Differential Equations Quiz: Transforms Of Common Functions

What this quiz covers

This quiz focuses on Transforms Of Common 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. Apply frequency shifting to $\mathcal{L}\{\sin(t)\}$, then apply the derivative property for the $t$ factor
2. Apply the derivative property to $\mathcal{L}\{e^{2t}\sin(t)\}$ first, then use linearity (correct answer)
3. Use convolution theorem since $f(t)$ is a product of three functions
4. Apply linearity first, then frequency shifting, then convolution for the remaining product

Explanation: The correct approach is to first find $\mathcal{L}\{e^{2t}\sin(t)\}$ using frequency shifting: $\mathcal{L}\{\sin(t)\} = \frac{1}{s^2+1}$, so $\mathcal{L}\{e^{2t}\sin(t)\} = \frac{1}{(s-2)^2+1}$. Then apply $\mathcal{L}\{tf(t)\} = -\frac{d}{ds}[F(s)]$ to get $\mathcal{L}\{te^{2t}\sin(t)\} = -\frac{d}{ds}\left[\frac{1}{(s-2)^2+1}\right] = \frac{2(s-2)}{[(s-2)^2+1]^2}$. Choice A applies operations in wrong order. Choice C incorrectly suggests convolution (this is multiplication by $t$ and exponential, not convolution). Choice D misapplies multiple theorems unnecessarily.

Question 2

1. $\frac{3}{s^2-4s-5}$ (correct answer)
2. $\frac{s-2}{s^2-4s-5}$
3. $\frac{3}{s^2-4s+13}$
4. $\frac{3}{s^2+4s-5}$

Explanation: This can be solved using the first shifting theorem, $\mathcal{L}\{e^{at}g(t)\} = G(s-a)$. Let $g(t) = \sinh(3t)$ and $a=2$. The transform of $g(t)$ is $G(s) = \mathcal{L}\{\sinh(3t)\} = \frac{3}{s^2-3^2} = \frac{3}{s^2-9}$. Now, apply the shift by replacing $s$ with $s-a = s-2$: $F(s) = G(s-2) = \frac{3}{(s-2)^2-9} = \frac{3}{s^2-4s+4-9} = \frac{3}{s^2-4s-5}$. Alternatively, one could use the definition $\sinh(3t) = \frac{e^{3t}-e^{-3t}}{2}$, so $f(t) = e^{2t} \frac{e^{3t}-e^{-3t}}{2} = \frac{1}{2}(e^{5t} - e^{-t})$. The transform is $\frac{1}{2}(\frac{1}{s-5} - \frac{1}{s+1}) = \frac{1}{2}\frac{(s+1)-(s-5)}{(s-5)(s+1)} = \frac{1}{2}\frac{6}{s^2-4s-5} = \frac{3}{s^2-4s-5}$.

Question 3

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

Explanation: When you encounter products of trigonometric and hyperbolic functions in Laplace transforms, the most efficient approach is to use the frequency shifting property along with standard transform formulas. Start by rewriting $\sinh(t) = \frac{e^t - e^{-t}}{2}$, so: $f(t) = \sinh(t)\cos(t) = \frac{e^t - e^{-t}}{2}\cos(t) = \frac{1}{2}[e^t\cos(t) - e^{-t}\cos(t)]$ Now apply the frequency shifting property: if $\mathcal{L}\{g(t)\} = G(s)$, then $\mathcal{L}\{e^{at}g(t)\} = G(s-a)$. Since $\mathcal{L}\{\cos(t)\} = \frac{s}{s^2+1}$:

• $\mathcal{L}\{e^t\cos(t)\} = \frac{s-1}{(s-1)^2+1} = \frac{s-1}{s^2-2s+2}$
• $\mathcal{L}\{e^{-t}\cos(t)\} = \frac{s+1}{(s+1)^2+1} = \frac{s+1}{s^2+2s+2}$

Question 4

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

Explanation: This problem requires the first shifting theorem (or s-shifting theorem), which states that $\mathcal{L}\{e^{at}g(t)\} = G(s-a)$, where $G(s) = \mathcal{L}\{g(t)\}$. Here, $g(t) = t^2$ and $a = -4$. First, find the transform of $g(t) = t^2$: $G(s) = \mathcal{L}\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}$. Next, apply the shifting theorem by replacing $s$ with $s-a = s-(-4) = s+4$. So, $F(s) = G(s+4) = \frac{2}{(s+4)^3}$.

