Tutorial Laplace 21 October 2009 1. (2) Use Laplace transform methods to solve the differential equation  1, 0 ≤ t < c 00 0 f + 2f − 3f = 0, t ≥ c

(1)

subject to the initial conditions f (0) = f 0 (0) = 0. (3)

Solution: Taking Laplace transforms of both sides and using the tables for the Laplace transform of the right hand side function, leads to 1 − e−cs s 1 − e−cs F = s(s2 + 2s − 3)

(s2 + 2s − 3)F =

1 s(s − 1)(s + 3)   A B C −cs + + = (1 − e ) s s−1 s+3 = (1 − e−cs )

(2)

Concentrating on the partial fractions part, we have

1 A B C = + + s(s − 1)(s + 3) s s−1 s+3 1 = A(s − 1)(s + 3) + Bs(s + 3) + Cs(s − 1) s=0: 1 = −3A 1 A = − 3 s=1: 1 = 0 + 4B + 0 1 B = 4 s = −3 : 1 = 0 + 012C 1 C = 12 Hence we have   11 1 1 1 1 −cs + + F = (1 − e ) − 3 s 4 s − 1 12 s + 3 1

UKSW, [email protected], see also http://laic.u-clermont1.fr/ tomasik/

1

(3)

From the tables, we know that   1 1 t 1 −3t 11 1 1 1 1 =− L − + e − e + + 3 4 12 3 s 4 s − 1 12 s + 3

(4)

and then using the second shift theorem   1 1 t 1 −3t 1 1 (t−c) 1 −3(t−c) + e f (t) = − + e + e − Hc (t) − + e 3 4 12 3 4 12 2. (3) Use Laplace transform methods to solve the differential equation   0, 0 ≤ t < 1 00 0 1, 1 ≤ t < 2 f + 2f − 3f =  0, t ≥ 2

(5)

(6)

subject to the initial conditions f (0) = 0 and f 0 (0) = 0.

Solution:So the thing here is to rewrite the right hand side of the equations in terms of Heaviside functions. Remember the definition of the Heaviside function:  0 t