ODEs with Discontinuous Inputs - Differential Equations

An RLC circuit with resistance $R=3\Omega$ , inductance $L=1\text{H}$ , and capacitance $C=0.5\text{F}$ is governed by the equation $Lq'' + Rq' + \frac{1}{C}q = E(t)$ . If the circuit starts with zero charge and zero current, and a constant voltage of $E(t) = 10\text{V}$ is applied at time $t=2\text{s}$ , what is the charge $q(t)$ on the capacitor for $t>0$ ?
Answer: $q(t) = u(t-2) [5 - 10e^{-(t-2)} + 5e^{-2(t-2)}]$ 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: $1 \cdot q'' + 3q' + \frac{1}{0.5}q = E(t)$ , which simplifies to $q'' + 3q' + 2q = E(t)$ . The characteristic equation $r^2 + 3r + 2 = 0$ factors as $(r+1)(r+2) = 0$ , giving roots $r = -1, -2$ . Since the voltage $E(t) = 10$ is applied only for $t \geq 2$ , you need $E(t) = 10u(t-2)$ . For $t < 2$ , the circuit remains at rest with zero charge and current. At $t = 2$ , the system "starts" responding to the constant voltage. The solution for $t \geq 2$ takes the form $q(t) = q_h + q_p$ , where the homogeneous solution is $q_h = c_1e^{-(t-2)} + c_2e^{-2(t-2)}$ (shifted to start at $t = 2$ ) and the particular solution is $q_p = 5$ (since $\frac{10}{2} = 5$ ). Applying initial conditions $q(2) = 0$ and $q'(2) = 0$ gives $c_1 = -10$ and $c_2 = 5$ , yielding $q(t) = u(t-2)[5 - 10e^{-(t-2)} + 5e^{-2(t-2)}]$ . 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.
A second-order system has the response $y(t) = e^{-t}\cos(2t) + \frac{1}{2}[u(t-\pi) - u(t-2\pi)]e^{-(t-\pi)}\sin(2(t-\pi))$ for $t > 0$ . What was the form of the discontinuous input that produced this response?
Answer: A rectangular pulse of magnitude $1$ from $t = \pi$ to $t = 2\pi$ The first term $e^{-t}\cos(2t)$ represents the homogeneous response to initial conditions. The second term has the form of a response to a rectangular pulse input. The coefficient $\frac{1}{2}$ comes from the system's transfer function, and the factor $[u(t-\pi) - u(t-2\pi)]$ indicates the pulse duration. Since the impulse response would be $\frac{1}{2}e^{-t}\sin(2t)$ for this underdamped system, a unit rectangular pulse from $t = \pi$ to $t = 2\pi$ produces exactly the observed forced response term.
The response of a mechanical system initially at rest to an external force is described by the displacement $y(t) = e^{-(t-2)}\sin(t-2)u_2(t)$ . What is the instantaneous change in velocity, $y'(2^+) - y'(2^-)$ , at time $t=2$ ?
Answer: 1 The velocity is $y'(t)$ . For $t<2$ , the system is at rest, so $y(t)=0$ and the velocity $y'(t)=0$ . Thus, $y'(2^-) = 0$ . For $t>2$ , $u_2(t)=1$ , so $y(t) = e^{-(t-2)}\sin(t-2)$ . We find the velocity by differentiating with respect to $t$ using the product rule: $y'(t) = -e^{-(t-2)}\sin(t-2) + e^{-(t-2)}\cos(t-2)$ . To find the velocity just after the impulse, we evaluate this expression at $t=2^+$ . $y'(2^+) = -e^{-(2-2)}\sin(2-2) + e^{-(2-2)}\cos(2-2) = -e^0\sin(0) + e^0\cos(0) = 0 + 1 = 1$ . The instantaneous change in velocity is $y'(2^+) - y'(2^-) = 1 - 0 = 1$ . This jump in the first derivative is characteristic of the response of a second-order system to a Dirac delta impulse.
