Differential Equations Quiz: Heaviside Step Functions

What this quiz covers

This quiz focuses on Heaviside Step Functions, 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. $\frac{s}{s^2+1} - e^{-\pi s} \left( \frac{s}{s^2+1} + \frac{1}{s} \right)$
2. $\frac{s}{s^2+1} + e^{-\pi s} \left( \frac{s}{s^2+1} - \frac{1}{s} \right)$ (correct answer)
3. $\frac{s}{s^2+1} - e^{-\pi s} \left( \frac{s}{s^2+1} - \frac{1}{s} \right)$
4. $\frac{s}{s^2+1} + e^{-\pi s} \left( \frac{s}{s^2+1} + \frac{1}{s} \right)$

Explanation: First, represent $f(t)$ using the Heaviside step function $u_\pi(t)$. $f(t) = \cos(t)(1 - u_\pi(t)) - 1 \cdot u_\pi(t) = \cos(t) - (\cos(t) + 1)u_\pi(t)$. To find the Laplace transform of the second term using the second shifting theorem, $\mathcal{L}\{g(t-c)u_c(t)\} = e^{-cs}G(s)$, we must express $\cos(t)+1$ as a function of $(t-\pi)$. Let $g(t) = \cos(t)+1$. We need to find $\mathcal{L}\{g(t)u_\pi(t)\} = e^{-\pi s}\mathcal{L}\{g(t+\pi)\}$. $g(t+\pi) = \cos(t+\pi)+1 = -\cos(t)+1$. So, $\mathcal{L}\{g(t+\pi)\} = \mathcal{L}\{-\cos(t)+1\} = -\frac{s}{s^2+1} + \frac{1}{s}$. Therefore, $\mathcal{L}\{(\cos(t)+1)u_\pi(t)\} = e^{-\pi s} \left(\frac{1}{s} - \frac{s}{s^2+1}\right)$. The complete transform is: $F(s) = \mathcal{L}\{\cos(t)\} - \mathcal{L}\{(\cos(t)+1)u_\pi(t)\} = \frac{s}{s^2+1} - e^{-\pi s} \left(\frac{1}{s} - \frac{s}{s^2+1}\right) = \frac{s}{s^2+1} + e^{-\pi s} \left(\frac{s}{s^2+1} - \frac{1}{s}\right)$.

Question 2

1. $t^2 + (\cos(3t) - \frac{1}{3}\sin(3t))u_4(t)$
2. $t^2 + (\cos(3(t-4)) - \frac{1}{3}\sin(3(t-4)))u_4(t)$ (correct answer)
3. $t^2 + (\cos(3(t-4)) - \sin(3(t-4)))u_4(t)$
4. $\frac{t^2}{2} + (\cos(3(t-4)) - \frac{1}{3}\sin(3(t-4)))u_4(t)$

Explanation: We find the inverse transform of each term separately. For the first term: $\mathcal{L}^{-1}\left\{\frac{2}{s^3}\right\} = \mathcal{L}^{-1}\left\{\frac{2!}{s^{2+1}}\right\} = t^2$. For the second term, we use the second shifting theorem, $\mathcal{L}^{-1}\{e^{-cs}F(s)\} = f(t-c)u_c(t)$. Here, $c=4$ and $F(s) = \frac{s-1}{s^2+9}$. We find $f(t) = \mathcal{L}^{-1}\{F(s)\}$ first. $F(s) = \frac{s}{s^2+9} - \frac{1}{s^2+9} = \frac{s}{s^2+3^2} - \frac{1}{3}\frac{3}{s^2+3^2}$. $f(t) = \cos(3t) - \frac{1}{3}\sin(3t)$. Now, we apply the shift: $f(t-4)u_4(t) = \left(\cos(3(t-4)) - \frac{1}{3}\sin(3(t-4))\right)u_4(t)$. Combining the two parts, the complete inverse transform is $y(t) = t^2 + \left(\cos(3(t-4)) - \frac{1}{3}\sin(3(t-4))\right)u_4(t)$.

