Question 5. Apply Laplace Transform to solve the following Differential Equation when x(0) = x'(0) = 0 and x" + x = f(t) where f is given as 0 ƒ (t) = { i-₁ in in 0 < t < 1, t>1
Question 5. Apply Laplace Transform to solve the following Differential Equation when x(0) = x'(0) = 0 and x" + x = f(t) where f is given as 0 ƒ (t) = { i-₁ in in 0 < t < 1, t>1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Question 5. Apply Laplace Transform to solve the following Differential Equation when x(0) = x'(0) = 0 and
x" + x = f(t)
where f is given as
f(t)
=
{}
t-1
in 0 < t < 1,
in
t>1
![(a) i = √-1 and eit = cost + i sin t.
(b) For all integer n ≥ 0 we have:
cos(nx) = (−1)ª, sin(nx) = 0, sin ((2n + 1)2) = (−1)”, cos ((2n
(2n +1)
- 1977)
= 0.
(c) Integration by Parts: fudv=uv - fvdu.
(d) Quadratic formula: If ar² + br+c=0 for a 0. Then,
(e) Laplace Transform: F(s) = f est f(t)dt.
(f) Convolution: (f* g)(t) = fő ƒ (0)g(t — 0)d0. Note (ƒ * g)(t) = (g * f)(t).
(g) LT of f(t) = sin at, f(t) = cos at and f(t) = t" for n ≥ 0:
S
L[sin at]
and L[cos at]
s² + a²
(h) Properties of LT/ILT:
u₁ (t) =
a
s² + a²
==
r =
(k) Change of variables:
-b ± √b² - 4ac
2a
(1) LT/ILT are linear operators.
(2) Shift/Scaling: L[eat f(t)] | = F(s − a) for constant a € R and L[f(at)] = F() for constant a > 0.
(3) LT of derivatives: L[f(n) (t)] = sn F (s) — :
· sn-¹ƒ(0) — sn-2 f'(0) - ... - fn-¹(0) for n ≥ 1.
● LT of integrals: L [Ső f(0)d0] = F(s) when f(0) = 0.
(4) Derivative of LT or LT of multiplication by polynomials: L[t" f(t)] = (-1)" de F(s) for n ≥ 1.
(5) Unit Step Function: L[He(t)f(t — c)] = F(s)e-sc.
(6) Convolution Property: L[(f* g)(t)] = F(s)G(s).
(i) Formula for Reduction of Order:
/ [₁
x₂ (t) = x₁(t)
√ [²
(j) Formula for Variation of Parameters: xp(t) = x₁(t)u₁(t) + x₂(t)u₂(t) when
x₂(t)g(t)
dt
= /[ W
e-Sp(t)dt
x²(t)
and L[t”] :
and u₂(t) =
• Euler equation s = lnt or t = e³,
• Bernoulli equation: z = x-k,
• k-homogeneous equation: zor x = zt.
dt
n!
gn+1
x₁
W(x1, x2)(t)
dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd76a2d57-02e6-4734-ba8a-6fadc8c476a5%2F6bcedb34-87de-45b0-bdbf-67a11bd5bf79%2Ftrp0b52_processed.png&w=3840&q=75)
Transcribed Image Text:(a) i = √-1 and eit = cost + i sin t.
(b) For all integer n ≥ 0 we have:
cos(nx) = (−1)ª, sin(nx) = 0, sin ((2n + 1)2) = (−1)”, cos ((2n
(2n +1)
- 1977)
= 0.
(c) Integration by Parts: fudv=uv - fvdu.
(d) Quadratic formula: If ar² + br+c=0 for a 0. Then,
(e) Laplace Transform: F(s) = f est f(t)dt.
(f) Convolution: (f* g)(t) = fő ƒ (0)g(t — 0)d0. Note (ƒ * g)(t) = (g * f)(t).
(g) LT of f(t) = sin at, f(t) = cos at and f(t) = t" for n ≥ 0:
S
L[sin at]
and L[cos at]
s² + a²
(h) Properties of LT/ILT:
u₁ (t) =
a
s² + a²
==
r =
(k) Change of variables:
-b ± √b² - 4ac
2a
(1) LT/ILT are linear operators.
(2) Shift/Scaling: L[eat f(t)] | = F(s − a) for constant a € R and L[f(at)] = F() for constant a > 0.
(3) LT of derivatives: L[f(n) (t)] = sn F (s) — :
· sn-¹ƒ(0) — sn-2 f'(0) - ... - fn-¹(0) for n ≥ 1.
● LT of integrals: L [Ső f(0)d0] = F(s) when f(0) = 0.
(4) Derivative of LT or LT of multiplication by polynomials: L[t" f(t)] = (-1)" de F(s) for n ≥ 1.
(5) Unit Step Function: L[He(t)f(t — c)] = F(s)e-sc.
(6) Convolution Property: L[(f* g)(t)] = F(s)G(s).
(i) Formula for Reduction of Order:
/ [₁
x₂ (t) = x₁(t)
√ [²
(j) Formula for Variation of Parameters: xp(t) = x₁(t)u₁(t) + x₂(t)u₂(t) when
x₂(t)g(t)
dt
= /[ W
e-Sp(t)dt
x²(t)
and L[t”] :
and u₂(t) =
• Euler equation s = lnt or t = e³,
• Bernoulli equation: z = x-k,
• k-homogeneous equation: zor x = zt.
dt
n!
gn+1
x₁
W(x1, x2)(t)
dt
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