A system is modeled by the initial value problem $y' + 2y = f(t)$ with $y(0) = 0$ , where the input $f(t)$ is a single rectangular pulse of magnitude 1 for $0 \le t < 1$ . What is the solution $y(t)$ ?
Answer: $y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u(t-1)(1 - e^{-2(t-1)})$ The forcing function can be written using Heaviside step functions as $f(t) = u(t) - u(t-1)$ , which is $1 - u(t-1)$ for $t \ge 0$ . Taking the Laplace transform of the ODE gives $sY(s) - y(0) + 2Y(s) = \mathcal{L}\{1 - u(t-1)\}$ . With $y(0)=0$ , we get $(s+2)Y(s) = \frac{1}{s} - \frac{e^{-s}}{s}$ . Solving for $Y(s)$ : $Y(s) = \frac{1}{s(s+2)} - \frac{e^{-s}}{s(s+2)}$ . Using partial fractions, $\frac{1}{s(s+2)} = \frac{1}{2}(\frac{1}{s} - \frac{1}{s+2})$ . So, $Y(s) = \frac{1}{2}(\frac{1}{s} - \frac{1}{s+2}) - \frac{1}{2}e^{-s}(\frac{1}{s} - \frac{1}{s+2})$ . Let $G(s) = \frac{1}{2}(\frac{1}{s} - \frac{1}{s+2})$ , whose inverse Laplace transform is $g(t) = \frac{1}{2}(1 - e^{-2t})$ . The solution is $y(t) = g(t) - u(t-1)g(t-1)$ . Substituting $g(t)$ gives $y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u(t-1)(1 - e^{-2(t-1)})$ .
The convolution integral $y(t) = \int_0^t h(t-\tau)f(\tau)d\tau$ is used to find the response of $y'' + 4y' + 4y = f(t)$ where $f(t) = \sum_{k=0}^{\infty}(-1)^k\delta(t-k)$ . What is the pattern of $y(t)$ for large $t$ ?
Answer: $y(t)$ oscillates with decreasing amplitude, approaching zero 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 $y'' + 4y' + 4y = 0$ . The characteristic equation $r^2 + 4r + 4 = 0$ gives $(r+2)^2 = 0$ , so $r = -2$ is a repeated root. This means the impulse response function is $h(t) = te^{-2t}u(t)$ , where $u(t)$ is the unit step function. The forcing function $f(t) = \sum_{k=0}^{\infty}(-1)^k\delta(t-k)$ represents alternating unit impulses at integer times: positive at $t = 0, 2, 4, ...$ and negative at $t = 1, 3, 5, ...$ Each impulse contributes a response that decays exponentially with the factor $e^{-2(t-k)}$ for $t > k$ . 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 $t$ increases. Choice A is wrong because the exponential decay $e^{-2t}$ 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.
The solution to the initial value problem $y' + 2y = f(t)$ with $y(0)=0$ is given by $y(t) = \int_0^t e^{-2(t-\tau)} f(\tau) d\tau$ . If the input is a delayed step function $f(t) = 3u_1(t)$ , what is the resulting output $y(t)$ ?
Answer: $y(t) = \frac{3}{2}(1 - e^{-2(t-1)})u_1(t)$ 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 $u_1(t)$ equals 0 for $t < 1$ and 1 for $t ≥ 1$ , meaning the input "turns on" at $t = 1$ . Since $f(τ) = 3u_1(τ)$ is zero when $τ < 1$ , the integral $y(t) = \int_0^t e^{-2(t-τ)} f(τ) dτ$ has no contribution until $τ ≥ 1$ . For $t ≥ 1$ , the effective lower limit becomes $τ = 1$ , giving us: $y(t) = \int_1^t 3e^{-2(t-τ)} dτ = 3e^{-2t} \int_1^t e^{2τ} dτ$ Evaluating this integral: $3e^{-2t} \cdot \frac{1}{2}[e^{2τ}]_1^t = \frac{3}{2}e^{-2t}(e^{2t} - e^2) = \frac{3}{2}(1 - e^{-2(t-1)})$ Since this solution is valid only for $t ≥ 1$ , we multiply by $u_1(t)$ , 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 $\frac{1}{2}$ 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.
