Differential Equations Quiz: Linearity And Shifting Theorems

What this quiz covers

This quiz focuses on Linearity And Shifting Theorems, 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. $\left(\frac{1}{2} - \frac{1}{2}e^{-2t}\right)u(t-3)$
2. $\left(\frac{1}{2} - \frac{1}{2}e^{-2(t-3)}\right)u(t-3)$ (correct answer)
3. $e^{-3t}\left(\frac{1}{2} - \frac{1}{2}e^{-2t}\right)$
4. $\left(1 - e^{-2(t-3)}\right)u(t-3)$

Explanation: This problem uses the second shifting theorem for inverse transforms: $\mathcal{L}^{-1}\{e^{-as}F(s)\} = f(t-a)u(t-a)$. First, identify $F(s) = \frac{1}{s(s+2)}$ and $a=3$. We need to find $f(t) = \mathcal{L}^{-1}\{F(s)\}$. Using partial fraction decomposition on $F(s)$: $\frac{1}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2} \implies 1 = A(s+2) + Bs$ Setting $s=0$ gives $A=1/2$. Setting $s=-2$ gives $B=-1/2$. So, $F(s) = \frac{1/2}{s} - \frac{1/2}{s+2}$. The inverse transform is $f(t) = \frac{1}{2} - \frac{1}{2}e^{-2t}$. Now, apply the shifting theorem: $\mathcal{L}^{-1}\{e^{-3s}F(s)\} = f(t-3)u(t-3)$. Replace every $t$ in $f(t)$ with $(t-3)$: $g(t) = \left(\frac{1}{2} - \frac{1}{2}e^{-2(t-3)}\right)u(t-3)$

Question 2

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

Explanation: When you encounter an inverse Laplace transform with an exponential factor like $e^{-2s}$, this signals a time-shifting problem. The exponential $e^{-as}$ in the Laplace domain corresponds to shifting the function by $a$ units in the time domain, creating a unit step function $u(t-a)$. To solve this, first work with $\frac{s}{s^2+4s+13}$ by completing the square in the denominator: $s^2+4s+13 = (s+2)^2+9$. Rewrite the numerator to match this form: $s = (s+2)-2$. This gives us $\frac{s}{s^2+4s+13} = \frac{(s+2)-2}{(s+2)^2+9} = \frac{s+2}{(s+2)^2+9} - \frac{2}{(s+2)^2+9}$. These correspond to standard forms: $\frac{s+2}{(s+2)^2+9} \rightarrow e^{-2t}\cos(3t)$ and $\frac{2}{(s+2)^2+9} \rightarrow \frac{2}{3}e^{-2t}\sin(3t)$. So the inverse of $\frac{s}{s^2+4s+13}$ is $e^{-2t}(\cos(3t) - \frac{2}{3}\sin(3t))$. The $e^{-2s}$ factor shifts this by 2 units: replace $t$ with $(t-2)$ and multiply by $u(t-2)$, giving answer A. Answer B incorrectly keeps the original time variable $t$ in the exponential and trigonometric functions instead of shifting to $(t-2)$. Answer C has the wrong coefficient $-2$ instead of $-\frac{2}{3}$ for the sine term. Answer D incorrectly uses $e^{2(t-2)}$ instead of $e^{-2(t-2)}$, missing the negative sign. Remember: $e^{-as}$ in the Laplace domain means shift everything by $a$ units in time and add $u(t-a)$.

Question 3

1. $\frac{1}{((s-2)^2+1)^2}$
2. $\frac{s^2-4s+6}{(s-2)((s-2)^2+4)}$
3. $e^{2s} \frac{2}{s(s^2+4)}$
4. $\frac{2}{(s-2)((s-2)^2+4)}$ (correct answer)

Explanation: This problem requires a trigonometric identity followed by the first shifting theorem. First, find the Laplace transform of $g(t) = \sin^2(t)$. Use the identity $\sin^2(t) = \frac{1}{2}(1 - \cos(2t))$. Using linearity: $G(s) = \mathcal{L}\{\sin^2(t)\} = \mathcal{L}\{\frac{1}{2} - \frac{1}{2}\cos(2t)\} = \frac{1}{2s} - \frac{1}{2}\frac{s}{s^2+4}$ Combining the terms gives: $G(s) = \frac{s^2+4-s^2}{2s(s^2+4)} = \frac{4}{2s(s^2+4)} = \frac{2}{s(s^2+4)}$ Now, apply the first shifting theorem, $\mathcal{L}\{e^{at}g(t)\} = G(s-a)$, with $a=2$. We replace every $s$ in $G(s)$ with $(s-2)$: $\mathcal{L}\{e^{2t}\sin^2(t)\} = G(s-2) = \frac{2}{(s-2)((s-2)^2+4)}$