Question 3

1. $y(t) = 3e^{-2t} + \frac{5}{2}(1 - e^{-2t})u_1(t)$
2. $y(t) = \frac{5}{2}(1 - e^{-2(t-1)})u_1(t)$
3. $y(t) = 3e^{-2t} - \frac{5}{2}(1 - e^{-2(t-1)})u_1(t)$
4. $y(t) = 3e^{-2t} + \frac{5}{2}(1 - e^{-2(t-1)})u_1(t)$ (correct answer)

Explanation: First, write the forcing function as $f(t) = 5u_1(t)$. Take the Laplace transform of the entire equation: $\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = \mathcal{L}\{5u_1(t)\}$ $[sY(s) - y(0)] + 2Y(s) = \frac{5e^{-s}}{s}$ Substitute $y(0)=3$: $(s+2)Y(s) - 3 = \frac{5e^{-s}}{s}$ $Y(s) = \frac{3}{s+2} + \frac{5e^{-s}}{s(s+2)}$. To find the inverse transform of the second term, use partial fraction decomposition for $\frac{5}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2}$. This gives $A=5/2$ and $B=-5/2$. So, $\frac{5}{s(s+2)} = \frac{5}{2}\left(\frac{1}{s} - \frac{1}{s+2}\right)$. Let $G(s) = \frac{5}{2}\left(\frac{1}{s} - \frac{1}{s+2}\right)$, then $g(t) = \mathcal{L}^{-1}\{G(s)\} = \frac{5}{2}(1 - e^{-2t})$. The second term in $Y(s)$ is $e^{-s}G(s)$, so its inverse transform is $g(t-1)u_1(t) = \frac{5}{2}(1 - e^{-2(t-1)})u_1(t)$. The inverse transform of the first term is $\mathcal{L}^{-1}\left\{\frac{3}{s+2}\right\} = 3e^{-2t}$. Combining the parts, the solution is $y(t) = 3e^{-2t} + \frac{5}{2}(1 - e^{-2(t-1)})u_1(t)$.

Question 4

1. $3e^{-4}$
2. $\frac{5}{2} + \frac{1}{2}e^{-4}$
3. $3e^{-4} + \frac{5}{2} - \frac{5}{2}e^{-2}$ (correct answer)
4. $3e^{-4} + \frac{5}{2} - \frac{5}{2}e^{-4}$

Explanation: The solution is given by $y(t) = 3e^{-2t} + \frac{5}{2}(1 - e^{-2(t-1)})u_1(t)$. To find $y(2)$, we substitute $t=2$ into the expression. Since $2 > 1$, the Heaviside function $u_1(2)$ is equal to 1. So, we evaluate the entire expression at $t=2$: $y(2) = 3e^{-2(2)} + \frac{5}{2}(1 - e^{-2(2-1)})u_1(2)$ $y(2) = 3e^{-4} + \frac{5}{2}(1 - e^{-2(1)})(1)$ $y(2) = 3e^{-4} + \frac{5}{2}(1 - e^{-2})$ $y(2) = 3e^{-4} + \frac{5}{2} - \frac{5}{2}e^{-2}$. Distractor A results from ignoring the forcing term (evaluating only the homogeneous part). Distractor B results from incorrectly using the unshifted function $g(t)$ instead of $g(t-1)$. Distractor D results from an algebraic error in the exponent.

Question 5

1. $\frac{1}{s(e^s-1)}$ (correct answer)
2. $\frac{1}{s(1-e^{-s})}$
3. $\frac{1}{e^s-1}$
4. $\frac{e^{-s}}{1-e^{-s}}$

