Page 81 - LAPLACE TRANSFORM

Example 3

2 + 2 = 24 0( =y ) , 10 y ' ) 0 ( =

Solve the differential equation + 3 0
2

STEP 1: Take the Laplace Transform for both sides.
24
' '
'
2
) (t +
3 yL
) (t y
=
y
L [ ] [ ] [ ] [ ]
) (t +
L

STEP 2: Substitude the theorem for Left Hand Side (LHS) .
Find the Laplace for Right Hand Side (RHS)

24
] 2
(
0
(
] [sY(3
)
(
0
'
[ Ys 2 ( s) − sy ) − y 0 − s) − y ) + [Y( s) ] =

s
’
STEP 3 : Substitude the values of y(0)=10, y (0)=0

)
)
'
(
0
] 2
] [sY(3
(
(
[ Ys 2 ( s) − s 10 − y 0 − s) − y ) + [Y( s) ] = 24
s
STEP 4 : Solve for Y(s)
24
s 2 Y (s ) − 10 − 0 − 3sY (s ) − 30 + 2Y (s ) = Partial
s
s fraction 1
24
Y (s )[s 2 + s ] 2 = + 10 + 30
s
3 +
s
+
24 10s 30
Y (s ) = +
s
)( +
s
)( +
s ( + 1 s ) 2 ( + 1 s ) 2
−
STEP 5 : Solve for the inverse , ( ) = [ ( )] using Partial Fraction.

24 A B C
= + +
s
s
s ( + 1 )( + ) 2 s ( + ) 1 ( + ) 2
s
s
24 = A (s + 1 )(s + ) 2 + B (s )(S + ) 2 + C ( )(ss + 1 ).....eq 1

s=-2
If s=0 if
24 = A 0 ( + 1 )( 0 + ) 2 + ) 0 ( B + C ) 0 ( 24 = )0()0 ( A + B + C (− 2 )(− 2 + ) 1
24 = ) 2 ( A 24 = C ) 2 (
24 24
A = = 12 C = = 12
2 2

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