University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2024
Views:

72

Pages:

2
Author:

Shahbaz A.

L { a12 (eax − 1 − ax)} shahbaz ahmed July 2024 Introduction to Laplace Transform Let f(x) be defined for 0 ≤ x < ∞ and let s denote an arbitrary real variable. The Laplace transform of f (x), designated by either L or F (s), is R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx for all values of s for which the improper integral converges. Convergence occur when the limit RR limR−→∞ 0 e−sx f (x)dx exists. 1 1 Formulas used L {1} = F (s) = 1s (s > 0) L {x} = F (s) = 1 s2 (s L {eax } = F (s) = > 0) 1 s−a (s > 0) L {[c1 f (x) + c2 g(x)]} = c1 L {f (x)} + c2 L {g(x)} Find L { a12 (eax − 1 − ax)} Solution Using the formula L {[c1 f (x) + c2 g(x)]} = c1 L {f (x)} + c2 L {g(x)} L { a12 (eax − 1 − ax)} = 1 ax a2 L {(e − 1 − ax)} L { a12 (eax − 1 − ax)} = 1 ax a2 L {e } − L { a12 (eax − 1 − ax)} = 1 1 a2 s−a L { a12 (eax − 1 − ax)} = s2 −(s−a)s−a(s−a) a2 (s−a)s2 L { a12 (eax − 1 − ax)} = s2 −s2 +as−as+a2 a2 (s−a)s2 L { a12 (eax − 1 − ax)} = a2 (s−a)s2 L { a12 (eax − 1 − ax)} = 1 s2 (s−a) − 1 1 a2 s a2 2 1 a2 L {1} − 1 1 a s2 − a a2 L {x}