The transfer function $H(s) = \frac{1}{s^2 + 2s + 2}$ represents a system subjected to the input $f(t) = u(t-1) - 2u(t-2) + u(t-3)$ . What is the steady-state behavior of the output as $t \to \infty$ ?
Answer: The output approaches zero asymptotically The poles of $H(s)$ are at $s = -1 \pm i$ , both having negative real parts, making the system stable. The input $f(t)$ has $\sum$ of step coefficients = $1 - 2 + 1 = 0$ , 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: $\lim_{t \to \infty} y(t) = \lim_{s \to 0} sY(s) = 0$ .
Consider the initial value problem $y' + y = f(t)$ , with $y(0)=1$ , where the forcing function $f(t)$ is a rectangular pulse of height 2 for $1 \le t < 2$ , and 0 otherwise. Which expression correctly describes the solution $y(t)$ ?
Answer: $y(t) = e^{-t} + 2(1 - e^{-(t-1)})u_1(t) - 2(1 - e^{-(t-2)})u_2(t)$ The forcing function can be written using Heaviside functions as $f(t) = 2(u_1(t) - u_2(t))$ . Taking the Laplace transform of the ODE: $sY(s) - y(0) + Y(s) = \mathcal{L}\{f(t)\}$ . With $y(0)=1$ , we get $(s+1)Y(s) - 1 = \frac{2e^{-s}}{s} - \frac{2e^{-2s}}{s}$ . Solving for $Y(s)$ : $Y(s) = \frac{1}{s+1} + e^{-s}\frac{2}{s(s+1)} - e^{-2s}\frac{2}{s(s+1)}$ . The term $\frac{1}{s+1}$ corresponds to $e^{-t}$ . For the other terms, let $H(s) = \frac{2}{s(s+1)} = \frac{2}{s} - \frac{2}{s+1}$ . Its inverse transform is $h(t) = 2 - 2e^{-t} = 2(1-e^{-t})$ . Applying the second shifting theorem, $\mathcal{L}^{-1}\{e^{-s}H(s)\} = h(t-1)u_1(t)$ and $\mathcal{L}^{-1}\{e^{-2s}H(s)\} = h(t-2)u_2(t)$ . Combining everything gives the solution $y(t) = e^{-t} + h(t-1)u_1(t) - h(t-2)u_2(t) = e^{-t} + 2(1 - e^{-(t-1)})u_1(t) - 2(1 - e^{-(t-2)})u_2(t)$ .
The impulse response of a linear time-invariant system is $h(t) = e^{-t}\sin(t)$ . This is the solution to $ay''+by'+cy=\delta(t)$ with zero initial conditions. What is the system's step response, i.e., the solution for an input of $u_0(t)$ ?
Answer: $y(t) = \frac{1}{2}(1 - e^{-t}(\cos(t) + \sin(t)))$ 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 $h(t) = e^{-t}\sin(t)$ , the step response is found by integrating: $y(t) = \int_0^t h(\tau) d\tau = \int_0^t e^{-\tau}\sin(\tau) d\tau$ To evaluate this integral, use integration by parts twice. Let $u = \sin(\tau)$ and $dv = e^{-\tau}d\tau$ . After applying integration by parts twice (since the sine function cycles through derivatives), you get: $\int e^{-\tau}\sin(\tau) d\tau = -\frac{1}{2}e^{-\tau}(\cos(\tau) + \sin(\tau)) + C$ Evaluating from 0 to $t$ : $y(t) = -\frac{1}{2}e^{-t}(\cos(t) + \sin(t)) + \frac{1}{2}(1) = \frac{1}{2}(1 - e^{-t}(\cos(t) + \sin(t)))$ 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 $e^{-t}\sin(t)$ . 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.
