1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

12.9 • I NVERTING THE LAPLACE TRANSFORM 567

and we obtain A = 4 and B = - 4. Therefore,

1 -2 - 2 2l(s- 0)- 2(- l)(1)
Y(s)=-+--+ + 2 2
s s + l (s+l)^2 (s- 0)^2 +12

1 2 2 s+l

=-- --- +--

s s + 1 (s + 1)^2 s^2 + 1 ·

Now we use Table 12.2 to get

y(t) = l -2e- t -2te-t +cost +sint.

• EXAMPLE 12.27 Use Laplace transforms to solve the system

y'(t) = y(t)- x(t),

x' (t) = 5y (t) - 3x (t),

with y (0) = l;

with x (0) = 2.

Solution We let Y (s) and X (s) denote the Laplace transforms of y (t) and
x (t), respectively. Taking the transforms of the two differential equations gives

sY(s)-1 = Y(s)- X(s) and

sX (s) - 2 = 5Y (s) -3X (s),

which can be written as

(s - 1) Y (s) + X (s) = 1 and

5Y (s) - (s + 3) X (s) = -2.

We use Cramer's rule to solve for Y (s) and X (s):

Y(s) =

X(s) =

I !2 / 31 - s - 3 + 2 s + 1.

l

s - 1 1 1-(s- l)(- s - 3)- 5 - (s+1)^2 +1'

5 - s - 3

ls;l !21 -2s+2- 5 2(s+l)+ l

l

s - 1 1 I= (s-l)(-s-3)-5 = (s +1)^2 + 1 ·

5 - s - 3

We obtain the desired solution by computing the inverse transforms:

y (t) = e - t cost and

x(t) = e-^1 (2cost+sint).