Explanation: The Laplace transform of the infinite sum is the sum of the Laplace transforms: $F(s) = \mathcal{L}\left\{\sum_{n=1}^{\infty} u_n(t)\right\} = \sum_{n=1}^{\infty} \mathcal{L}\{u_n(t)\}$. The Laplace transform of a single Heaviside function is $\mathcal{L}\{u_n(t)\} = \frac{e^{-ns}}{s}$. So, $F(s) = \sum_{n=1}^{\infty} \frac{e^{-ns}}{s} = \frac{1}{s} \sum_{n=1}^{\infty} e^{-ns}$. The sum is a geometric series with first term $a = e^{-s}$ and common ratio $r = e^{-s}$. Since $|r| = e^{-s} < 1$ for $s>0$, the sum converges to $\frac{a}{1-r} = \frac{e^{-s}}{1-e^{-s}}$. Therefore, $F(s) = \frac{1}{s} \left( \frac{e^{-s}}{1-e^{-s}} \right)$. To match the answer choices, we can multiply the numerator and denominator by $e^s$: $F(s) = \frac{1}{s} \left( \frac{e^{-s}e^s}{(1-e^{-s})e^s} \right) = \frac{1}{s} \left( \frac{1}{e^s-1} \right) = \frac{1}{s(e^s-1)}$.

Question 6

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

Explanation: This problem requires using the second shifting theorem, $\mathcal{L}^{-1}\{e^{-cs}G(s)\} = g(t-c)u_c(t)$. Here, $c=\pi$ and $G(s) = \frac{s+3}{s(s^2+1)}$. First, we find $g(t) = \mathcal{L}^{-1}\{G(s)\}$ using partial fraction decomposition. $\frac{s+3}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1}$. Solving for the coefficients gives $A(s^2+1) + (Bs+C)s = s+3$, which leads to $A=3, C=1, B=-3$. So, $G(s) = \frac{3}{s} - \frac{3s}{s^2+1} + \frac{1}{s^2+1}$. The inverse transform is $g(t) = 3 - 3\cos(t) + \sin(t)$. Now, we must find $g(t-\pi)$. $g(t-\pi) = 3 - 3\cos(t-\pi) + \sin(t-\pi)$. Using the identities $\cos(t-\pi) = -\cos(t)$ and $\sin(t-\pi) = -\sin(t)$, we get: $g(t-\pi) = 3 - 3(-\cos(t)) + (-\sin(t)) = 3 + 3\cos(t) - \sin(t)$. Finally, the solution is $y(t) = g(t-\pi)u_\pi(t) = (3 + 3\cos(t) - \sin(t))u_\pi(t)$.

Question 7

1. $\frac{s}{s^2+4}(e^{-\pi s/2} - e^{-3\pi s/2})$
2. $\frac{s}{s^2+4}(e^{-3\pi s/2} - e^{-\pi s/2})$ (correct answer)
3. $\frac{s}{s^2+4}(e^{-\pi s/2} + e^{-3\pi s/2})$
4. $-\frac{s}{s^2+4}(e^{-\pi s/2} + e^{-3\pi s/2})$

Explanation: We need to compute $\mathcal{L}\{\cos(2t)u_{\pi/2}(t)\} - \mathcal{L}\{\cos(2t)u_{3\pi/2}(t)\}$. For the first term, with $c=\pi/2$, we must write $\cos(2t)$ as a function of $(t-\pi/2)$. $2t = 2(t-\pi/2) + \pi$. Using the identity $\cos(x+\pi) = -\cos(x)$, we have $\cos(2t) = \cos(2(t-\pi/2)+\pi) = -\cos(2(t-\pi/2))$. So, $\mathcal{L}\{\cos(2t)u_{\pi/2}(t)\} = \mathcal{L}\{-\cos(2(t-\pi/2))u_{\pi/2}(t)\} = -e^{-\pi s/2}\mathcal{L}\{\cos(2t)\} = -e^{-\pi s/2}\frac{s}{s^2+4}$. For the second term, with $c=3\pi/2$, we write $\cos(2t)$ as a function of $(t-3\pi/2)$. $2t = 2(t-3\pi/2) + 3\pi$. Using the identity $\cos(x+3\pi) = -\cos(x)$, we have $\cos(2t) = \cos(2(t-3\pi/2)+3\pi) = -\cos(2(t-3\pi/2))$. So, $\mathcal{L}\{\cos(2t)u_{3\pi/2}(t)\} = -e^{-3\pi s/2}\mathcal{L}\{\cos(2t)\} = -e^{-3\pi s/2}\frac{s}{s^2+4}$. Combining the results: $F(s) = -e^{-\pi s/2}\frac{s}{s^2+4} - \left(-e^{-3\pi s/2}\frac{s}{s^2+4}\right) = \frac{s}{s^2+4}(e^{-3\pi s/2} - e^{-\pi s/2})$.

