is found by integrating: y(t)=∫0th(τ)dτ=∫0te−τsin⁡(τ)dτy(t) = \int_0^t h(\tau) d\tau = \int_0^t e^{-\tau}\sin(\tau) d\tau To evaluate this integral, use integration by parts twice. Let u=sin⁡(τ)u = \sin(\tau) and dv=e−τdτdv = e^{-\tau}d\tau. After applying integration by parts twice (since the sine function cycles through derivatives), you get: ∫e−τsin⁡(τ)dτ=−12e−τ(cos⁡(τ)+sin⁡(τ))+C\int e^{-\tau}\sin(\tau) d\tau = -\frac{1}{2}e^{-\tau}(\cos(\tau) + \sin(\tau)) + C Evaluating from 0 to tt: y(t)=−12e−t(cos⁡(t)+sin⁡(t))+12(1)=12(1−e−t(cos⁡(t)+sin⁡(t)))y(t) = -\frac{1}{2}e^{-t}(\cos(t) + \sin(t)) + \frac{1}{2}(1) = \frac{1}{2}(1 - e^{-t}(\cos(t) + \sin(t))) This confirms answer B is correct. A represents the derivative of the impulse response, not its integral. C is missing the sine term that appears when integrating e−tsin⁡(t)e^{-t}\sin(t). D simply multiplies the impulse response by the unit step, which doesn't give the step response—that's a common misconception about how these responses relate. Study tip: Remember the key relationship: step response = integral of impulse response. Practice integration by parts with exponential-trigonometric products, as they appear frequently in control systems problems. The pattern always yields both sine and cosine terms in the result.

Question 6

The solution to the initial value problem y′+2y=f(t)y' + 2y = f(t) with y(0)=0y(0)=0 is given by y(t)=∫0te−2(t−τ)f(τ)dτy(t) = \int_0^t e^{-2(t-\tau)} f(\tau) d\tau. If the input is a delayed step function f(t)=3u1(t)f(t) = 3u_1(t), what is the resulting output y(t)y(t)?

  1. y(t)=32(1−e−2t)u1(t)y(t) = \frac{3}{2}(1 - e^{-2t})u_1(t)
  2. y(t)=3(1−e−2(t−1))u1(t)y(t) = 3(1 - e^{-2(t-1)})u_1(t)
  3. y(t)=32(1−e−2(t−1))u1(t)y(t) = \frac{3}{2}(1 - e^{-2(t-1)})u_1(t) (correct answer)
  4. y(t)=3e−2(t−1)u1(t)y(t) = 3e^{-2(t-1)}u_1(t)
Explanation: When you encounter a differential equation with a delayed input function, you're working with convolution integrals where the limits of integration are crucial. The unit step function u1(t)u_1(t) equals 0 for t<1t < 1 and 1 for t≥1t ≥ 1, meaning the input "turns on" at t=1t = 1. Since f(τ)=3u1(τ)f(τ) = 3u_1(τ) is zero when τ<1τ < 1, the integral y(t)=∫0te−2(t−τ)f(τ)dτy(t) = \int_0^t e^{-2(t-τ)} f(τ) dτ has no contribution until τ≥1τ ≥ 1. For t≥1t ≥ 1, the effective lower limit becomes τ=1τ = 1, giving us: y(t)=∫1t3e−2(t−τ)dτ=3e−2t∫1te2τdτy(t) = \int_1^t 3e^{-2(t-τ)} dτ = 3e^{-2t} \int_1^t e^{2τ} dτ Evaluating this integral: 3e−2t⋅12[e2τ]1t=32e−2t(e2t−e2)=32(1−e−2(t−1))3e^{-2t} \cdot \frac{1}{2}[e^{2τ}]_1^t = \frac{3}{2}e^{-2t}(e^{2t} - e^2) = \frac{3}{2}(1 - e^{-2(t-1)}) Since this solution is valid only for t≥1t ≥ 1, we multiply by u1(t)u_1(t), giving us answer C. Answer A incorrectly uses the undelayed step response formula. Answer B has the correct exponential form but missing the factor of 12\frac{1}{2} that comes from the integration. Answer D represents only the homogeneous solution term and ignores the steady-state response. Key strategy: When dealing with delayed inputs in convolution integrals, always adjust your integration limits to account for when the input function is actually non-zero. The delay appears in both the exponential term and the unit step function.

Question 7

A system is described by y′′+y=f(t)y'' + y = f(t) with y(0)=0,y′(0)=0y(0)=0, y'(0)=0. The input is a square wave starting at t=0t=0, given by f(t)=u0(t)−2uπ(t)+2u2π(t)−…f(t) = u_0(t) - 2u_{\pi}(t) + 2u_{2\pi}(t) - \dots. What is the solution y(t)y(t) on the interval π≤t<2π\pi \le t < 2\pi?

