PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

362 Practical MATLAB® Applications for Engineers

Ys s()[ ]

(^2) 
48
Ys
ss
()


8
4
4
2
(^222)  (^2)
and using Table 1.2
yt()
£ [ ()]Ys
 1
yt( )
^42 sinh( ) ( )tut
c. Example (#3)
Let
dyt
dt
yt
2
2 9

()

 ()

with the IC given by

y

dy t

dt t

()

()

00 15

0

and

ANALYTICAL Solution

Taking the LT of the given equation

sYs sy

dy t

dt

Ys

t

2

0

() ( ) 09

()

 ()

sYs sy Ys

(^2) ()( )015 9
()
Ys s()( )
(^2) 
915
Ys
ss
()





15
9
5
3
(^2223)




then
yt()
£ [ ()]Ys
 1
yt()
^53 sin( ) ()tut
d. Example (#4)
Let
2330
dy t
dt
yt

()


()

with the IC given by y(0) = 0.