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

2024
Views:

476

Pages:

3
Author:

Shahbaz A.

Laplace Transform of ebx sin ax + ebx cos ax shahbaz ahmed June 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 Formulas used Prove that If L {f (x)} = F (s), L {g(x)} = G(s), then for any two constants c1 and c2 L {c1 f (x) + c2 g(x)} = c1 L {f (x)} + c2 L {g(x)} Proof L {c1 f (x) + c2 g(x)} = R∞ e−sx [c1 f (x) + c2 g(x)] dx R∞ R∞ L {c1 f (x) + c2 g(x)} = 0 e−sx [c1 f (x)] dx + 0 e−sx [c2 g(x)] dx R∞ R∞ L {c1 f (x) + c2 g(x)} = c1 0 e−sx f (x) dx + c2 0 e−sx g(x) dx 0 L {c1 f (x) + c2 g(x)} = c1 L {f (x)} + c2 L {g(x)} Putting c1 = c2 = 1 L {f (x) + g(x)} = L {f (x)} + L {g(x)} L {ebx sin ax} = a (s−b)2 +a2 L {ebx cos ax} = s−b (s−b)2 +a2 Find L {ebx sin ax + ebx cos ax} Solution Since 2 L {f (x) + g(x)} = L {f (x)} + L {g(x)} Putting f (x) = ebx sin ax, g(x) = ebx cos ax L {ebx sin ax + ebx cos ax} = L {ebx sin ax} + L {ebx cos ax} Putting L {ebx sin ax} = a (s−b)2 +a2 L {ebx cos ax} = s−b (s−b)2 +a2 L {ebx sin ax + ebx cos ax} = a (s−b)2 +a2 L {ebx sin ax + ebx cos ax} = s+a−b (s−b)2 +a2 3 + s−b (s−b)2 +a2

Laplace Transform of e^(bx) sin(ax)+e^(bx) cos(ax)

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