Laplace Transform of sinax+xcosax
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)}
Also L {x sin ax} =
L {x cos ax} =
2as
(s2 +a2 )2
s2 −a2
(s2 +a2 )2
(s > 0)
(s > 0)}
L {sin ax} =
a
s2 +a2
(s > 0)
L {cos ax} =
s
s2 +a2
(s > 0)
Find L {sin ax + x cos ax}
Solution
Using the formula
2 L {c1 f (x) + c2 g(x)} = c1 L {f (x)} + c2 L {g(x)}
Putting
c1 = c2 = 1, f (x) = sinax, g(x) = x cos ax
L {sinax + x cos ax} = L {sinax} + L {x cos ax}
Putting
L {sinax} =
a
s2 +a2
L {x cos ax} =
s2 −a2
(s2 +a2 )2
(s > 0)}
s2 −a2
(s2 +a2 )2
L {sinax + x cos ax} =
s2 +a2
L {sinax + x cos ax} =
a(s2 +a2 )+s2 −a2
(s2 +a2 )2
L {sinax + x cos ax} =
s2 (a+1)+a2 (a−1)
(s2 +a2 )2
a
+
Applications
Find area under the curve
y = e−4x (sin2x + x cos 2x)
Solution
Here
s = 4, a = 2
2
Area= 4
(2+1)+22 (2−1)
(42 +22 )2
=
13
100
3
Laplace Transform of sin(ax)+xcos(ax)
of 3
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