Page 111 - Advanced engineering mathematics

3.3 Shifting and the Heaviside Function 91

First apply L to the differential equation, using equations (3.1) and (3.3):
2

L[y ]+ 4L[y]=[s − sy(0) − y (0)]Y(s) + 4Y(s)
2
2
= s Y(s) + 4Y(s) = (s + 4)Y(s) = L[ f ].
To compute L[ f ], use the second shifting theorem. Since f (t) = H(t − 3)t, we can write
L[ f ]= L[H(t − 3)t]
= L[H(t − 3)(t − 3 + 3)]
= L[H(t − 3)(t − 3)]+ 3L[H(t − 3)]
e −3s 3e −3s
= + .
s 2 s
In summary, we have
1 3 3s + 1
2 −3s −3s −3s
(s + 4)Y(s) = e + e = e .
s 2 s s 2
The transform of the solution is therefore
3s + 1
Y(s) = e −3s .
2
s (s + 4)
2
The solution is the inverse transform of Y(s). To take this inverse, use a partial fractions
decomposition, writing
3s + 1 A B Cs + D
= + + .
2
2
s (s + 4) s s 2 s + 4
2
After solving for A, B,C, and D, we obtain
3s + 1
Y(s) = e −3s
2
2
s (s + 4)
3 1 3 s 1 1 1 1
= e −3s − e −3s + e −3s − e −3s .
2
2
4 s 4 s + 4 4 s 2 4 s + 4
Now apply the second shifting theorem to write the solution
3 3
y(t) = H(t − 3) − H(t − 3)cos(2(t − 3))
4 4
1 1
+ H(t − 3)(t − 3) − H(t − 3)sin(2(t − 3)).
4 8
This solution is 0 until time t = 3. Since H(t − 3) = 1for t ≥ 3, then for these times,
3 3 1
y(t) = − cos(2(t − 3)) + (t − 3)
4 4 4
1
− sin(2(t − 3)).
8
Upon combining terms, the solution is

0 for t < 3
y(t) = 1
8 [2t − 6cos(2(t − 3)) − sin(2(t − 3))] for t ≥ 3.
Figure 3.16 shows part of the graph of this solution.

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October 14, 2010 14:14 THM/NEIL Page-91 27410_03_ch03_p77-120