To find L[e^(3t) sin(2t)], which approach is most efficient?
ACompute ∫₀^∞ e^(−st) e^(3t) sin(2t) dt directly from the definition
BUse L[sin(2t)] = 2/(s²+4) and apply the first shifting theorem, replacing s with s−3 to get 2/((s−3)²+4)
CMultiply the separate transforms: L[e^(3t)] · L[sin(2t)] = (1/(s−3)) · (2/(s²+4))
DApply integration by parts with u = e^(3t) and dv = sin(2t) dt twice
The first shifting (s-shift) theorem states L[e^(at)f(t)] = F(s−a). Since L[sin(2t)] = 2/(s²+4), multiplying by e^(3t) simply shifts s to s−3, giving 2/((s−3)²+4) immediately. Option C is a common error: the Laplace transform of a product is NOT the product of transforms (that would be convolution in the time domain). The direct integral (A) works but is inefficient; the shifting theorem exists precisely to avoid repeating that computation.
Question 2 Multiple Choice
Which best explains why the transform pair L[t^n] = n!/s^(n+1) holds?
ABecause t^n is always non-negative for even n, guaranteeing the integral converges
BIt is an empirical result confirmed numerically and must be taken from the table
CRepeated integration by parts reduces the integral — each step pulls down a factor of the current power and gains an extra 1/s — yielding n! in the numerator after n steps
DIt follows directly from the s-shift theorem applied to L[1] = 1/s
The result is derived by computing ∫₀^∞ t^n e^(−st) dt via integration by parts (or by recognizing it as the gamma function). Each application of integration by parts reduces the power of t by 1 while contributing a factor of 1/s to the denominator. After n steps the integral reduces to ∫₀^∞ e^(−st) dt = 1/s, and the accumulated prefactor is n · (n−1) · ... · 1 = n!. The s-shift theorem shifts s but does not create polynomial factors, so option D is incorrect.
Question 3 True / False
Because the Laplace transform is linear, L[3cos(2t) − 5sin(t)] can be computed by transforming each term separately and combining the results.
TTrue
FFalse
Answer: True
Linearity states L[af + bg] = aL[f] + bL[g], mirroring the linearity of integration. So L[3cos(2t) − 5sin(t)] = 3·L[cos(2t)] − 5·L[sin(t)] = 3·s/(s²+4) − 5·1/(s²+1). This is one of the two most powerful tools for extending the basic table — linearity plus the shifting theorems handle the vast majority of functions encountered in ODE applications.
Question 4 True / False
The Laplace transform of a product of two functions equals the product of their individual Laplace transforms: L[f(t)g(t)] = F(s)G(s).
TTrue
FFalse
Answer: False
This is a very common and consequential error. The Laplace transform does NOT distribute over products: L[f(t)g(t)] ≠ F(s)G(s) in general. The correct relationship for products involves convolution: L[f*g] = F(s)G(s), where (f*g)(t) = ∫₀^t f(τ)g(t−τ) dτ is the convolution of f and g. To handle products like e^(at)sin(bt), use the s-shift theorem — not multiplication of transforms.
Question 5 Short Answer
Explain why the first shifting theorem (s-shift) is useful in practice. What does it let you do that you couldn't easily do otherwise?
Think about your answer, then reveal below.
Model answer: The s-shift theorem states that multiplying a time-domain function f(t) by e^(at) shifts the transform variable: L[e^(at)f(t)] = F(s−a). This means you can immediately write down transforms of exponentially modulated functions — like e^(2t)cos(3t) or e^(−t)t² — by taking the known basic transform and replacing s with s−a, without computing a new integral. It dramatically extends the table with no additional computation.
Without the shifting theorem, every new exponentially modulated function would require computing a fresh integral from the definition. The theorem recognizes a structural pattern — multiplying by e^(at) in time corresponds to a horizontal shift in s-space — and packages that pattern as a reusable rule. Combined with linearity and the basic pairs, it covers the overwhelming majority of functions arising in ODE initial-value problems.