Question 8

1. $-\frac{8}{3}\sin(2t)$ (correct answer)
2. $\frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t)$
3. $0$
4. $-\frac{8}{3}\cos(t) - \frac{4}{3}\cos(2t)$

Explanation: First, take the Laplace transform of the IVP. The forcing function is $f(t) = 8\sin(t) - 8\sin(t)u_\pi(t)$. We rewrite the second term: $\sin(t) = \sin((t-\pi)+\pi) = -\sin(t-\pi)$. So $f(t) = 8\sin(t) + 8\sin(t-\pi)u_\pi(t)$. \{L}\{f(t)\} = \frac{8}{s^2+1} + \frac{8e^{-\pi s}}{s^2+1}. The transformed DE is $(s^2+4)Y(s) = \frac{8}{s^2+1}(1+e^{-\pi s})$. $Y(s) = \frac{8}{(s^2+1)(s^2+4)}(1+e^{-\pi s})$. Using partial fractions, $\frac{8}{(s^2+1)(s^2+4)} = \frac{8/3}{s^2+1} - \frac{8/3}{s^2+4}$. Let $G(s) = \frac{8/3}{s^2+1} - \frac{8/3}{s^2+4}$. Its inverse transform is $g(t) = \frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t)$. Then $y(t) = g(t) + g(t-\pi)u_\pi(t)$. $g(t-\pi) = \frac{8}{3}\sin(t-\pi) - \frac{4}{3}\sin(2(t-\pi)) = -\frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t)$. For $t \ge \pi$, $u_\pi(t)=1$, so the solution is $y(t) = g(t) + g(t-\pi)$. $y(t) = \left(\frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t)\right) + \left(-\frac{8}{3}\sin(t) - \frac{4}{3}\sin(2t)\right) = -\frac{8}{3}\sin(2t)$.

Question 9

1. $e^{-2s}\left( \frac{2}{s^3} + \frac{1}{s} \right)$
2. $e^{-2s}\left( \frac{2}{s^3} + \frac{5}{s} \right)$
3. $e^{-2s}\left( \frac{2}{s^3} + \frac{4}{s^2} + \frac{5}{s} \right)$ (correct answer)
4. $e^{-2s}\left( \frac{2}{s^3} + \frac{4}{s^2} + \frac{4}{s} \right)$

Explanation: To find the Laplace transform of $f(t)=g(t)u_c(t)$, we must use the second shifting theorem, $\mathcal{L}\{g(t)u_c(t)\} = e^{-cs}\mathcal{L}\{g(t+c)\}$. In this case, $g(t)=t^2+1$ and $c=2$. We need to find $\mathcal{L}\{g(t+2)\}$. $g(t+2) = (t+2)^2 + 1 = (t^2 + 4t + 4) + 1 = t^2 + 4t + 5$. Now, we take the Laplace transform of this new polynomial: $\mathcal{L}\{t^2 + 4t + 5\} = \mathcal{L}\{t^2\} + 4\mathcal{L}\{t\} + \mathcal{L}\{5\} = \frac{2}{s^3} + \frac{4}{s^2} + \frac{5}{s}$. Finally, we multiply by $e^{-cs} = e^{-2s}$ to get the final answer: $F(s) = e^{-2s}\left( \frac{2}{s^3} + \frac{4}{s^2} + \frac{5}{s} \right)$. Distractor A represents the common mistake of applying $e^{-cs}\mathcal{L}\{g(t)\}$ instead of $e^{-cs}\mathcal{L}\{g(t+c)\}$.

