Page 233 - Engineering Mathematics Workbook_Final

Laplace Transforms

19. Let ‘y’ be the solution of If ( ) S = 2 then Lt f t =
( )
2
d y = 6cos2t , ( ) 0 = 3, ( ) 0 = 1 23. F S (1 S ) x→ 0
+
1
y
y
dt 2 __________ where ( L f t F ( ) S
( )) =
. Let laplace transform of y be F(S),
the value of F(1) = (a) 0 (b) 1
17 13 (c) 2 (d) 
(a) (b)


5 5  sin , t 0 t 

f
24. If ( ) t =  then
26 11  0,   t  2 
(c) (d)
( )) =
5 5 ( L f t
20. The solution of initial value problem 1
1
y + 2y + 10y = 6 ( ) t , ( ) 0 = , (a) 1 e −  s ( s + 2 ) 1
11
0
y
−
f
0
y 1 ( ) 0 = where ( ) t is direct delta 1
function is (b) − 
+
1 e s ( s + 2 ) 1
t
t
e
e
(a) 2 sin3t (b) 6 sin3t
+
1 e −  s
(c) 2e t − sin3t (d) 6e t − sin3t (c) s + 1
2
−
21. Consider the initial value problems (d) 1 e −  s
2
=
1
y + 2y + y  ( ) t with s + 1
11
dy LAPLACE TRANSFORM
y ( ) t = − 2and = 0. The
t=
0
dt t= 0 USING MAIN DEFINITION
dy  1
numerical value of is  , where  0 t  k

dt t= 0 + 25. If ( ) t =  k then
f


(a) -2 (b) -1  1 when t  k
( L f t
( )) =
(c) 0 (d) 1
−
1 e − ks
5s + 2 23s + 6 (a)
22. If ( ) f = L F ( ) S = ks 2
( s s + 2 2s + ) 2 − ks
+
( )
then Lt f t = (b) 1 e
x→ 0 ks 2
(a) 3 (b) 5 k − e − ks
(c) 2
17 s
(c) (d) 0 −
2 1 + (k − ) 1 e ks
(d)
ks

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