Differential Equations Quiz: Odes With Discontinuous Inputs

What this quiz covers

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.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $y(t) = 3(t-1)e^{-(t-1)}u_1(t)$ (correct answer)
2. $y(t) = 3te^{-t}u_1(t)$
3. $y(t) = 3(t-1)e^{-t}u_1(t)$
4. $y(t) = 3(t-1)e^{-(t-1)}$

Explanation: 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 $y(t) = h(t-1)u_1(t) = 3(t-1)e^{-(t-1)}u_1(t)$.

Question 2

1. 1 (correct answer)
2. 0
3. $e^{-2}(\cos(2) - \sin(2))$
4. -1

Explanation: 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.

Question 3

1. $q(t) = 1 + (1+t-\pi)e^{-(t-\pi)}$
2. $q(t) = 1 - (1+t)e^{-t}$
3. $q(t) = (1 - e^{-(t-\pi)} - (t-\pi)e^{-(t-\pi)})u_0(t)$
4. $q(t) = 1 - (1+t-\pi)e^{-(t-\pi)}$ (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 $L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = V(t)$. 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 $t = 0$.

Question 4

1. $y(t) = \frac{1}{2}(e^{t-2} - e^{-(t+2)}) u_2(t)$
2. $y(t) = \cosh(t) u_2(t)$
3. $y(t) = \sinh(t-2) u_2(t)$
4. $y(t) = \cosh(t-2) u_2(t)$ (correct answer)

Explanation: 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 $\mathcal{L}^{-1}\{\frac{1}{s^2-a^2}\} = \frac{\sinh(at)}{a}$.

Question 5

1. $y(t) = e^{-2(t-c)}\cos(t-c)u_c(t)$
2. $y(t) = e^{-2t}\sin(t)u_c(t)$
3. $y(t) = e^{-2(t-c)}\sin(t-c)u_c(t)$ (correct answer)
4. $y(t) = (e^{-(t-c)}\sin(2(t-c)))u_c(t)$

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 $t = c$. 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.

Question 6

1. $y(t) = e^{-t}(\cos(t) - \sin(t))$
2. $y(t) = \frac{1}{2}(1 - e^{-t}(\cos(t) + \sin(t)))$ (correct answer)
3. $y(t) = 1 - e^{-t}\cos(t)$
4. $y(t) = e^{-t}\sin(t)u_0(t)$

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 $h(t) = e^{-t}\sin(t)$, 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.

Question 7

1. $y(t) = \frac{3}{2}(1 - e^{-2t})u_1(t)$
2. $y(t) = 3(1 - e^{-2(t-1)})u_1(t)$
3. $y(t) = \frac{3}{2}(1 - e^{-2(t-1)})u_1(t)$ (correct answer)
4. $y(t) = 3e^{-2(t-1)}u_1(t)$

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 $u_1(t)$ 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.

Question 8

1. $y(t) = \left( \frac{2}{9} + \frac{1}{9}(t-2) - \frac{2}{9}\cos(3(t-2)) - \frac{1}{27}\sin(3(t-2)) \right) u_2(t)$ (correct answer)
2. $y(t) = \left( \frac{1}{9}(t-2) - \frac{1}{27}\sin(3(t-2)) \right) u_2(t)$
3. $y(t) = \left( \frac{1}{9}t - \frac{1}{27}\sin(3t) \right) u_2(t)$
4. $y(t) = \left( \frac{2}{9} + \frac{1}{9}t - \frac{2}{9}\cos(3t) - \frac{1}{27}\sin(3t) \right) u_2(t)$

Explanation: To apply the second shifting theorem, we must write the forcing function $g(t) = t u_2(t)$ 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 $y(t) = h(t-2)u_2(t)$.

Question 9

1. $y(t) = -2(1 - \cos(t-\pi))$
2. $y(t) = 1 - \cos(t) - 2(1 - \cos(t-\pi))$ (correct answer)
3. $y(t) = 1 - \cos(t)$
4. $y(t) = 1 - \cos(t) + 2(1 - \cos(t-\pi))$

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 $\pi \le t < 2\pi$, 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.

Question 10

1. $y(t) = e^{-t} + 2(1 - e^{-(t-1)})u_1(t) - 2(1 - e^{-(t-2)})u_2(t)$ (correct answer)
2. $y(t) = 2(1 - e^{-(t-1)})u_1(t) - 2(1 - e^{-(t-2)})u_2(t)$
3. $y(t) = e^{-t} + 2(1 - e^{-t})u_1(t) - 2(1 - e^{-t})u_2(t)$
4. $y(t) = e^{-t} + 2(1 - e^{-(t-1)})u_1(t) + 2(1 - e^{-(t-2)})u_2(t)$

Explanation: The forcing function can be written using Heaviside functions as $f(t) = 2(u_1(t) - u_2(t))$. 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 $y(t) = e^{-t} + h(t-1)u_1(t) - h(t-2)u_2(t) = e^{-t} + 2(1 - e^{-(t-1)})u_1(t) - 2(1 - e^{-(t-2)})u_2(t)$.

Question 11

1. $y(t) = -e^{-t} + 2e^{-2t}$
2. $y(t) = e^{-t} - e^{-2t}$
3. $y(t) = 3e^{-t} - 2e^{-2t}$ (correct answer)
4. $y(t) = 2e^{-t} - e^{-2t}$

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 $y'' + 3y' + 2y = 0$. 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.

Question 12

1. $q(t) = u(t-2) [5 - 10e^{-t} + 5e^{-2t}]$
2. $q(t) = u(t-2) [5 - 10e^{-(t-2)} + 5e^{-2(t-2)}]$ (correct answer)
3. $q(t) = u(t-2) [10 - 10e^{-(t-2)} + 5e^{-2(t-2)}]$
4. $q(t) = u(t-2) [5 + 5e^{-(t-2)} - 10e^{-2(t-2)}]$

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: $1 \cdot q'' + 3q' + \frac{1}{0.5}q = E(t)$, 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.

Question 13

1. $y(t) = u(t-1)[1 - e^{-t}] + u(t-2)e^{-t}$
2. $y(t) = u(t-1)[1 - e^{-(t-1)}] + u(t-2)e^{-(t-2)}$ (correct answer)
3. $y(t) = u(t-1)[1 - e^{-(t-1)}] + u(t-2)e^{-(t-1)}$
4. $y(t) = [1 - e^{-(t-1)}] + e^{-(t-2)}$

Explanation: 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.

Question 14

1. $y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u(t-1)(1 - e^{-2(t-1)})$ (correct answer)
2. $y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u(t-1)(1 - e^{-2t})$
3. $y(t) = 1 - e^{-2t} - u(t-1)(1 - e^{-2(t-1)})$
4. $y(t) = \frac{1}{2}(1 - e^{-2t}) + \frac{1}{2}u(t-1)(1 - e^{-2(t-1)})$

Explanation: The forcing function can be written using Heaviside step functions as $f(t) = u(t) - u(t-1)$, 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 $y(t) = \frac{1}{2}(1 - e^{-2t}) - \frac{1}{2}u(t-1)(1 - e^{-2(t-1)})$.

Question 15

1. $y(t) = e^{-t} + u(t-1)t$
2. $y(t) = e^{-t} + u(t-1)(t-2+e^{-(t-1)})$
3. $y(t) = e^{-t} + u(t-1)(t-1)$ (correct answer)
4. $y(t) = e^{-t} + u(t-1)(1 - e^{-(t-1)})$

Explanation: 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.