Question 5

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

Explanation: When you encounter a Laplace transform of a trigonometric function with a phase shift, you need to use trigonometric identities to break it down before applying standard transform formulas. For $f(t) = \sin(2t + \pi/4)$, use the sine addition formula: $\sin(A + B) = \sin A \cos B + \cos A \sin B$ So: $\sin(2t + \pi/4) = \sin(2t)\cos(\pi/4) + \cos(2t)\sin(\pi/4)$ Since $\cos(\pi/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2}$: $f(t) = \frac{\sqrt{2}}{2}\sin(2t) + \frac{\sqrt{2}}{2}\cos(2t) = \frac{\sqrt{2}}{2}[\sin(2t) + \cos(2t)]$ Now apply the linearity of Laplace transforms and standard formulas: $\mathcal{L}\{\sin(at)\} = \frac{a}{s^2+a^2}$ and $\mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2}$ $\mathcal{L}\{f(t)\} = \frac{\sqrt{2}}{2}\left[\frac{2}{s^2+4} + \frac{s}{s^2+4}\right] = \frac{\sqrt{2}}{2}\left(\frac{s+2}{s^2+4}\right)$ This matches answer choice D. Answer A gives only the sine component without the phase shift factor. Answer B has the wrong sign in the numerator ($s-2$ instead of $s+2$), which would result from incorrectly handling the trigonometric identity. Answer C lacks the $\frac{\sqrt{2}}{2}$ coefficient that comes from the phase shift. Study tip: Always expand trigonometric functions with phase shifts using addition formulas before taking Laplace transforms. The phase shift will always introduce coefficients like $\frac{\sqrt{2}}{2}$ or $\frac{1}{2}$ that you can't ignore.

Question 6

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

Explanation: When you encounter Laplace transforms of polynomial expressions like $(t-1)^3$, you have two main approaches: expand the polynomial first, or use shifting properties. The expansion method is more straightforward here. First, expand $(t-1)^3$ using the binomial theorem: $(t-1)^3 = t^3 - 3t^2 + 3t - 1$ Now apply the Laplace transform to each term using the basic formula $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$:

• $\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}$
• $\mathcal{L}\{3t^2\} = 3 \cdot \frac{2!}{s^3} = \frac{6}{s^3}$
• $\mathcal{L}\{3t\} = 3 \cdot \frac{1!}{s^2} = \frac{3}{s^2}$
• $\mathcal{L}\{1\} = \frac{1}{s}$

Question 7

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

Explanation: When finding the Laplace transform of a squared expression like $(t+e^{-t})^2$, you need to expand it algebraically first rather than trying to use transform properties directly on the squared form. Let's expand: $(t+e^{-t})^2 = t^2 + 2te^{-t} + e^{-2t}$ Now we can find the Laplace transform of each term separately. For $t^2$, we use $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$, giving us $\frac{2!}{s^3} = \frac{2}{s^3}$. For $2te^{-t}$, we need the shifting property: $\mathcal{L}\{te^{-at}\} = \frac{1}{(s+a)^2}$. With $a=1$, we get $2 \cdot \frac{1}{(s+1)^2} = \frac{2}{(s+1)^2}$. For $e^{-2t}$, we use $\mathcal{L}\{e^{-at}\} = \frac{1}{s+a}$ with $a=2$, giving us $\frac{1}{s+2}$. Adding these together: $\frac{2}{s^3} + \frac{2}{(s+1)^2} + \frac{1}{s+2}$, which matches answer B. Answer A incorrectly suggests you can square the individual transforms, but $\mathcal{L}\{(f+g)^2\} \neq (\mathcal{L}\{f\} + \mathcal{L}\{g\})^2$. Answer C is missing the cross-term $\frac{2}{(s+1)^2}$ from $2te^{-t}$. Answer D has sign errors in the denominators—it shows $(s-1)$ and $(s-2)$ instead of the correct $(s+1)$ and $(s+2)$. Strategy tip: Always expand polynomial expressions before taking Laplace transforms, and remember that exponential terms with negative exponents create positive terms in the denominator of your transform.