  1. y(t)=−2(1−cos⁡(t−π))y(t) = -2(1 - \cos(t-\pi))
  2. y(t)=1−cos⁡(t)−2(1−cos⁡(t−π))y(t) = 1 - \cos(t) - 2(1 - \cos(t-\pi)) (correct answer)
  3. y(t)=1−cos⁡(t)y(t) = 1 - \cos(t)
  4. y(t)=1−cos⁡(t)+2(1−cos⁡(t−π))y(t) = 1 - \cos(t) + 2(1 - \cos(t-\pi))
Explanation: When you encounter a differential equation with discontinuous forcing functions like unit step functions, you need to solve it piecewise and apply the principle of superposition. The key insight is that each unit step function "turns on" a new forcing term at specific times. For the interval π≤t<2π\pi \le t < 2\pi, the forcing function f(t)=u0(t)−2uπ(t)+2u2π(t)−…f(t) = u_0(t) - 2u_{\pi}(t) + 2u_{2\pi}(t) - \dots simplifies to f(t)=1−2=−1f(t) = 1 - 2 = -1 since u0(t)=1u_0(t) = 1 and uπ(t)=1u_{\pi}(t) = 1 while u2π(t)=0u_{2\pi}(t) = 0 in this interval. The solution consists of two parts: the response to u0(t)u_0(t) from t=0t = 0 and the response to −2uπ(t)-2u_{\pi}(t) starting at t=πt = \pi. For y′′+y=1y'' + y = 1 with zero initial conditions, the solution is 1−cos⁡(t)1 - \cos(t). When the forcing changes to −2-2 at t=πt = \pi, this adds a response −2(1−cos⁡(t−π))-2(1 - \cos(t-\pi)) with the time shift reflecting when this forcing began. Therefore, y(t)=1−cos⁡(t)−2(1−cos⁡(t−π))y(t) = 1 - \cos(t) - 2(1 - \cos(t-\pi)) for π≤t<2π\pi \le t < 2\pi. Choice A omits the initial response from t=0t = 0. Choice C ignores the effect of the step change at t=πt = \pi. Choice D incorrectly adds rather than subtracts the second term, missing that −2uπ(t)-2u_{\pi}(t) creates a negative contribution. Remember: with piecewise forcing functions, track each "switching point" separately and use superposition to combine all active responses, being careful with the signs of each contribution.

Question 8

Consider the initial value problem y′+y=f(t)y' + y = f(t), with y(0)=1y(0)=1, where the forcing function f(t)f(t) is a rectangular pulse of height 2 for 1≤t<21 \le t < 2, and 0 otherwise. Which expression correctly describes the solution y(t)y(t)?

  1. y(t)=e−t+2(1−e−(t−1))u1(t)−2(1−e−(t−2))u2(t)y(t) = e^{-t} + 2(1 - e^{-(t-1)})u_1(t) - 2(1 - e^{-(t-2)})u_2(t) (correct answer)
  2. y(t)=2(1−e−(t−1))u1(t)−2(1−e−(t−2))u2(t)y(t) = 2(1 - e^{-(t-1)})u_1(t) - 2(1 - e^{-(t-2)})u_2(t)
  3. y(t)=e−t+2(1−e−t)u1(t)−2(1−e−t)u2(t)y(t) = e^{-t} + 2(1 - e^{-t})u_1(t) - 2(1 - e^{-t})u_2(t)
  4. y(t)=e−t+2(1−e−(t−1))u1(t)+2(1−e−(t−2))u2(t)y(t) = e^{-t} + 2(1 - e^{-(t-1)})u_1(t) + 2(1 - e^{-(t-2)})u_2(t)
Explanation: The forcing function can be written using Heaviside functions as f(t)=2(u1(t)−u2(t))f(t) = 2(u_1(t) - u_2(t)). Taking the Laplace transform of the ODE: sY(s)−y(0)+Y(s)=L{f(t)}sY(s) - y(0) + Y(s) = \mathcal{L}\{f(t)\}. With y(0)=1y(0)=1, we get (s+1)Y(s)−1=2e−ss−2e−2ss(s+1)Y(s) - 1 = \frac{2e^{-s}}{s} - \frac{2e^{-2s}}{s}. Solving for Y(s)Y(s): Y(s)=1s+1+e−s2s(s+1)−e−2s2s(s+1)Y(s) = \frac{1}{s+1} + e^{-s}\frac{2}{s(s+1)} - e^{-2s}\frac{2}{s(s+1)}. The term 1s+1\frac{1}{s+1} corresponds to e−te^{-t}. For the other terms, let H(s)=2s(s+1)=2s−2s+1H(s) = \frac{2}{s(s+1)} = \frac{2}{s} - \frac{2}{s+1}. Its inverse transform is h(t)=2−2e−t=2(1−e−t)h(t) = 2 - 2e^{-t} = 2(1-e^{-t}). Applying the second shifting theorem, L−1{e−sH(s)}=h(t−1)u1(t)\mathcal{L}^{-1}\{e^{-s}H(s)\} = h(t-1)u_1(t) and L−1{e−2sH(s)}=h(t−2)u2(t)\mathcal{L}^{-1}\{e^{-2s}H(s)\} = h(t-2)u_2(t). Combining everything gives the solution y(t)=e−t+h(t−1)u1(t)−h(t−2)u2(t)=e−t+2(1−e−(t−1))u1(t)−2(1−e−(t−2))u2(t)y(t) = e^{-t} + h(t-1)u_1(t) - h(t-2)u_2(t) = e^{-t} + 2(1 - e^{-(t-1)})u_1(t) - 2(1 - e^{-(t-2)})u_2(t).

