En anglais

Turning calculus into algebra

Mathematics · Differential Equations · ages 20-21

Name ______________________ Date ____________

The Laplace transform turns a function of time t into a function of a new variable s.
Circle one: True False
True or false: for f(t) = e^(-t), the transform 1/(s + 1) holds for every s > -1.
Circle one: True False
True or false: the integral that defines L{e^(3t)} converges when s = 2.
Circle one: True False
Using the defining integral, what is the transform of the constant function f(t) equal to 1, for s above zero?
- 1 over s
- s
- 1 over s squared
Transform y' + 2y = 0 with y(0) = 3 using L{y'} = sY(s) - y(0). What is Y(s)?
- Y(s) = 3/(s - 2)
- Y(s) = 3(s + 2)
- Y(s) = 3/(s + 2)
- Y(s) = 1/(s + 2)
f(t) = e^(2t), so F(s) = 1/(s - 2) and f(0) = 1. Using L{f'} = sF(s) - f(0), what is L{f'} at s = 4?
Answer: ______________
L{e^(5t)} = 1/(s - 5). The defining integral converges only when s is greater than what number?
Answer: ______________
The rule says L of f prime equals s times F(s) minus f(0). If f(0) equals 1 and F(s) equals 1 over s, what is L of f prime?
Answer: ______________
F(s) equals 1 over ((s minus 2)(s plus 3)), which splits into one fifth over (s minus 2) minus one fifth over (s plus 3). Reading the table backward, what is f(t)?
- One fifth e to the 2t minus one fifth e to the minus 3t
- One fifth e to the minus 2t minus one fifth e to the 3t
- 5 e to the 2t minus 5 e to the minus 3t
F(s) = 2/((s - 1)(s + 1)) splits into 1/(s - 1) - 1/(s + 1). What is f(t)?
- e^(-t) - e^t
- 2e^t - 2e^(-t)
- e^t + e^(-t)
- e^t - e^(-t)

Answer key

For grown-ups. Fold this page away before handing over the rest.

True · The integral takes in f(t), and after integrating over t only s remains.
True · Here a = -1, and 1/(s - a) holds for s > a, so for s > -1.
False · With s = 2 the integrand is e^((3 - 2)t) = e^t, which grows without bound, so the integral diverges. It converges only for s > 3.
1 over s · Putting f(t) equal to 1 in the integral and working it out gives 1 over s.
Y(s) = 3/(s + 2) · sY - 3 + 2Y = 0, so (s + 2)Y = 3 and Y = 3/(s + 2).
1 · s F(s) - f(0) = 4 × 1/2 - 1 = 1. Check: f' = 2e^(2t), whose transform 2/(s - 2) is also 1 at s = 4.
5 · e^((5 - s)t) dies away only when 5 - s < 0, that is s > 5.
0 · Multiplying F(s) by s and subtracting f(0): s times 1 over s is 1, minus 1 is 0.
One fifth e to the 2t minus one fifth e to the minus 3t · Each piece matches an exponential entry, keeping the one fifth weights: 1 over (s minus a) goes back to e to the a t.
e^t - e^(-t) · 1/(s - 1) goes back to e^t and 1/(s + 1) to e^(-t), so f(t) = e^t - e^(-t).