Lecture Notes in Differential Equations - Bruce E. Shapiro

337
apply this result iteratively. For example,
] d
L[f ′′ (t)] = L[
dt f ′ (t)
(32.36)
= −f ′ (0) + sL[f ′ (t)] (32.37)
= −f ′ (0) + s(−f(0) + sF (s)) (32.38)
= −f ′ (0) − sf(0) + s 2 F (s) (32.39)
Theorem 32.8. Let f, f ′ , . . . , f (n−1) be continuous on [0, ∞), of exponential
order, and suppose that f (n) (t) is piecewise continuous on [0, ∞).
Then
[ ]
L f (n) (t) = s n F (s) − s n−1 f(0) − s n−2 f ′ (0) − · · · − f (n−1) (0) (32.40)
where F (s) = L[f(t)].
Proof. This can be proven by mathematical induction. We have already
proven the cases for n = 1 and n = 2. As an inductive hypothesis we will
assume (32.40) is true for general n ≥ 1. We must prove that
[ ]
L f (n+1) (t) = s n+1 F (s) − s n f(0) − s n−1 f ′ (0) − · · · − f (n) (0) (32.41)
follows directly from (32.40).
Let g(t) = f (n) . Then g ′ (t) = f (n+1) (t) by definition of derivative notation.
Hence
[ ]
L f (n+1) (t) = L[g ′ (t)] (32.42)
Substitution of (32.40) yields (32.41) as required.
= sG(s) − g(0) (32.43)
[ ]
= sL f (n) (t) − f (n+1) (0) (32.44)