Question 9

What is the solution to the initial value problem y′′+3y′+2y=δ(t)y'' + 3y' + 2y = \delta(t) with initial conditions y(0)=1y(0)=1 and y′(0)=0y'(0)=0?

  1. y(t)=−e−t+2e−2ty(t) = -e^{-t} + 2e^{-2t}
  2. y(t)=e−t−e−2ty(t) = e^{-t} - e^{-2t}
  3. y(t)=3e−t−2e−2ty(t) = 3e^{-t} - 2e^{-2t} (correct answer)
  4. y(t)=2e−t−e−2ty(t) = 2e^{-t} - e^{-2t}
Explanation: When you encounter a differential equation with a delta function (impulse) and initial conditions, you're dealing with a problem that requires finding both the homogeneous solution and accounting for the impulse response. First, solve the homogeneous equation y′′+3y′+2y=0y'' + 3y' + 2y = 0. The characteristic equation is r2+3r+2=0r^2 + 3r + 2 = 0, which factors as (r+1)(r+2)=0(r+1)(r+2) = 0, giving roots r=−1r = -1 and r=−2r = -2. So the homogeneous solution is yh(t)=c1e−t+c2e−2ty_h(t) = c_1e^{-t} + c_2e^{-2t}. For the delta function δ(t)\delta(t), the solution has a jump discontinuity in the first derivative at t=0t = 0. The jump magnitude equals the coefficient of δ(t)\delta(t) (which is 1), so y′(0+)−y′(0−)=1y'(0^+) - y'(0^-) = 1. Since y′(0−)=0y'(0^-) = 0, we have y′(0+)=1y'(0^+) = 1. Using initial conditions: y(0)=c1+c2=1y(0) = c_1 + c_2 = 1 and y′(0+)=−c1−2c2=1y'(0^+) = -c_1 - 2c_2 = 1. Solving this system: from the first equation, c1=1−c2c_1 = 1 - c_2. Substituting: −(1−c2)−2c2=1-(1-c_2) - 2c_2 = 1, so −1+c2−2c2=1-1 + c_2 - 2c_2 = 1, giving c2=−2c_2 = -2 and c1=3c_1 = 3. Therefore, y(t)=3e−t−2e−2ty(t) = 3e^{-t} - 2e^{-2t}, which is answer C. Answer A uses the wrong initial condition for y′y'. Answer B ignores the original initial conditions entirely. Answer D has the coefficients backwards. Key strategy: With impulse functions, always account for the jump discontinuity in the derivative—this changes your effective initial conditions for the homogeneous solution.

Question 10

An RLC circuit with resistance R=3ΩR=3\Omega, inductance L=1HL=1\text{H}, and capacitance C=0.5FC=0.5\text{F} is governed by the equation Lq′′+Rq′+1Cq=E(t)Lq'' + Rq' + \frac{1}{C}q = E(t). If the circuit starts with zero charge and zero current, and a constant voltage of E(t)=10VE(t) = 10\text{V} is applied at time t=2st=2\text{s}, what is the charge q(t)q(t) on the capacitor for t>0t>0?

