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

2024
Views:

482

Pages:

3
Author:

Shahbaz A.

Laplace Transform x x − ae b −be− a ab(a−b) 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 −x −x b −be a } Find L { aeab(a−b) Solution x x ae− b −be− a ab(a−b) Putting f (x) = in x x R∞ −x −x b −be− a b −be− a L { aeab(a−b) }=F (s)= 0 e−sx [ aeab(a−b) ]dx x x R ∞ −sx ae− xb −be− xa ae− b −be− a L { ab(a−b) } = 0 e [ ab(a−b) ] dx x R ∞ −( 1+bs R ∞ −( 1+as −x )x a )x b −be− a b L { aeab(a−b) } = 0 aeab(a−b) dx − 0 beab(a−b) dx x x R ∞ e−( 1+bs R ∞ −( 1+as )x a )x b ae− b −be− a dx − 0 e a(a−b) dx L { ab(a−b) } = 0 b(a−b) x R ∞ −( 1+bs )x R ∞ −( 1+as )x −x b −be− a 1 1 b a } = b(a−b) e L { aeab(a−b) dx − dx 0 a(a−b) 0 e 1+bs b )x d(e−( 1+bs b )x ) = e−( d(e−( 1+bs b )x −( ) = −( 1+bs b )e b d(e−( − 1+bs 1+bs b )x d[−( 1+bs b )x] ) = e−( 1+bs b )x 1+bs b )x dx dx Similarly a − 1+as d(e−( −x 1+as a )x −x ) = e−( b −be a }= L { aeab(a−b) 1+as a )x RR 1 b −( 1+bs b )x ) [− ]lim R−→∞ 0 d(e b(a−b) 1+bs 1 a − a(a−b) [− 1+as ]limR−→∞ −x RR 0 d(e−( 1+as a )x ) −x b −be a 1 L { aeab(a−b) } = [− (a−b)(1+bs) ]limR−→∞ [ 1 + (a−b)(1+as) limR−→∞ [ −x 1 e( 1+as )x a 1 e( 1+bs )x b ]R 0 ]R 0 −x b −be a 1 } = [− (a−b)(1+bs) ]limR−→∞ [ L { aeab(a−b) 2 1 1+bs e( b )R − 1] 1 + (a−b)(1+as) limR−→∞ [ x x − b −be− a L { aeab(a−b) } −x −x x x −x −x x x 1 e( 1+as )R a − 1] 1 = [− (a−b)(1+bs) ][0 − 1] + b −be a 1 L { aeab(a−b) } = [− (a−b)(1+bs) ][−1] + − b −be− a L { aeab(a−b) } 1 = [ (a−b)(1+bs) ]− b −be a L { aeab(a−b) }= − b −be− a } L { aeab(a−b) 1 1 a−b [ 1+bs − 1 1+as ] 1 1+as−1−bs a−b [ (1+bs)(1+as) ] b −be a L { aeab(a−b) }= (a−b)s 1 a−b [ (1+bs)(1+as) ] −x x x − b −be− a L { aeab(a−b) } = 1 (a−b)(1+as) [−1] 1 (a−b)(1+as) = −x 1 (a−b)(1+as) [0 s (1+bs)(1+as) 3 − 1]