1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

12.6 • LAPLACE TRANSFORMS OF DERIVATIVES AND INTEGRALS 551

2 6

• EXAMPLE 12.14 Show that .C (t^2 ) = 3 and .C (t^3 ) = 4.

s s

Solution Using Theorem 12.14 and the fact that £. (2t) = ~,we obtain
s

.c (t)^2 = £. (lt 2-r d-r ) = -£.(2t)^1 = --1 2 = -.^2
o s s s^2 s3

Now we can use this first result,£. (t^2 ) = ~, to establish the second result:

s

One of the main uses of the Laplace transform is its role in the solution
of differential equations. The utility of the Laplace transform lies in the fact
that the transform of the derivative f' (t) corresponds to multiplication of the
transform F (s) by s and then the subtraction of f (0). This permits us to
replace the calculus operation of differentiation with simple algebraic operations
on transforms.
This idea is used to develop a method for solving linear differential equations
with constant coefficients. Let's consider the initial value problem

y" (t) + ay' (t) +by (t) = f (t)

with initial conditions y (0) = y 0 and y' (0) = d 0. We can use the linearity

property of the Laplace transform to obtain

£. (y" (t)) +a£. (y' (t)) + b£. (y (t)) = £. ( f (t)).

If we let Y (s) = .C (y (t)) and F (s) = £. (f (t)) and apply Theorem 12.13 and

Corollary 12.1 in the form .C(y'(t)) = sY(s)- y(O) and .C(y"(t)) = s^2 Y(s)-

sy(O)- y' (0), then we can rewrite the preceding equation in the form

s^2 Y (s) + asY (s) + bY (s) = F (s) + sy (0) + y' (0) + ay (0). (12-30)