Find the solution to the initial value problem $y'' + 4y = \delta(t - \pi/2)$ , with initial conditions $y(0)=0$ and $y'(0)=0$ .
Answer: $y(t) = -\frac{1}{2}u(t-\pi/2)\sin(2t)$ 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 $y'' + 4y = 0$ has characteristic equation $r^2 + 4 = 0$ , giving $r = \pm 2i$ . This means the natural frequency is $\omega = 2$ , and the homogeneous solution involves $\sin(2t)$ and $\cos(2t)$ . For an impulse $\delta(t - \pi/2)$ applied to this second-order system, the response is the impulse response function multiplied by the unit step function $u(t - \pi/2)$ . The impulse response for $y'' + \omega^2 y = \delta(t - a)$ with zero initial conditions is $\frac{1}{\omega}\sin(\omega(t-a))u(t-a)$ . With $\omega = 2$ and $a = \pi/2$ , this gives $\frac{1}{2}\sin(2(t-\pi/2))u(t-\pi/2)$ . Using the identity $\sin(2(t-\pi/2)) = \sin(2t - \pi) = -\sin(2t)$ , the solution becomes $-\frac{1}{2}\sin(2t)u(t-\pi/2)$ . Choice A incorrectly uses coefficient 1 instead of $\frac{1}{2}$ 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.
What is the solution to the initial value problem $y'' + 3y' + 2y = \delta(t)$ with initial conditions $y(0)=1$ and $y'(0)=0$ ?
Answer: $y(t) = 3e^{-t} - 2e^{-2t}$ 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 $y'' + 3y' + 2y = 0$ . The characteristic equation is $r^2 + 3r + 2 = 0$ , which factors as $(r+1)(r+2) = 0$ , giving roots $r = -1$ and $r = -2$ . So the homogeneous solution is $y_h(t) = c_1e^{-t} + c_2e^{-2t}$ . For the delta function $\delta(t)$ , the solution has a jump discontinuity in the first derivative at $t = 0$ . The jump magnitude equals the coefficient of $\delta(t)$ (which is 1), so $y'(0^+) - y'(0^-) = 1$ . Since $y'(0^-) = 0$ , we have $y'(0^+) = 1$ . Using initial conditions: $y(0) = c_1 + c_2 = 1$ and $y'(0^+) = -c_1 - 2c_2 = 1$ . Solving this system: from the first equation, $c_1 = 1 - c_2$ . Substituting: $-(1-c_2) - 2c_2 = 1$ , so $-1 + c_2 - 2c_2 = 1$ , giving $c_2 = -2$ and $c_1 = 3$ . Therefore, $y(t) = 3e^{-t} - 2e^{-2t}$ , which is answer C. Answer A uses the wrong initial condition for $y'$ . 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.
Determine the solution to the initial value problem $y' + y = u(t-1) + \delta(t-2)$ , given $y(0)=0$ .
Answer: $y(t) = u(t-1)[1 - e^{-(t-1)}] + u(t-2)e^{-(t-2)}$ When you encounter a differential equation with unit step functions $u(t-a)$ and Dirac delta functions $\delta(t-a)$ , 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 $y' + y = u(t-1) + \delta(t-2)$ , the homogeneous solution has integrating factor $e^t$ . The unit step $u(t-1)$ acts like a constant forcing function starting at $t=1$ , producing a particular solution of the form $1 - e^{-(t-1)}$ that's "turned on" by multiplying by $u(t-1)$ . The delta function $\delta(t-2)$ creates an instantaneous impulse at $t=2$ , contributing a decaying exponential $e^{-(t-2)}$ that begins at that moment, activated by $u(t-2)$ . Choice B correctly captures both behaviors: $u(t-1)[1 - e^{-(t-1)}] + u(t-2)e^{-(t-2)}$ . Choice A incorrectly uses $e^{-t}$ 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 $e^{-(t-2)}$ , not $e^{-(t-1)}$ . Choice D omits the unit step functions entirely, meaning the solution would be active for all $t$ 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.