  1. q(t)=u(t−2)[5−10e−t+5e−2t]q(t) = u(t-2) [5 - 10e^{-t} + 5e^{-2t}]
  2. q(t)=u(t−2)[5−10e−(t−2)+5e−2(t−2)]q(t) = u(t-2) [5 - 10e^{-(t-2)} + 5e^{-2(t-2)}] (correct answer)
  3. q(t)=u(t−2)[10−10e−(t−2)+5e−2(t−2)]q(t) = u(t-2) [10 - 10e^{-(t-2)} + 5e^{-2(t-2)}]
  4. q(t)=u(t−2)[5+5e−(t−2)−10e−2(t−2)]q(t) = u(t-2) [5 + 5e^{-(t-2)} - 10e^{-2(t-2)}]
Explanation: When you encounter RLC circuit problems with delayed inputs, you're dealing with shifted differential equations that require careful attention to initial conditions and the timing of the applied voltage. First, substitute the given values into the governing equation: 1⋅q′′+3q′+10.5q=E(t)1 \cdot q'' + 3q' + \frac{1}{0.5}q = E(t), which simplifies to q′′+3q′+2q=E(t)q'' + 3q' + 2q = E(t). The characteristic equation r2+3r+2=0r^2 + 3r + 2 = 0 factors as (r+1)(r+2)=0(r+1)(r+2) = 0, giving roots r=−1,−2r = -1, -2. Since the voltage E(t)=10E(t) = 10 is applied only for t≥2t \geq 2, you need E(t)=10u(t−2)E(t) = 10u(t-2). For t<2t < 2, the circuit remains at rest with zero charge and current. At t=2t = 2, the system "starts" responding to the constant voltage. The solution for t≥2t \geq 2 takes the form q(t)=qh+qpq(t) = q_h + q_p, where the homogeneous solution is qh=c1e−(t−2)+c2e−2(t−2)q_h = c_1e^{-(t-2)} + c_2e^{-2(t-2)} (shifted to start at t=2t = 2) and the particular solution is qp=5q_p = 5 (since 102=5\frac{10}{2} = 5). Applying initial conditions q(2)=0q(2) = 0 and q′(2)=0q'(2) = 0 gives c1=−10c_1 = -10 and c2=5c_2 = 5, yielding q(t)=u(t−2)[5−10e−(t−2)+5e−2(t−2)]q(t) = u(t-2)[5 - 10e^{-(t-2)} + 5e^{-2(t-2)}]. Choice A incorrectly uses unshifted exponentials. Choice C has the wrong particular solution coefficient. Choice D has incorrect signs in the exponential terms, which would violate the initial conditions. Strategy tip: In delayed differential equation problems, always shift your time variable in both the exponentials and initial conditions to match when the input actually begins.

Question 11

Determine the solution to the initial value problem y′+y=u(t−1)+δ(t−2)y' + y = u(t-1) + \delta(t-2), given y(0)=0y(0)=0.

  1. y(t)=u(t−1)[1−e−t]+u(t−2)e−ty(t) = u(t-1)[1 - e^{-t}] + u(t-2)e^{-t}
  2. y(t)=u(t−1)[1−e−(t−1)]+u(t−2)e−(t−2)y(t) = u(t-1)[1 - e^{-(t-1)}] + u(t-2)e^{-(t-2)} (correct answer)
  3. y(t)=u(t−1)[1−e−(t−1)]+u(t−2)e−(t−1)y(t) = u(t-1)[1 - e^{-(t-1)}] + u(t-2)e^{-(t-1)}
  4. y(t)=[1−e−(t−1)]+e−(t−2)y(t) = [1 - e^{-(t-1)}] + e^{-(t-2)}
Explanation: When you encounter a differential equation with unit step functions u(t−a)u(t-a) and Dirac delta functions δ(t−a)\delta(t-a), you're dealing with piecewise forcing functions that "turn on" at specific times. The key insight is that each term contributes to the solution only after its activation time, and the form of each contribution depends on the type of function. For this first-order linear ODE y′+y=u(t−1)+δ(t−2)y' + y = u(t-1) + \delta(t-2), the homogeneous solution has integrating factor ete^t. The unit step u(t−1)u(t-1) acts like a constant forcing function starting at t=1t=1, producing a particular solution of the form 1−e−(t−1)1 - e^{-(t-1)} that's "turned on" by multiplying by u(t−1)u(t-1). The delta function δ(t−2)\delta(t-2) creates an instantaneous impulse at t=2t=2, contributing a decaying exponential e−(t−2)e^{-(t-2)} that begins at that moment, activated by u(t−2)u(t-2). Choice B correctly captures both behaviors: u(t−1)[1−e−(t−1)]+u(t−2)e−(t−2)u(t-1)[1 - e^{-(t-1)}] + u(t-2)e^{-(t-2)}. Choice A incorrectly uses e−te^{-t} instead of the shifted exponentials, missing that solutions should start fresh at their activation times. Choice C has the wrong decay rate for the delta impulse term—it should decay as e−(t−2)e^{-(t-2)}, not e−(t−1)e^{-(t-1)}. Choice D omits the unit step functions entirely, meaning the solution would be active for all tt rather than starting at the proper times. Remember: unit steps and delta functions create solutions that begin at their shift points, so your exponential decay should always be measured from those activation times.

