Gautam Buddha University
Engineering Mathematics-III (MA-201)
Second semester (2022-2023)
Tutorial Sheet-7
Differentiation and Integration for Laplace Transform, Convolution Theorem, Applications of
Laplace Transform
Convolution Theorem: If F (p) = L f (x) and G(p) = L g(x) exist then
Z x
L (f ∗ g)(x) = F (p) · G(p), where f ∗ g (x) = f (x − t)g(t) dt.
0
Q.1 Prove the following differentiation and Integration formulae for Laplace Transform :
′
(a) L f (x) = pL f (x) − f (0). Generalize this result.
Z x L f (x)
(b) L f (t) dt = .
0 p
d
(c) F (p) = −L xf (x) , where F (p) = L f (x) . Generalize this result.
dp
Z ∞
f (x)
(d) F (u) du = L , where F (p) = L f (x) .
p x
Q.2 Using convolution theorem, find the inverse Laplace transform of the following functions:
p 1
(a) ,
(p2 + 1)(p2 + 9) p(p + 1)(p + 2)
1 1
(b) ,
(p2 + 1)2 (p2 + a2 )2
1
(c)
(p4 + 4)
Q.3 (Applications of Laplace Transform to differential and integral equations) Use the Laplace transform to
solve the following equations:
(a) y ′′ + 4y = 4x, y(0) = 1, y ′ (0) = 5.
(b) xy ′′ + (2x + 3)y ′ + (x + 3)y = 3e−x , y(0) = 0.
0, 0≤x<5
(c) 2y ′′ + y ′ + 2y = f (x), y(0) = y ′ (0) = 0; where f (x) = 1, 5 ≤ x < 20
0, x ≥ 20
x
(d) y ′ + 3y + 2 0 y dx = x, y(0) = 0.
R
Rx
(e) y ′ + 4y + 5 0 y dx = e−x , y(0) = 0.
Q.4 (Applications of Laplace Transform to evaluate some improper integrals) Find the values of following
integrals
Z ∞ −2x Z ∞ −αx
e − e−4x e − e−βx
(a) dx. In general, dx
0 x 0 x
Z ∞ −x Z ∞ −αx
e sin x e sin βx
(b) dx. In general, dx.
0 x 0 x
Z ∞ Z ∞ −x 2
e sin x
(c) e−5x x sin x dx, dx
0 0 x
Z ∞ Z ∞
sin x sin αx
(d) dx. In general, dx for α > 0
0 x 0 x
Q.3 (Applications of Laplace Transform to system of differential equations) Use the Laplace transform to
solve the following system:
(a) (D + 1)y1 − y2 = 6 and y1 − (D + 1)y2 = 0 with y1 (0) = 0, y2 (0) = 1.
d
where y1 , y2 are two dependent variables and D ≡ dx .
Q.2 (Miscellaneous)
Z x
3/2 αx sin βx
(a) Find L x L x − [x] and L e dx .
x 0
1 p+α
(b) Find L−1 ln 1 + 2 and L−1 ln
p p+β
p+1 1
(c) Find L−1 cot−1 and L−1 tan−1
2 p2
(d) y ′ + y = 1 + xex , y(0) = 1.
(e) xy ′′ + (1 − x)y ′ + ny = 0, y(0) = 1.
Rx
(f) y ′ + 2 0 y dx = xu1 (x), y(0) = 2.
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