The Laplace transform of the solution to an initial value problem is given by $Y(s) = \frac{s e^{-2s}}{s^2-1}$ . What is the solution $y(t)$ ?
Answer: $y(t) = \cosh(t-2) u_2(t)$ When you encounter a Laplace transform with an exponential factor like $e^{-2s}$ , 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 $y(t)$ , you need to use the time-shifting property: if $\mathcal{L}^{-1}\{F(s)\} = f(t)$ , then $\mathcal{L}^{-1}\{e^{-as}F(s)\} = f(t-a)u_a(t)$ , where $u_a(t)$ is the unit step function. First, identify $F(s) = \frac{s}{s^2-1}$ . The inverse Laplace transform of this is $\cosh(t)$ , since $\mathcal{L}^{-1}\{\frac{s}{s^2-1}\} = \cosh(t)$ . With the factor $e^{-2s}$ , this becomes $\cosh(t-2)u_2(t)$ 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 $\cosh(t)u_2(t)$ , which ignores the time shift—this would be correct if there were no $e^{-2s}$ factor. Choice C gives $\sinh(t-2)u_2(t)$ , which would correspond to $\mathcal{L}^{-1}\{\frac{e^{-2s}}{s^2-1}\}$ , but our numerator has $s$ , not 1. Study tip: Always decompose problems with $e^{-as}$ factors into two steps: find the inverse transform of the rational part first, then apply the time-shifting property. Remember that $\mathcal{L}^{-1}\{\frac{s}{s^2-a^2}\} = \cosh(at)$ and $\mathcal{L}^{-1}\{\frac{1}{s^2-a^2}\} = \frac{\sinh(at)}{a}$ .
A system is modeled by the differential equation $y'' + 4y' + 5y = f(t)$ with $y(0)=y'(0)=0$ . If the input is an impulse at time $t=c$ , $f(t) = \delta(t-c)$ , what is the impulse response $y(t)$ ?
Answer: $y(t) = e^{-2(t-c)}\sin(t-c)u_c(t)$ 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 $t = c$ . To solve $y'' + 4y' + 5y = \delta(t-c)$ with zero initial conditions, start by finding the characteristic equation: $r^2 + 4r + 5 = 0$ . Using the quadratic formula: $r = \frac{-4 \pm \sqrt{16-20}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i$ . This gives us complex roots with real part $\alpha = -2$ and imaginary part $\beta = 1$ . For an impulse at $t = c$ , the response has the form of the homogeneous solution shifted by $c$ and multiplied by the unit step function $u_c(t)$ . With complex roots $-2 \pm i$ , the impulse response is: $y(t) = e^{-2(t-c)}\sin(t-c)u_c(t)$ . 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 $t = 0$ rather than $t = c$ , ignoring the time shift in the impulse. Choice D has the wrong exponential decay rate ( $-1$ instead of $-2$ ) and wrong frequency (argument $2(t-c)$ instead of $(t-c)$ ). Study tip: For impulse responses, remember that the time shift in the input ( $t-c$ ) must appear consistently throughout the solution, and complex roots $\alpha \pm \beta i$ give responses involving $e^{\alpha(t-c)}\sin(\beta(t-c))$ for zero initial conditions.
A system is modeled by the initial value problem $y'' + 2y' + y = 3\delta(t-1)$ , with initial conditions $y(0)=0$ and $y'(0)=0$ . What is the system's response $y(t)$ for $t>0$ ?