Question 12

The equation y′′−2y′+y=g(t)y'' - 2y' + y = g(t) has initial conditions y(0)=1y(0) = 1 and y′(0)=0y'(0) = 0, where g(t)={tif 0≤t<24−tif 2≤t<40if t≥4g(t) = \begin{cases} t & \text{if } 0 \leq t < 2 \\ 4-t & \text{if } 2 \leq t < 4 \\ 0 & \text{if } t \geq 4 \end{cases} . Which expression correctly represents L{g(t)}\mathcal{L}\{g(t)\}?

  1. 1s2−e−2s(2s+1)s2+e−4s(4s+1)s2\frac{1}{s^2} - \frac{e^{-2s}(2s+1)}{s^2} + \frac{e^{-4s}(4s+1)}{s^2}
  2. 1s2+e−2s(2s−1)s2−e−4s(4s−1)s2\frac{1}{s^2} + \frac{e^{-2s}(2s-1)}{s^2} - \frac{e^{-4s}(4s-1)}{s^2}
  3. 1s2−e−2s(2s−1)s2+e−4s(4s−1)s2\frac{1}{s^2} - \frac{e^{-2s}(2s-1)}{s^2} + \frac{e^{-4s}(4s-1)}{s^2} (correct answer)
  4. 1s2+e−2s(2s+1)s2−e−4s(4s+1)s2\frac{1}{s^2} + \frac{e^{-2s}(2s+1)}{s^2} - \frac{e^{-4s}(4s+1)}{s^2}
Explanation: We rewrite g(t)=t[u(t)−u(t−2)]+(4−t)[u(t−2)−u(t−4)]g(t) = t[u(t) - u(t-2)] + (4-t)[u(t-2) - u(t-4)]. Using the shifting property and the fact that L{t}=1s2\mathcal{L}\{t\} = \frac{1}{s^2}, we get L{g(t)}=1s2−e−2sL{(t+2)}+e−4sL{(t+4−4)}\mathcal{L}\{g(t)\} = \frac{1}{s^2} - e^{-2s}\mathcal{L}\{(t+2)\} + e^{-4s}\mathcal{L}\{(t+4-4)\}. After simplification, this yields choice C.

Question 13

The transfer function H(s)=1s2+2s+2H(s) = \frac{1}{s^2 + 2s + 2} represents a system subjected to the input f(t)=u(t−1)−2u(t−2)+u(t−3)f(t) = u(t-1) - 2u(t-2) + u(t-3). What is the steady-state behavior of the output as t→∞t \to \infty?

  1. The output oscillates with exponentially decreasing amplitude
  2. The output approaches a non-zero constant value
  3. The output approaches zero asymptotically (correct answer)
  4. The output grows without bound
Explanation: The poles of H(s)H(s) are at s=−1±is = -1 \pm i, both having negative real parts, making the system stable. The input f(t)f(t) has ∑\sum of step coefficients = 1−2+1=01 - 2 + 1 = 0, meaning the final value of the input is zero. For a stable system with zero final input value, the steady-state output approaches zero. This is confirmed by the final value theorem: lim⁡t→∞y(t)=lim⁡s→0sY(s)=0\lim_{t \to \infty} y(t) = \lim_{s \to 0} sY(s) = 0.

Question 14

A second-order system has the response y(t)=e−tcos⁡(2t)+12[u(t−π)−u(t−2π)]e−(t−π)sin⁡(2(t−π))y(t) = e^{-t}\cos(2t) + \frac{1}{2}[u(t-\pi) - u(t-2\pi)]e^{-(t-\pi)}\sin(2(t-\pi)) for t>0t > 0. What was the form of the discontinuous input that produced this response?

  1. A unit impulse at t=πt = \pi followed by a negative unit impulse at t=2πt = 2\pi
  2. A unit step at t=πt = \pi followed by a negative unit step at t=2πt = 2\pi
  3. A rectangular pulse of magnitude 12\frac{1}{2} from t=πt = \pi to t=2πt = 2\pi
  4. A rectangular pulse of magnitude 11 from t=πt = \pi to t=2πt = 2\pi (correct answer)
Explanation: The first term e−tcos⁡(2t)e^{-t}\cos(2t) represents the homogeneous response to initial conditions. The second term has the form of a response to a rectangular pulse input. The coefficient 12\frac{1}{2} comes from the system's transfer function, and the factor [u(t−π)−u(t−2π)][u(t-\pi) - u(t-2\pi)] indicates the pulse duration. Since the impulse response would be 12e−tsin⁡(2t)\frac{1}{2}e^{-t}\sin(2t) for this underdamped system, a unit rectangular pulse from t=πt = \pi to t=2πt = 2\pi produces exactly the observed forced response term.