Question 8

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

Explanation: When you encounter Laplace transforms of expressions like $(e^t + 1)^2$, the key strategy is to expand the function algebraically first, then apply linearity properties rather than trying to work with the squared form directly. Let's expand $f(t) = (e^t + 1)^2$: $f(t) = (e^t + 1)^2 = (e^t)^2 + 2e^t \cdot 1 + 1^2 = e^{2t} + 2e^t + 1$ Now we can find the Laplace transform of each term using standard formulas. Recall that $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$ and $\mathcal{L}\{1\} = \frac{1}{s}$: $\mathcal{L}\{f(t)\} = \mathcal{L}\{e^{2t}\} + 2\mathcal{L}\{e^t\} + \mathcal{L}\{1\} = \frac{1}{s-2} + \frac{2}{s-1} + \frac{1}{s}$ This matches answer D. Answer A uses incorrect signs in the denominators - it has $s+2$ and $s+1$ instead of $s-2$ and $s-1$, which would correspond to transforms of $e^{-2t}$ and $e^{-t}$. Answer B represents the transform of $(e^t + 1)$ squared, not the transform of $(e^t + 1)^2$. The Laplace transform doesn't distribute over multiplication this way. Answer C is missing the middle term $\frac{2}{s-1}$ that comes from the cross-product $2e^t$ when expanding the square. Study tip: Always expand polynomial expressions algebraically before taking Laplace transforms. The linearity property $\mathcal{L}\{af + bg\} = a\mathcal{L}\{f\} + b\mathcal{L}\{g\}$ is your friend, but it doesn't apply to products or powers directly.

Question 9

1. $\frac{s^2+18}{s(s^2+36)}$ (correct answer)
2. $\frac{18}{s(s^2+36)}$
3. $\frac{s^2}{(s^2+9)^2}$
4. $\frac{2s^2+9}{2s(s^2+9)}$

Explanation: To find the Laplace transform of $f(t) = \cos^2(3t)$, first use the half-angle identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$. Applying this identity, we get $f(t) = \frac{1}{2}(1 + \cos(6t))$. By linearity of the Laplace transform, $\mathcal{L}\{f(t)\} = \frac{1}{2}\mathcal{L}\{1\} + \frac{1}{2}\mathcal{L}\{\cos(6t)\}$. Using the standard transforms $\mathcal{L}\{1\} = \frac{1}{s}$ and $\mathcal{L}\{\cos(bt)\} = \frac{s}{s^2+b^2}$, we have: $\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \frac{1}{s} + \frac{s}{s^2+36} \right)$. To combine the terms, find a common denominator: $\frac{1}{2} \left( \frac{s^2+36+s^2}{s(s^2+36)} \right) = \frac{1}{2} \frac{2s^2+36}{s(s^2+36)} = \frac{s^2+18}{s(s^2+36)}$.