Question 10

1. $t^2 + (1-t^2)u_1(t) + (3-t)u_3(t)$ (correct answer)
2. $t^2 u_0(t) + u_1(t) + (4-t)u_3(t)$
3. $t^2(u_0(t) - u_1(t)) + u_1(t) + (4-t)u_3(t)$
4. $t^2 + (1-t^2)u_1(t) + (5-t)u_3(t)$

Explanation: A standard way to write a piecewise function $f(t) = \begin{cases} g_0(t) & 0 \le t < c_1 \\ g_1(t) & c_1 \le t < c_2 \\ g_2(t) & t \ge c_2 \end{cases}$ is $f(t) = g_0(t) + [g_1(t) - g_0(t)]u_{c_1}(t) + [g_2(t) - g_1(t)]u_{c_2}(t)$. In this case, $g_0(t) = t^2$, $g_1(t) = 1$, and $g_2(t) = 4-t$. The points of discontinuity are $c_1=1$ and $c_2=3$. Applying the formula: $f(t) = t^2 + (1-t^2)u_1(t) + ((4-t) - 1)u_3(t)$ $f(t) = t^2 + (1-t^2)u_1(t) + (3-t)u_3(t)$. Let's verify: For $0 \le t < 1$, $u_1=0, u_3=0$, so $f(t)=t^2$. Correct. For $1 \le t < 3$, $u_1=1, u_3=0$, so $f(t)=t^2+(1-t^2)=1$. Correct. For $t \ge 3$, $u_1=1, u_3=1$, so $f(t)=t^2+(1-t^2)+(3-t)=1+3-t=4-t$. Correct.

Question 11

1. $e^{-5} + e^{-3} - 1$
2. $1 - e^{-3}$ (correct answer)
3. $e^{-2} - e^{-5}$
4. $1 - e^{-3} + e^{-5}$

Explanation: For $t = 5$, we evaluate $\int_0^5 e^{-(5-\tau)} u_2(\tau) d\tau$. Since $u_2(\tau) = 0$ for $\tau < 2$ and $u_2(\tau) = 1$ for $\tau \geq 2$, the integral becomes $\int_2^5 e^{-(5-\tau)} d\tau = \int_2^5 e^{\tau-5} d\tau = e^{-5}\int_2^5 e^{\tau} d\tau = e^{-5}[e^{\tau}]_2^5 = e^{-5}(e^5 - e^2) = 1 - e^{-3}$. Choice A incorrectly includes extra exponential terms that don't arise from the integration. Choice C gives the wrong limits and signs. Choice D adds an incorrect $e^{-5}$ term that would come from mishandling the exponential factor.

Question 12

1. $y(t)$ oscillates between $\frac{1}{2}$ and $-\frac{1}{2}$ with decreasing amplitude
2. $y(t)$ approaches $\frac{1}{2}$ asymptotically with exponential decay of oscillations
3. $y(t)$ oscillates between $\frac{1}{2}$ and $0$ with period approaching 1 (correct answer)
4. $y(t)$ grows without bound due to the infinite sum of step functions

Explanation: The forcing function $g(t) = u_1(t) - u_2(t) + u_3(t) - u_4(t) + \ldots$ creates a square wave that alternates between 1 and 0. For $t \in [n, n+1)$, we have $g(t) = (-1)^{n+1}$. The solution to $y' + 2y = g(t)$ with $y(0) = 0$ is found using integrating factors. For each interval $[n, n+1)$, the solution is $y(t) = \frac{(-1)^{n+1}}{2}(1 - e^{-2(t-n)}) + y(n)e^{-2(t-n)}$. As $t$ increases through each interval, $y(t)$ exponentially approaches $\frac{(-1)^{n+1}}{2}$ from its value at the start of the interval. Since the forcing alternates sign every unit of time, and the exponential decay is rapid (time constant $\frac{1}{2}$), the solution oscillates between approximately $\frac{1}{2}$ and $0$, spending most of each interval close to the steady-state value $\frac{1}{2}$ or $-\frac{1}{2}$ corresponding to the current forcing level. However, since the forcing switches between 1 and 0 (not 1 and -1), the oscillation is between $\frac{1}{2}$ and $0$. Choice A incorrectly assumes symmetric oscillation about zero. Choice B misses the periodic switching. Choice D ignores the stabilizing effect of the decay term.