Question 15

The convolution integral y(t)=∫0th(t−τ)f(τ)dτy(t) = \int_0^t h(t-\tau)f(\tau)d\tau is used to find the response of y′′+4y′+4y=f(t)y'' + 4y' + 4y = f(t) where f(t)=∑k=0∞(−1)kδ(t−k)f(t) = \sum_{k=0}^{\infty}(-1)^k\delta(t-k). What is the pattern of y(t)y(t) for large tt?

  1. y(t)y(t) oscillates with increasing amplitude proportional to tt
  2. y(t)y(t) oscillates with decreasing amplitude, approaching zero (correct answer)
  3. y(t)y(t) approaches a periodic steady state with period 22
  4. y(t)y(t) grows monotonically without bound
Explanation: When you encounter convolution problems involving impulse functions and differential equations, you're dealing with system response analysis. The key insight is understanding how the system's impulse response function behaves and how repeated impulses affect the overall response. First, solve the homogeneous equation y′′+4y′+4y=0y'' + 4y' + 4y = 0. The characteristic equation r2+4r+4=0r^2 + 4r + 4 = 0 gives (r+2)2=0(r+2)^2 = 0, so r=−2r = -2 is a repeated root. This means the impulse response function is h(t)=te−2tu(t)h(t) = te^{-2t}u(t), where u(t)u(t) is the unit step function. The forcing function f(t)=∑k=0∞(−1)kδ(t−k)f(t) = \sum_{k=0}^{\infty}(-1)^k\delta(t-k) represents alternating unit impulses at integer times: positive at t=0,2,4,...t = 0, 2, 4, ... and negative at t=1,3,5,...t = 1, 3, 5, ... Each impulse contributes a response that decays exponentially with the factor e−2(t−k)e^{-2(t-k)} for t>kt > k. Since the exponential decay dominates and the system is stable (both characteristic roots have negative real parts), the overall response oscillates but with decreasing amplitude as tt increases. Choice A is wrong because the exponential decay e−2te^{-2t} dominates any polynomial growth. Choice C is incorrect because the responses from previous impulses don't completely die out between impulses, preventing true periodicity. Choice D fails because the system is stable—the negative characteristic roots ensure bounded responses. Remember: for stable linear systems (negative characteristic roots), bounded inputs always produce bounded outputs that eventually decay, regardless of oscillatory behavior in the forcing function.

Question 16

The equation y′′+y=f(t)y'' + y = f(t) where f(t)=sin⁡(t)[u(t)−u(t−π)]f(t) = \sin(t)[u(t) - u(t-\pi)] represents a resonance condition with finite duration input. With initial conditions y(0)=0y(0) = 0, y′(0)=1y'(0) = 1, what is the maximum value of ∣y(t)∣|y(t)| for t>2πt > 2\pi?

  1. π2\frac{\pi}{2} (correct answer)
  2. π\pi
  3. π2+1\frac{\pi}{2} + 1
  4. π+1\pi + 1
Explanation: For t∈[0,π]t \in [0,\pi], we have resonance since the input frequency equals the natural frequency. The particular solution during this interval grows as −t2cos⁡(t)-\frac{t}{2}\cos(t). The complete solution for t∈[0,π]t \in [0,\pi] is y(t)=sin⁡(t)−t2cos⁡(t)y(t) = \sin(t) - \frac{t}{2}\cos(t). At t=πt = \pi: y(π)=0−π2(−1)=π2y(\pi) = 0 - \frac{\pi}{2}(-1) = \frac{\pi}{2} and y′(π)=0y'(\pi) = 0. For t>πt > \pi, only the homogeneous solution exists: y(t)=π2cos⁡(t−π)=−π2cos⁡(t)y(t) = \frac{\pi}{2}\cos(t-\pi) = -\frac{\pi}{2}\cos(t). The maximum value of ∣y(t)∣|y(t)| for t>2πt > 2\pi is π2\frac{\pi}{2}.

Question 17

A system is modeled by the initial value problem y′′+2y′+y=3δ(t−1)y'' + 2y' + y = 3\delta(t-1), with initial conditions y(0)=0y(0)=0 and y′(0)=0y'(0)=0. What is the system's response y(t)y(t) for t>0t>0?

