Lecture Notes in Differential Equations - Bruce E. Shapiro

343
Derivatives of the Laplace Transform
Now let us see what happens when we differentiate the Laplace Transform.
Let F (s) be the transform of f(t).
d
ds F (s) = d ds
=
∫ ∞
0
∫ ∞
= −
0
∫ ∞
0
e −st f(t)dt (32.95)
d
ds e−st f(t)dt (32.96)
e −st tf(t)dt (32.97)
= −L[tf(t)] (32.98)
By a similar argument, we get additional factors of −t for each order of
derivative, so in general we have
Thus we have the results
F (n) (s) = d(n)
ds (n) L[f(t)] = (−1)n L[t n f(t)] (32.99)
Example 32.12. Find F (s) for f(t) = t cos kt.
L[tf(t)] = F ′ (s) (32.100)
L[t n f(t)] = (−1) n F (n)(s) (32.101)
From (32.19) we have L[cos kt] = s/(s 2 + k 2 ). Hence
L[t cos kt] = d L[cos kt]
ds
(32.102)
= − d s
ds s 2 + k 2 (32.103)
= − (s2 + k 2 )(1) − s(2s)
(s 2 + k 2 ) 2 (32.104)
= s2 − k 2
(s 2 + k 2 ) 2 (32.105)
Step Functions and Translations in the Time
Variable
Step functions are special cases of piecewise continuous functions, where the
function “steps” between two different constant values like the treads on