Laplace Transforms: Example Important Solved Problems with formula, steps, derivation and answer based on Transforms of Derivatives and Integrals
TRANSFORMS OF
DERIVATIVES AND INTEGRALS
Result :
Prove that L[f '' (t)] = s L[ f(t)] ‒ f(0)
Proof:
We
know that,

= ‒f(0)
+ sL[f(t)] = s L [ft] ‒ f(0)
Result :
Prove that L [f '' (t)] = s2
L [f(t)] − sf(0) – f '(0)
Proof:
We know that,

=
[0 ‒ f '(0)] + s 0ʃ∞e‒st
f ' (t) dt
=
‒f ' (0) + s L [f ' (t)]
=
‒f '(0) + s[sL [f(t)] ‒ [f(0)] by result
(15)
L [f ''(t)] = s2L[f(t)] − s f (0) ‒ f'(0)
Note:
(15)
L[F(n)(t)] = sn L[f(t)]
‒ sn‒1f(0) ‒ sn‒2 f '(0) ‒ ... ‒f(n‒1)(0)
Result :
If L[f(t)] = F(s), then L [ 0ʃt f(u) du ] = 1/s L[f(t)]

Proof:
Given:
L [f(t)] = F(s)
Let
ϕ(t) = 0ʃt f(u)
du,
then ϕ'(t) = f(t) and ϕ(0) = 0
….(1)
L
[ϕ'(t)] = sL [ϕ(t)] ‒ ϕ(t) = sL[ϕ(t)]
L [ϕ(t)] = 1/s L [ϕ'(t)]

Transforms and its Applications: UNIT 1: Laplace Transforms : Tag: Engineering mathematics, Maths : Laplace Transforms - Transforms of Derivatives and Integrals
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