  1. y(t)=3(t−1)e−(t−1)u1(t)y(t) = 3(t-1)e^{-(t-1)}u_1(t) (correct answer)
  2. y(t)=3te−tu1(t)y(t) = 3te^{-t}u_1(t)
  3. y(t)=3(t−1)e−tu1(t)y(t) = 3(t-1)e^{-t}u_1(t)
  4. y(t)=3(t−1)e−(t−1)y(t) = 3(t-1)e^{-(t-1)}
Explanation: Taking the Laplace transform of the equation gives (s2+2s+1)Y(s)=3e−s(s^2+2s+1)Y(s) = 3e^{-s}. This simplifies to (s+1)2Y(s)=3e−s(s+1)^2 Y(s) = 3e^{-s}, so Y(s)=3e−s(s+1)2Y(s) = \frac{3e^{-s}}{(s+1)^2}. Let H(s)=3(s+1)2H(s) = \frac{3}{(s+1)^2}. The inverse Laplace transform of H(s)H(s) is h(t)=3te−th(t) = 3te^{-t}. According to the second shifting theorem, the inverse transform of e−csH(s)e^{-cs}H(s) is h(t−c)uc(t)h(t-c)u_c(t). Here, c=1c=1, so y(t)=h(t−1)u1(t)=3(t−1)e−(t−1)u1(t)y(t) = h(t-1)u_1(t) = 3(t-1)e^{-(t-1)}u_1(t).

Question 18

A system is modeled by the differential equation y′′+4y′+5y=f(t)y'' + 4y' + 5y = f(t) with y(0)=y′(0)=0y(0)=y'(0)=0. If the input is an impulse at time t=ct=c, f(t)=δ(t−c)f(t) = \delta(t-c), what is the impulse response y(t)y(t)?

  1. y(t)=e−2(t−c)cos⁡(t−c)uc(t)y(t) = e^{-2(t-c)}\cos(t-c)u_c(t)
  2. y(t)=e−2tsin⁡(t)uc(t)y(t) = e^{-2t}\sin(t)u_c(t)
  3. y(t)=e−2(t−c)sin⁡(t−c)uc(t)y(t) = e^{-2(t-c)}\sin(t-c)u_c(t) (correct answer)
  4. y(t)=(e−(t−c)sin⁡(2(t−c)))uc(t)y(t) = (e^{-(t-c)}\sin(2(t-c)))u_c(t)
Explanation: When you encounter a differential equation with an impulse function (Dirac delta), you're dealing with finding the system's impulse response - how it reacts to a sudden, instantaneous input at time t=ct = c. To solve y′′+4y′+5y=δ(t−c)y'' + 4y' + 5y = \delta(t-c) with zero initial conditions, start by finding the characteristic equation: r2+4r+5=0r^2 + 4r + 5 = 0. Using the quadratic formula: r=−4±16−202=−4±2i2=−2±ir = \frac{-4 \pm \sqrt{16-20}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i. This gives us complex roots with real part α=−2\alpha = -2 and imaginary part β=1\beta = 1. For an impulse at t=ct = c, the response has the form of the homogeneous solution shifted by cc and multiplied by the unit step function uc(t)u_c(t). With complex roots −2±i-2 \pm i, the impulse response is: y(t)=e−2(t−c)sin⁡(t−c)uc(t)y(t) = e^{-2(t-c)}\sin(t-c)u_c(t). Looking at the wrong answers: Choice A uses cosine instead of sine - but impulse responses for second-order systems with zero initial conditions always involve sine, not cosine. Choice B has the exponential decay starting from t=0t = 0 rather than t=ct = c, ignoring the time shift in the impulse. Choice D has the wrong exponential decay rate (−1-1 instead of −2-2) and wrong frequency (argument 2(t−c)2(t-c) instead of (t−c)(t-c)). Study tip: For impulse responses, remember that the time shift in the input (t−ct-c) must appear consistently throughout the solution, and complex roots α±βi\alpha \pm \beta i give responses involving eα(t−c)sin⁡(β(t−c))e^{\alpha(t-c)}\sin(\beta(t-c)) for zero initial conditions.

Question 19

Find the solution to the initial value problem y′′+9y=tu2(t)y'' + 9y = t u_2(t), with initial conditions y(0)=0y(0)=0 and y′(0)=0y'(0)=0.