Question 10

1. $\frac{3}{s-2} - \frac{24}{(s-1)^4}$
2. $\frac{3}{s-2} - \frac{24}{(s+1)^4}$ (correct answer)
3. $\frac{3}{s+2} - \frac{12}{(s+1)^3}$
4. $\frac{3}{s-2} - \frac{24}{s^4(s+1)}$

Explanation: When finding Laplace transforms of composite functions, you need to apply the fundamental transform formulas along with key properties like the frequency shifting theorem. To solve $\mathcal{L}\{3e^{2t} - 4t^3e^{-t}\}$, handle each term separately. For the first term, use the basic exponential formula: $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$. So $\mathcal{L}\{3e^{2t}\} = \frac{3}{s-2}$. For the second term, you need the frequency shifting theorem: if $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$. Since $\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}$, we get $\mathcal{L}\{t^3e^{-t}\} = \frac{6}{(s-(-1))^4} = \frac{6}{(s+1)^4}$. Therefore, $\mathcal{L}\{4t^3e^{-t}\} = \frac{24}{(s+1)^4}$. Combining both terms: $\mathcal{L}\{f(t)\} = \frac{3}{s-2} - \frac{24}{(s+1)^4}$, which is answer B. Answer A incorrectly uses $(s-1)^4$ instead of $(s+1)^4$ in the denominator—this comes from misapplying the frequency shift with $e^{-t}$. Answer C has the wrong coefficient (12 instead of 24) and wrong power (3 instead of 4). Answer D completely mishandles the polynomial-exponential term, failing to apply the frequency shifting theorem properly. Remember: when you see $t^n e^{at}$, always use the frequency shifting property. Replace $s$ with $(s-a)$ in the transform of $t^n$, and watch your signs carefully—$e^{-t}$ means $a = -1$.

Question 11

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

Explanation: When you encounter Laplace transforms of products involving exponential and trigonometric functions, the key is to use the frequency shifting property and trigonometric identities to simplify before transforming. To find the Laplace transform of $f(t) = e^{-t}\cos^2(t)$, first use the identity $\cos^2(t) = \frac{1 + \cos(2t)}{2}$. This gives us: $f(t) = e^{-t} \cdot \frac{1 + \cos(2t)}{2} = \frac{e^{-t}}{2} + \frac{e^{-t}\cos(2t)}{2}$ Now apply the Laplace transform to each term. For the first term, $\mathcal{L}\{e^{-t}\} = \frac{1}{s+1}$. For the second term, use the frequency shifting property: $\mathcal{L}\{e^{-at}g(t)\} = G(s+a)$. Since $\mathcal{L}\{\cos(2t)\} = \frac{s}{s^2+4}$, we have $\mathcal{L}\{e^{-t}\cos(2t)\} = \frac{s+1}{(s+1)^2+4}$. Therefore: $F(s) = \frac{1}{2(s+1)} + \frac{s+1}{2((s+1)^2+4)}$ To match the given options, find a common denominator: $F(s) = \frac{(s+1)^2+4 + (s+1)^2}{2(s+1)((s+1)^2+4)} = \frac{2(s+1)^2+4}{2(s+1)((s+1)^2+4)} = \frac{(s+1)^2+2}{(s+1)((s+1)^2+4)}$ Wait - this matches option A, but expanding $(s+1)^2+4 = s^2+2s+5$, so option C is equivalent: $\frac{(s+1)^2+2}{(s+1)(s^2+2s+5)}$. Option A uses the unexpanded form, option B ignores the exponential shift, and option D represents the intermediate step before combining fractions. Study tip: Always use trigonometric identities to eliminate squares before applying Laplace transforms, and remember that frequency shifting replaces $s$ with $s+a$ throughout the transform.

Question 12

1. -3
2. 3 (correct answer)
3. 6
4. 5

Explanation: When you encounter Laplace transforms involving exponential and trigonometric functions, you're dealing with a frequency shift property. The key insight is recognizing how the exponential factor $e^{at}$ modifies the standard cosine transform. The standard Laplace transform of $\cos(bt)$ is $\frac{s}{s^2 + b^2}$. When multiplied by $e^{at}$, the frequency shift property tells us that $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$, where $F(s)$ is the transform of $f(t)$. For $f(t) = e^{at}\cos(5t)$, we start with $\mathcal{L}\{\cos(5t)\} = \frac{s}{s^2 + 25}$. Applying the shift property with parameter $a$, we get: $\mathcal{L}\{e^{at}\cos(5t)\} = \frac{s-a}{(s-a)^2 + 25}$ Expanding the denominator: $(s-a)^2 + 25 = s^2 - 2as + a^2 + 25$ Comparing with the given result $F(s) = \frac{s-3}{s^2-6s+34}$, we need:

• Numerator: $s-a = s-3$, so $a = 3$
• Denominator: $s^2-2as+a^2+25 = s^2-6s+34$

Question 13

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

Explanation: First find the transforms of the polynomial terms: $\mathcal{L}\{2t^2\} = \frac{4}{s^3}$, $\mathcal{L}\{3t\} = \frac{3}{s^2}$, $\mathcal{L}\{1\} = \frac{1}{s}$. So $\mathcal{L}\{2t^2-3t+1\} = \frac{4}{s^3} - \frac{3}{s^2} + \frac{1}{s}$. Now apply frequency shifting theorem for $e^{-t}$ ($a=-1$): replace $s$ with $(s+1)$ to get $\frac{4}{(s+1)^3} - \frac{3}{(s+1)^2} + \frac{1}{s+1}$. Choice B has wrong coefficient for $t^2$ term. Choice C uses $(s-1)$ which corresponds to $e^{t}$, not $e^{-t}$. Choice D has wrong signs for the middle terms.

Question 14

1. First apply the frequency shifting theorem, then the second derivative formula for polynomials, then linearity
2. First apply linearity to separate the coefficient, then the time shifting theorem for the exponential, then the transform of cosine
3. First apply linearity to separate the coefficient, then the frequency shifting theorem for the exponential, then the second derivative property for $t^2$ (correct answer)
4. First apply the convolution theorem to handle the product, then linearity, then the basic cosine transform

Explanation: The correct approach is: (1) Use linearity to factor out the constant 3, (2) Apply the frequency shifting theorem (first shifting property) to handle $e^{-2t}$ which shifts $s$ to $(s+2)$, (3) Use the derivative property $\mathcal{L}\{t^n F(t)\} = (-1)^n F^{(n)}(s)$ where $F(s) = \mathcal{L}\{\cos(4t)\} = \frac{s}{s^2+16}$, requiring the second derivative. Choice A incorrectly mentions polynomial derivative formulas. Choice B incorrectly refers to time shifting (which involves unit step functions). Choice D incorrectly suggests convolution, but this is a product involving $t^n$ and exponential terms, not a convolution integral.

Question 15

1. $\frac{48s(s^2-4)}{(s^2+4)^4}$ (correct answer)
2. $\frac{48(s^2-4)}{(s^2+4)^4}$
3. $\frac{24s(s^2-4)}{(s^2+4)^4}$
4. $\frac{48s(3s^2-4)}{(s^2+4)^4}$

Explanation: Starting with $\mathcal{L}\{\sin(2t)\} = \frac{2}{s^2+4}$, we need the third derivative: $\frac{d}{ds}\left(\frac{2}{s^2+4}\right) = \frac{-4s}{(s^2+4)^2}$, $\frac{d^2}{ds^2}\left(\frac{2}{s^2+4}\right) = \frac{-4(s^2+4)^2 + 4s \cdot 2(s^2+4) \cdot 2s}{(s^2+4)^4} = \frac{4(3s^2-4)}{(s^2+4)^3}$, $\frac{d^3}{ds^3}\left(\frac{2}{s^2+4}\right) = \frac{48s(s^2-4)}{(s^2+4)^4}$. Since $n=3$, we have $(-1)^3 = -1$, so $\mathcal{L}\{t^3\sin(2t)\} = -\left(-\frac{48s(s^2-4)}{(s^2+4)^4}\right) = \frac{48s(s^2-4)}{(s^2+4)^4}$. Choice B is missing the $s$ factor. Choice C has the wrong coefficient. Choice D has an incorrect numerator polynomial.