Question 4

1. $\frac{1}{2}t^2e^{-3t}$ (correct answer)
2. $t^2e^{-3t}$
3. $\frac{1}{6}t^3e^{-3t}$
4. $\frac{1}{2}t^2e^{3t}$

Explanation: When you encounter inverse Laplace transforms with powers in the denominator like $(s+a)^n$, you're dealing with a combination of shifting and polynomial multiplication patterns. The key is recognizing that this involves both the shifting property and the transform of polynomial functions. To find $\mathcal{L}^{-1}\left\{\frac{1}{(s+3)^3}\right\}$, start with the basic transform $\mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} = \frac{t^2}{2!} = \frac{t^2}{2}$. Then apply the first shifting theorem: if $\mathcal{L}^{-1}\{F(s)\} = f(t)$, then $\mathcal{L}^{-1}\{F(s+a)\} = e^{-at}f(t)$. Since $\frac{1}{(s+3)^3} = F(s+3)$ where $F(s) = \frac{1}{s^3}$, we get: $\mathcal{L}^{-1}\left\{\frac{1}{(s+3)^3}\right\} = e^{-3t} \cdot \frac{t^2}{2} = \frac{1}{2}t^2e^{-3t}$ This confirms answer A is correct. Answer B ($t^2e^{-3t}$) omits the factorial correction factor $\frac{1}{2!}$ that comes from the $\frac{1}{s^3}$ transform. Answer C ($\frac{1}{6}t^3e^{-3t}$) incorrectly uses the transform for $\frac{1}{s^4}$, which would give $\frac{t^3}{3!}$. Answer D ($\frac{1}{2}t^2e^{3t}$) has the wrong sign in the exponential—confusing the shifting direction. Study tip: Memorize that $\mathcal{L}^{-1}\left\{\frac{1}{s^n}\right\} = \frac{t^{n-1}}{(n-1)!}$ and remember the first shifting theorem always produces $e^{-at}$ when you see $(s+a)$ terms.

Question 5

1. $G(s) = F(s-a)$
2. $G(s) = e^{-a}F(s)$
3. $G(s) = e^{-as}F(s)/s$
4. $G(s) = e^{-as}F(s)$ (correct answer)

Explanation: The piecewise definition of $g(t)$ is the standard way to represent a time-shifted function that is zero before the shift. This definition is equivalent to writing $g(t) = f(t-a)u(t-a)$, where $u(t-a)$ is the Heaviside step function. The Laplace transform of this expression is given by the second shifting theorem (or t-shifting theorem), which states that $\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)$, where $F(s) = \mathcal{L}\{f(t)\}$. Therefore, the correct relationship is $G(s) = e^{-as}F(s)$.

Question 6

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

Explanation: We use the linearity of the Laplace transform to transform each term separately. $\mathcal{L}\{g(t)\} = \mathcal{L}\{3f(t-2)u(t-2)\} + \mathcal{L}\{e^{-t}f(t)\}$ For the first term, we use the property of linearity with constants and the second shifting theorem, $\mathcal{L}\{h(t-a)u(t-a)\} = e^{-as}H(s)$. Here, $a=2$ and $h(t)=f(t)$. So, $\mathcal{L}\{3f(t-2)u(t-2)\} = 3e^{-2s}F(s)$. For the second term, we use the first shifting theorem, $\mathcal{L}\{e^{at}h(t)\} = H(s-a)$. Here, $a=-1$ and $h(t)=f(t)$. So, $\mathcal{L}\{e^{-t}f(t)\} = F(s-(-1)) = F(s+1)$. Combining the results gives $\mathcal{L}\{g(t)\} = 3e^{-2s}F(s) + F(s+1)$.

