1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

576 CHAPTER 12 • FOURIER SERIES AND THE LAPLACE TRANSFORM

y

0.4

0.2

--0.5

-0.I

Figure 12.31 The solution y = y (t).

Solution Taking transforms results in (s^2 + 4s + 13) Y (s) = 3.C (6 (t)) = 3 so

that

3 3

Y(s)= s^2 +4s+ 13 = (s+2)^2 + 32'

and the solution is

y (t) = e-^2 t sin 3t.

Remark 12.2 The condition y' (o-) = 0 is not satisfied by the "solution"

y (t). Recall that all solutions involving the use of the Laplace transform are

to be considered zero for va lues oft < ~hence the graph of y (t) as given in

Figure 12.31. Note that y' (t) has a jump discontinuity of magnitude + 3 at the

origin. This discontinuity occurs because either y (t) or y' (t) must have a j ump

discontinuity at t he origin whenever the Dirac delta function occurs as part of

the input or driving function. •

The convolution method can be used to solve initial value problems. The

tedious mechanical details of problem solving can be facilitated with computer

software such as MapleTM, Matlab™, or Mathematica TM.