Question 16

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

Explanation: When you encounter a Laplace transform of a function involving both polynomial and exponential terms, the key is recognizing how to apply the linearity property and the frequency shifting theorem effectively. Since $g(t) = (t^2 - 2t + 1)e^{3t}$, you can use linearity to write this as $\mathcal{L}\{t^2 e^{3t}\} - 2\mathcal{L}\{t e^{3t}\} + \mathcal{L}\{e^{3t}\}$. The frequency shifting theorem states that $\mathcal{L}\{f(t)e^{at}\} = F(s-a)$ where $F(s) = \mathcal{L}\{f(t)\}$. Using the standard transforms $\mathcal{L}\{1\} = \frac{1}{s}$, $\mathcal{L}\{t\} = \frac{1}{s^2}$, and $\mathcal{L}\{t^2\} = \frac{2}{s^3}$, applying the frequency shifting theorem with $a = 3$ gives:

• $\mathcal{L}\{e^{3t}\} = \frac{1}{s-3}$
• $\mathcal{L}\{te^{3t}\} = \frac{1}{(s-3)^2}$
• $\mathcal{L}\{t^2e^{3t}\} = \frac{2}{(s-3)^3}$

Question 17

1. $\frac{s}{s^2+6} + \frac{s}{s^2+36}$
2. $\frac{1}{2}\left[\frac{s}{s^2+25} + \frac{s}{s^2+1}\right]$
3. $\frac{s^2+13}{2(s^2+1)(s^2+25)}$
4. $\frac{1}{2}\left[\frac{s}{s^2+1} + \frac{s}{s^2+25}\right]$ (correct answer)

Explanation: When you encounter a product of trigonometric functions in a Laplace transform problem, the key insight is to use trigonometric identities to simplify before transforming. The product-to-sum identity is your best tool here. To find $\mathcal{L}\{\cos(2t)\cos(3t)\}$, first apply the product-to-sum identity: $\cos(A)\cos(B) = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$. This gives us: $\cos(2t)\cos(3t) = \frac{1}{2}[\cos(2t-3t) + \cos(2t+3t)] = \frac{1}{2}[\cos(-t) + \cos(5t)]$ Since $\cos(-t) = \cos(t)$, we have: $\cos(2t)\cos(3t) = \frac{1}{2}[\cos(t) + \cos(5t)]$ Now apply the Laplace transform using linearity and the standard formula $\mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2}$: $\mathcal{L}\{\cos(2t)\cos(3t)\} = \frac{1}{2}\left[\mathcal{L}\{\cos(t)\} + \mathcal{L}\{\cos(5t)\}\right] = \frac{1}{2}\left[\frac{s}{s^2+1} + \frac{s}{s^2+25}\right]$ This confirms answer D is correct. Answer A incorrectly uses frequencies 6 and 36, which don't correspond to any meaningful decomposition of the original product. Answer B has the right structure and factor of $\frac{1}{2}$, but the denominators should be $s^2+1$ and $s^2+25$, not $s^2+25$ and $s^2+1$. Answer C attempts to combine the fractions but makes algebraic errors in the process. Study tip: Always use trigonometric identities to break down products before applying Laplace transforms—it's much easier than trying to use convolution properties.

Question 18

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

Explanation: The function equals $t^2$ for $t < 2$ and $4e^{t-2}$ for $t \geq 2$. We can write this as: start with $t^2$ everywhere, then at $t=2$, subtract $t^2$ and add $4e^{t-2}$. This gives $f(t) = t^2 + [4e^{t-2} - t^2]u(t-2)$. When $t < 2$, $u(t-2) = 0$, so $f(t) = t^2$. When $t \geq 2$, $u(t-2) = 1$, so $f(t) = t^2 + 4e^{t-2} - t^2 = 4e^{t-2}$. Choice A has redundant notation. Choice B is identical to A. Choice D incorrectly uses $u(t)$ and has wrong logic for the intervals.