The Laplace transform turns a function of time t into a function of a new variable s.

Circle one:   True   False

True or false: because L{f'(t)} = s*F(s) - f(0), taking a derivative in the t-world becomes just multiplying by s in the s-world.

Circle one:   True   False

True or false: the Laplace transform turns a function of time t into a function of a new variable s.

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

Why does the rule L{f'(t)} = s*F(s) - f(0) make the Laplace transform useful for solving differential equations?

• It turns a derivative in the t-world into ordinary multiplication by s in the s-world, so the whole differential equation becomes an algebra equation
• It turns the function into a trigonometric identity
• It removes the need to know any initial conditions
• It converts s back into t automatically

The derivative rule says L{f'(t)} = s*F(s) - f(0), where F(s) = L{f(t)}. If f(0) = 5 and F(s) = 1/s, what is L{f'(t)}?

Answer: ______________

Using the defining integral, the Laplace transform of f(t) = e^(-4t) works out to 1/(s + a). What is the value of a?

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 1)(s plus 2)), which splits into one third over (s minus 1) minus one third over (s plus 2). Reading the table backward, what is f(t)?

• One third e to the t minus one third e to the minus 2t
• One third e to the minus t plus one third e to the 2t
• 3 e to the t minus 3 e to the minus 2t

F(s) = 1/((s-1)(s+2)) splits by partial fractions into A/(s-1) + B/(s+2), with A = 1/3 and B = -1/3. Reading the table backward, what is f(t)?

• (1/3)e^t - (1/3)e^(-2t)
• (1/3)e^(-t) + (1/3)e^(2t)
• 3e^t - 3e^(-2t)
• (1/3)t e^t