Answer: $y(t) = 3(t-1)e^{-(t-1)}u_1(t)$ Taking the Laplace transform of the equation gives $(s^2+2s+1)Y(s) = 3e^{-s}$ . This simplifies to $(s+1)^2 Y(s) = 3e^{-s}$ , so $Y(s) = \frac{3e^{-s}}{(s+1)^2}$ . Let $H(s) = \frac{3}{(s+1)^2}$ . The inverse Laplace transform of $H(s)$ is $h(t) = 3te^{-t}$ . According to the second shifting theorem, the inverse transform of $e^{-cs}H(s)$ is $h(t-c)u_c(t)$ . Here, $c=1$ , so $y(t) = h(t-1)u_1(t) = 3(t-1)e^{-(t-1)}u_1(t)$ .
The equation $y'' + y = f(t)$ where $f(t) = \sin(t)[u(t) - u(t-\pi)]$ represents a resonance condition with finite duration input. With initial conditions $y(0) = 0$ , $y'(0) = 1$ , what is the maximum value of $|y(t)|$ for $t > 2\pi$ ?
Answer: $\frac{\pi}{2}$ For $t \in [0,\pi]$ , we have resonance since the input frequency equals the natural frequency. The particular solution during this interval grows as $-\frac{t}{2}\cos(t)$ . The complete solution for $t \in [0,\pi]$ is $y(t) = \sin(t) - \frac{t}{2}\cos(t)$ . At $t = \pi$ : $y(\pi) = 0 - \frac{\pi}{2}(-1) = \frac{\pi}{2}$ and $y'(\pi) = 0$ . For $t > \pi$ , only the homogeneous solution exists: $y(t) = \frac{\pi}{2}\cos(t-\pi) = -\frac{\pi}{2}\cos(t)$ . The maximum value of $|y(t)|$ for $t > 2\pi$ is $\frac{\pi}{2}$ .
Solve the initial value problem $y' + y = u(t-1)t$ with the initial condition $y(0)=1$ .
Answer: $y(t) = e^{-t} + u(t-1)(t-1)$ When you encounter a differential equation with a unit step function like $u(t-1)$ , you need to solve it piecewise since the forcing function changes at $t = 1$ . First, solve for $t < 1$ where $u(t-1) = 0$ . The equation becomes $y' + y = 0$ , giving $y = Ce^{-t}$ . Using the initial condition $y(0) = 1$ , we get $C = 1$ , so $y(t) = e^{-t}$ for $t < 1$ . For $t \geq 1$ , where $u(t-1) = 1$ , the equation becomes $y' + y = t$ . This is a first-order linear equation. The homogeneous solution is $e^{-t}$ , and using the method of undetermined coefficients, the particular solution is $y_p = t - 1$ . So the general solution is $y = Ae^{-t} + t - 1$ . To find $A$ , use continuity at $t = 1$ . From the left: $y(1^-) = e^{-1}$ . From the right: $y(1^+) = Ae^{-1} + 1 - 1 = Ae^{-1}$ . Therefore $A = 1$ . This gives us $y(t) = e^{-t} + u(t-1)(t-1)$ , which is answer C. Answer A incorrectly uses $t$ instead of $(t-1)$ in the step function term. Answer B includes an unnecessary exponential term $e^{-(t-1)}$ that would come from solving the shifted equation incorrectly. Answer D represents $(1 - e^{-(t-1)})$ , which suggests confusion with the integrating factor method. Study tip: For piecewise differential equations, always check continuity at the switching point and remember that unit step functions "turn on" their argument at the specified time.
A system is described by $y'' + y = f(t)$ with $y(0)=0, y'(0)=0$ . The input is a square wave starting at $t=0$ , given by $f(t) = u_0(t) - 2u_{\pi}(t) + 2u_{2\pi}(t) - \dots$ . What is the solution $y(t)$ on the interval $\pi \le t < 2\pi$ ?
