Page 103 - Advanced engineering mathematics

3.2 Solution of Initial Value Problems 83

There is no derivative in this equation! L has converted the differential equation into an algebraic
equation for the transform Y(s) of the unknown function y(t). Solve for Y(s) to obtain
1 1
Y(s) = + .
s − 4 s(s − 4)
This is the transform of the solution of the initial value problem. The solution is y(t), which we
obtain by applying the inverse transform:
−1
y = L [Y]

1 1
−1 −1
= L + L .
s − 4 s(s − 4)
From entry (3) of the table with a = 4,
1

−1 4t
L = e
s − 4
and from entry (5) with a = 0 and b = 4,

1 1
−1 0t 4t
L = (e − e )
s(s − 4) −4
1
4t
= (e − 1).
4
The solution is
1 5 1
y(t) = e + (e − 1) = e − .
4t
4t
4t
4 4 4
EXAMPLE 3.3
Solve
t
y + 4y + 3y = e ; y(0) = 0, y (0) = 2.

Using the linearity of L and equations (3.1) and (3.3), we obtain

L[y ]+ 4L[y ]+ 3L[y]
2
=[s Y − sy(0) − y (0)]+ 4[sY − y(0)]+ 3Y

2
=[s Y − 2]− 4sY + 3Y
1
t
= L[e ]= .
s − 1
Solve for Y to get
2s − 1
Y(s) =
2
(s − 1)(s + 4s + 3)
2s − 1
= .
(s − 1)(s + 1)(s + 3)
To read the inverse transform from the table, use a partial fractions decomposition to write
the quotient on the right as a sum of simpler quotients. We will carry out the algebra of this
decomposition. First write
2s − 1 A B C
= + + .
(s − 1)(s + 1)(s + 3) s − 1 s + 1 s + 3

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October 14, 2010 14:14 THM/NEIL Page-83 27410_03_ch03_p77-120