Question

Points Use the integral definition of Laplace transform to show that L(f

′
(t))=sF(s)−f(0) 4 Points Use the above Laplace transform formula to show that L(f
′′
(t))=s
2
F(s)−as−b for some constants a,b. What are the values of a and b ?

Answer 1

Using the Laplace transform definition, the transform of a function's first derivative is L(f'(t)) = sF(s) - f(0), and the transform of the second derivative is L(f''(t)) = s^2F(s) - as - b, with a and b being the initial value of the function and its first derivative respectively.

Step-by-step explanation:

The Laplace transform of the derivative of a function can be shown by using the integral definition of the Laplace transform. Given f(t), the Laplace transform of its first derivative, f', is L(f'(t)) = sF(s) - f(0), where F(s) is the Laplace transform of f(t). To find the transform of the second derivative, f'', we apply the definition again and get L(f''(t)) = s^2F(s) - sf(0) - f'(0), which is the Laplace transform formula for a second derivative. Here, the constants a and b correspond to f(0) and f'(0) respectively, so a = f(0) and b = f'(0). The formula for the second derivative thus becomes L(f''(t)) = s^2F(s) - as - b.