Answer: $y(t) = 1 - \cos(t) - 2(1 - \cos(t-\pi))$ 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 $\pi \le t < 2\pi$ , the forcing function $f(t) = u_0(t) - 2u_{\pi}(t) + 2u_{2\pi}(t) - \dots$ simplifies to $f(t) = 1 - 2 = -1$ since $u_0(t) = 1$ and $u_{\pi}(t) = 1$ while $u_{2\pi}(t) = 0$ in this interval. The solution consists of two parts: the response to $u_0(t)$ from $t = 0$ and the response to $-2u_{\pi}(t)$ starting at $t = \pi$ . For $y'' + y = 1$ with zero initial conditions, the solution is $1 - \cos(t)$ . When the forcing changes to $-2$ at $t = \pi$ , this adds a response $-2(1 - \cos(t-\pi))$ with the time shift reflecting when this forcing began. Therefore, $y(t) = 1 - \cos(t) - 2(1 - \cos(t-\pi))$ for $\pi \le t < 2\pi$ . Choice A omits the initial response from $t = 0$ . Choice C ignores the effect of the step change at $t = \pi$ . Choice D incorrectly adds rather than subtracts the second term, missing that $-2u_{\pi}(t)$ 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.
An RLC circuit with inductance $L=1$ H, resistance $R=2$ $\Omega$ , and capacitance $C=1$ F is initially inert (zero charge and current). At time $t=\pi$ seconds, a constant voltage of 1 V is applied. What is the charge $q(t)$ on the capacitor for $t > \pi$ ?
Answer: $q(t) = 1 - (1+t-\pi)e^{-(t-\pi)}$ 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 $L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = V(t)$ . With the given values ( $L=1$ , $R=2$ , $C=1$ ) and voltage applied at $t=\pi$ , this becomes $\frac{d^2q}{dt^2} + 2\frac{dq}{dt} + q = 1$ for $t > \pi$ . The solution has the form $q(t) = q_h(t) + q_p(t)$ . The characteristic equation $r^2 + 2r + 1 = 0$ gives $(r+1)^2 = 0$ , so $r = -1$ (repeated root). This yields $q_h(t) = (A + Bt)e^{-t}$ . The particular solution for the constant forcing function is $q_p = 1$ . Since the voltage starts at $t = \pi$ with zero initial charge and current, we use shifted conditions: $q(\pi) = 0$ and $q'(\pi) = 0$ . Substituting $u = t - \pi$ transforms this to a problem starting at $u = 0$ , giving us $q(u) = 1 - (1 + u)e^{-u}$ . Converting back: $q(t) = 1 - (1 + t - \pi)e^{-(t-\pi)}$ . 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 $t = 0$ . 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 $t = 0$ .
Find the solution to the initial value problem $y'' + 9y = t u_2(t)$ , with initial conditions $y(0)=0$ and $y'(0)=0$ .
Answer: $y(t) = \left( \frac{2}{9} + \frac{1}{9}(t-2) - \frac{2}{9}\cos(3(t-2)) - \frac{1}{27}\sin(3(t-2)) \right) u_2(t)$ To apply the second shifting theorem, we must write the forcing function $g(t) = t u_2(t)$ in the form $f(t-2)u_2(t)$ . Let $t-2 = v$ , so $t = v+2$ . Thus, $t u_2(t) = ((t-2)+2)u_2(t) = (t-2)u_2(t) + 2u_2(t)$ . The Laplace transform is $\mathcal{L}\{t u_2(t)\} = e^{-2s}\mathcal{L}\{t+2\} = e^{-2s}(\frac{1}{s^2} + \frac{2}{s}) = e^{-2s}\frac{1+2s}{s^2}$ . The transformed ODE is $(s^2+9)Y(s) = e^{-2s}\frac{2s+1}{s^2}$ . So, $Y(s) = e^{-2s} H(s)$ where $H(s) = \frac{2s+1}{s^2(s^2+9)}$ . Partial fraction decomposition gives $H(s) = \frac{2/9}{s} + \frac{1/9}{s^2} - \frac{(2/9)s + 1/9}{s^2+9}$ . The inverse transform is $h(t) = \frac{2}{9} + \frac{1}{9}t - \frac{2}{9}\cos(3t) - \frac{1}{27}\sin(3t)$ . The final solution is $y(t) = h(t-2)u_2(t)$ .

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