  1. y(t)=(29+19(t−2)−29cos⁡(3(t−2))−127sin⁡(3(t−2)))u2(t)y(t) = \left( \frac{2}{9} + \frac{1}{9}(t-2) - \frac{2}{9}\cos(3(t-2)) - \frac{1}{27}\sin(3(t-2)) \right) u_2(t) (correct answer)
  2. y(t)=(19(t−2)−127sin⁡(3(t−2)))u2(t)y(t) = \left( \frac{1}{9}(t-2) - \frac{1}{27}\sin(3(t-2)) \right) u_2(t)
  3. y(t)=(19t−127sin⁡(3t))u2(t)y(t) = \left( \frac{1}{9}t - \frac{1}{27}\sin(3t) \right) u_2(t)
  4. y(t)=(29+19t−29cos⁡(3t)−127sin⁡(3t))u2(t)y(t) = \left( \frac{2}{9} + \frac{1}{9}t - \frac{2}{9}\cos(3t) - \frac{1}{27}\sin(3t) \right) u_2(t)
Explanation: To apply the second shifting theorem, we must write the forcing function g(t)=tu2(t)g(t) = t u_2(t) in the form f(t−2)u2(t)f(t-2)u_2(t). Let t−2=vt-2 = v, so t=v+2t = v+2. Thus, tu2(t)=((t−2)+2)u2(t)=(t−2)u2(t)+2u2(t)t u_2(t) = ((t-2)+2)u_2(t) = (t-2)u_2(t) + 2u_2(t). The Laplace transform is L{tu2(t)}=e−2sL{t+2}=e−2s(1s2+2s)=e−2s1+2ss2\mathcal{L}\{t u_2(t)\} = e^{-2s}\mathcal{L}\{t+2\} = e^{-2s}(\frac{1}{s^2} + \frac{2}{s}) = e^{-2s}\frac{1+2s}{s^2}. The transformed ODE is (s2+9)Y(s)=e−2s2s+1s2(s^2+9)Y(s) = e^{-2s}\frac{2s+1}{s^2}. So, Y(s)=e−2sH(s)Y(s) = e^{-2s} H(s) where H(s)=2s+1s2(s2+9)H(s) = \frac{2s+1}{s^2(s^2+9)}. Partial fraction decomposition gives H(s)=2/9s+1/9s2−(2/9)s+1/9s2+9H(s) = \frac{2/9}{s} + \frac{1/9}{s^2} - \frac{(2/9)s + 1/9}{s^2+9}. The inverse transform is h(t)=29+19t−29cos⁡(3t)−127sin⁡(3t)h(t) = \frac{2}{9} + \frac{1}{9}t - \frac{2}{9}\cos(3t) - \frac{1}{27}\sin(3t). The final solution is y(t)=h(t−2)u2(t)y(t) = h(t-2)u_2(t).

Question 20

A system is modeled by the initial value problem y′+2y=f(t)y' + 2y = f(t) with y(0)=0y(0) = 0, where the input f(t)f(t) is a single rectangular pulse of magnitude 1 for 0≤t<10 \le t < 1. What is the solution y(t)y(t)?

  1. y(t)=12(1−e−2t)−12u(t−1)(1−e−2(t−1))y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u(t-1)(1 - e^{-2(t-1)}) (correct answer)
  2. y(t)=12(1−e−2t)−12u(t−1)(1−e−2t)y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u(t-1)(1 - e^{-2t})
  3. y(t)=1−e−2t−u(t−1)(1−e−2(t−1))y(t) = 1 - e^{-2t} - u(t-1)(1 - e^{-2(t-1)})
  4. y(t)=12(1−e−2t)+12u(t−1)(1−e−2(t−1))y(t) = \frac{1}{2}(1 - e^{-2t}) + \frac{1}{2}u(t-1)(1 - e^{-2(t-1)})
Explanation: The forcing function can be written using Heaviside step functions as f(t)=u(t)−u(t−1)f(t) = u(t) - u(t-1), which is 1−u(t−1)1 - u(t-1) for t≥0t \ge 0. Taking the Laplace transform of the ODE gives sY(s)−y(0)+2Y(s)=L{1−u(t−1)}sY(s) - y(0) + 2Y(s) = \mathcal{L}\{1 - u(t-1)\}. With y(0)=0y(0)=0, we get (s+2)Y(s)=1s−e−ss(s+2)Y(s) = \frac{1}{s} - \frac{e^{-s}}{s}. Solving for Y(s)Y(s): Y(s)=1s(s+2)−e−ss(s+2)Y(s) = \frac{1}{s(s+2)} - \frac{e^{-s}}{s(s+2)}. Using partial fractions, 1s(s+2)=12(1s−1s+2)\frac{1}{s(s+2)} = \frac{1}{2}(\frac{1}{s} - \frac{1}{s+2}). So, Y(s)=12(1s−1s+2)−12e−s(1s−1s+2)Y(s) = \frac{1}{2}(\frac{1}{s} - \frac{1}{s+2}) - \frac{1}{2}e^{-s}(\frac{1}{s} - \frac{1}{s+2}). Let G(s)=12(1s−1s+2)G(s) = \frac{1}{2}(\frac{1}{s} - \frac{1}{s+2}), whose inverse Laplace transform is g(t)=12(1−e−2t)g(t) = \frac{1}{2}(1 - e^{-2t}). The solution is y(t)=g(t)−u(t−1)g(t−1)y(t) = g(t) - u(t-1)g(t-1). Substituting g(t)g(t) gives y(t)=12(1−e−2t)−12u(t−1)(1−e−2(t−1))y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u(t-1)(1 - e^{-2(t-1)}).