Question 7

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

Explanation: First, express the piecewise function $f(t)$ using Heaviside step functions $u(t-a)$. $f(t) = 2[u(t)-u(t-1)] + t[u(t-1)-u(t-3)]$ $f(t) = 2 - 2u(t-1) + tu(t-1) - tu(t-3)$ $f(t) = 2 + (t-2)u(t-1) - tu(t-3)$ To use the second shifting theorem, $\mathcal{L}\{g(t-a)u(t-a)\} = e^{-as}G(s)$, we rewrite the terms: For $(t-2)u(t-1)$, let $t-1 = \tau$, so $t-2 = \tau - 1$. This term becomes $( (t-1) - 1 )u(t-1)$. Here $g(t) = t-1$, so $G(s) = \frac{1}{s^2}-\frac{1}{s}$. The transform is $e^{-s}(\frac{1}{s^2}-\frac{1}{s})$. For $-tu(t-3)$, let $t-3 = \tau$, so $-t = -(\tau+3)$. This term is $-((t-3)+3)u(t-3)$. Here $g(t) = -(t+3)$, so $G(s) = -(\frac{1}{s^2}+\frac{3}{s})$. The transform is $-e^{-3s}(\frac{1}{s^2}+\frac{3}{s})$. Combining all parts: $\mathcal{L}\{f(t)\} = \mathcal{L}\{2\} + \mathcal{L}\{(t-2)u(t-1)\} + \mathcal{L}\{-tu(t-3)\} = \frac{2}{s} + e^{-s}(\frac{1}{s^2}-\frac{1}{s}) - e^{-3s}(\frac{1}{s^2}+\frac{3}{s})$. Distributing gives $\frac{2}{s} - \frac{e^{-s}}{s} + \frac{e^{-s}}{s^2} - \frac{3e^{-3s}}{s} - \frac{e^{-3s}}{s^2}$. This corresponds to choice B. Choice C had a sign error in the setup.

Question 8

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

Explanation: This question requires applying both the linearity property and the first shifting theorem. First, use the linearity property: $\mathcal{L}\{2f(t) - e^{3t}f(t)\} = \mathcal{L}\{2f(t)\} - \mathcal{L}\{e^{3t}f(t)\} = 2\mathcal{L}\{f(t)\} - \mathcal{L}\{e^{3t}f(t)\}$ The first term is $2F(s)$. For the second term, apply the first shifting theorem, $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$. Here, $a=3$, so $\mathcal{L}\{e^{3t}f(t)\} = F(s-3)$. Combining the two parts, the expression is $2F(s) - F(s-3)$.

Question 9

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

Explanation: To find the inverse Laplace transform, first complete the square in the denominator: $s^2+2s+5 = (s^2+2s+1) + 4 = (s+1)^2 + 2^2$. This suggests using the first shifting theorem with $a=-1$. Rewrite the numerator in terms of $(s+1)$: $s+3 = (s+1) + 2$. Now, split the fraction: $F(s) = \frac{s+1}{(s+1)^2+2^2} + \frac{2}{(s+1)^2+2^2}$ The inverse transform is found using linearity and the first shifting theorem, $\mathcal{L}^{-1}\{G(s-a)\} = e^{at}g(t)$. The first term corresponds to the shifted cosine transform: $\mathcal{L}^{-1}\left\{\frac{s+1}{(s+1)^2+2^2}\right\} = e^{-t}\cos(2t)$. The second term corresponds to the shifted sine transform: $\mathcal{L}^{-1}\left\{\frac{2}{(s+1)^2+2^2}\right\} = e^{-t}\sin(2t)$. Summing these results gives $e^{-t}(\cos(2t) + \sin(2t))$.

Question 10

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

Explanation: The denominator $s^2-6s+13$ suggests completing the square. $s^2-6s+13 = (s^2-6s+9) + 4 = (s-3)^2 + 2^2$. This indicates a shift of $a=3$. We must rewrite the numerator, $2s-5$, in terms of $(s-3)$: $2s-5 = 2(s-3) + 6 - 5 = 2(s-3) + 1$ Now, substitute this back into the fraction and split it: $G(s) = \frac{2(s-3) + 1}{(s-3)^2+2^2} = \frac{2(s-3)}{(s-3)^2+2^2} + \frac{1}{(s-3)^2+2^2}$ We find the inverse transform of each part using linearity and the first shifting theorem. For the first term: $\mathcal{L}^{-1}\left\{ \frac{2(s-3)}{(s-3)^2+2^2} \right\} = 2e^{3t}\cos(2t)$ For the second term, we need a 2 in the numerator for the sine transform: $\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)^2+2^2} \right\} = \frac{1}{2}\mathcal{L}^{-1}\left\{ \frac{2}{(s-3)^2+2^2} \right\} = \frac{1}{2}e^{3t}\sin(2t)$ Combining these gives the final answer: $g(t) = 2e^{3t}\cos(2t) + \frac{1}{2}e^{3t}\sin(2t) = e^{3t}(2\cos(2t) + \frac{1}{2}\sin(2t))$.

