Fourier and Laplace 363
ANALYTICAL Solution
Taking the LT of the given equation
203
30
sY s y Y s
s
() ( )()
Ys s
s
()[ 23 ]
30
then
Ys s
s
()( 23 )
30
Ys
ss ss
A
s
B
s
()
()(.).
30
23
15
(^1515)
by partial fraction expansion
Ys
ss
()
.
10 10
15
then
yt()£ [ ()]Ys
1
and
yt ut e ut
() 10 () 10 ^15 .t()
e. Example (#5)
Let
dy t
dt
yt
()
40 ()
with y(0) = 10.
ANALYTICAL Solution
Taking the LT of the preceding equation
sY s()^104 Y s()^0
Ys s Ys
s
()() , ()
410
10
(^4)
and
yt()£ [ ()]Ys
1
then
yt e ut
() 10 ^4 t()
R.4.114 The MATLAB function F = laplace(f) returns the LT of ƒ, denoted by f(t) → F(s),
where f is a symbolic object with independent variable t.