Question 11

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

Explanation: This problem requires the second shifting theorem, $\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)$. The given function is $g(t) = t^2 u(t-2)$. We must express the function $t^2$ in terms of $(t-2)$. Let $\tau = t-2$, which means $t = \tau+2$. Then $t^2 = (\tau+2)^2 = \tau^2+4\tau+4$. So, $g(t) = ((t-2)^2 + 4(t-2) + 4)u(t-2)$. This is in the form $f(t-2)u(t-2)$ where $f(t) = t^2+4t+4$. First, we find the Laplace transform of $f(t)$: $F(s) = \mathcal{L}\{t^2+4t+4\} = \frac{2!}{s^3} + \frac{4(1!)}{s^2} + \frac{4}{s} = \frac{2}{s^3} + \frac{4}{s^2} + \frac{4}{s}$ Now, applying the second shifting theorem with $a=2$, we get $\mathcal{L}\{g(t)\} = e^{-2s}F(s) = e^{-2s}\left(\frac{2}{s^3} + \frac{4}{s^2} + \frac{4}{s}\right)$.

Question 12

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

Explanation: For the Dirac delta function $\delta(t-2)$, we use the property $\mathcal{L}\{\delta(t-a)\} = e^{-as}$, so $\mathcal{L}\{\delta(t-2)\} = e^{-2s}$. For the term $e^{-t}u(t-1)$, we need to apply the second shifting theorem carefully. We have $e^{-t}u(t-1) = e^{-(t-1+1)}u(t-1) = e^{-1}e^{-(t-1)}u(t-1)$. Using the second shifting theorem, $\mathcal{L}\{g(t-a)u(t-a)\} = e^{-as}G(s)$, where $g(t) = e^{-t}$, so $G(s) = \frac{1}{s+1}$. Therefore, $\mathcal{L}\{e^{-(t-1)}u(t-1)\} = e^{-s} \cdot \frac{1}{s+1} = \frac{e^{-s}}{s+1}$. The factor $e^{-1}$ remains, giving us $\mathcal{L}\{e^{-t}u(t-1)\} = e^{-1} \cdot \frac{e^{-s}}{s+1} = \frac{e^{-s}e^{-1}}{s+1}$. Therefore, the total transform of the right-hand side is $e^{-2s} + \frac{e^{-s}e^{-1}}{s+1}$.

Question 13

1. $\frac{e^{-3s}}{s(s+2)}$ (correct answer)
2. $\frac{e^{-3s}(1-e^{-6})}{s(s+2)}$
3. $\frac{e^{-3s}}{2s(s+2)}$
4. $\frac{e^{-3s}(1-e^{-2t})}{s(s+2)}$

Explanation: First, let's evaluate the integral: $\int_0^{t-3} e^{-2\tau} d\tau = \left[-\frac{1}{2}e^{-2\tau}\right]_0^{t-3} = -\frac{1}{2}e^{-2(t-3)} + \frac{1}{2} = \frac{1}{2}(1 - e^{-2(t-3)})$. So our function becomes $\frac{1}{2}(1 - e^{-2(t-3)})u(t-3)$. This can be written as $\frac{1}{2}u(t-3) - \frac{1}{2}e^{-2(t-3)}u(t-3)$. Using the second shifting theorem: $\mathcal{L}\{u(t-3)\} = \frac{e^{-3s}}{s}$ and $\mathcal{L}\{e^{-2(t-3)}u(t-3)\} = e^{-3s} \cdot \frac{1}{s+2}$. Therefore, $\mathcal{L}\left\{\frac{1}{2}(1 - e^{-2(t-3)})u(t-3)\right\} = \frac{1}{2} \cdot \frac{e^{-3s}}{s} - \frac{1}{2} \cdot \frac{e^{-3s}}{s+2} = \frac{e^{-3s}}{2}\left(\frac{1}{s} - \frac{1}{s+2}\right) = \frac{e^{-3s}}{2} \cdot \frac{(s+2)-s}{s(s+2)} = \frac{e^{-3s}}{2} \cdot \frac{2}{s(s+2)} = \frac{e^{-3s}}{s(s+2)}$.