Question 3. Apply Laplace Transform to solve the following Integral Equation + 2√² e²-² x(t) = et +2 et-x(0) do
Question 3. Apply Laplace Transform to solve the following Integral Equation + 2√² e²-² x(t) = et +2 et-x(0) do
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 3. Apply Laplace Transform to solve the following Integral Equation
2 fr e²-
x(t) = et + 2
et-ox(0)do
![(a) i = √-1 and eit = cost + i sin t.
(b) For all integer n ≥ 0 we have:
cos(NT)
(c) Integration by Parts: fudv=uv - -Svdu.
(d) Quadratic formula: If ar² + br+c=0 for a 0. Then,
-b ± √b² - 4ac
2a
=
(-1)", sin(n)
L[sin at]
(e) Laplace Transform: F(s) =
e st f(t) dt.
(f) Convolution: (f * g)(t) = ff 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) = tn for n ≥ 0:
=
(h) Properties of LT/ILT:
) = 0, sin ((2n + 1) 7) = (−1)”, cos (
; ((2n +1) 7/2) =
= 0.
a
s² + a²
(1) LT/ILT are linear operators.
u₁(t): ==
(k) Change of variables:
T =
and L[cos at]
(j) Formula for Variation of Parameters: xp(t) =
- / [w
x2 (t)g(t)
W(x₁, x₂)(t)]
dt
(2) Shift/Scaling: L[eat f(t)] = F(s — a) for constant a R and L[f(at)] = F(²) for constant a > 0.
(3) LT of derivatives: L[f(n) (t)] = s¹ F(s) — sn−¹ƒ(0) — sn-2 f'(0) — ... — fn−¹(0) for n ≥ 1.
• LT of integrals: L [ f(0)d0] = F(s) when ƒ(0) = 0.
(4) Derivative of LT or LT of multiplication by polynomials: L[t" f(t)] = (−1)nd F(s) for n ≥ 1.
(5) Unit Step Function: L[H₂(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:
x₂(t) = x₁(t)
• Euler equation s = lnt or t = es,
• Bernoulli equation: z = x-k,
• k-homogeneous equation: z =
=
S
s² + a²
√ [²
x₁(t)u₁(t) + x2(t)u₂(t) when
x₁ (t)g(t)
or x = zt.
e-S p(t)dt
and L[t"]
x² (t)
and u₂(t) =
n!
sn+1
dt
- / [W ()] de
dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd76a2d57-02e6-4734-ba8a-6fadc8c476a5%2F9a219e77-4ac4-4b26-8b89-37a774d57782%2Fr4sptb_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(NT)
(c) Integration by Parts: fudv=uv - -Svdu.
(d) Quadratic formula: If ar² + br+c=0 for a 0. Then,
-b ± √b² - 4ac
2a
=
(-1)", sin(n)
L[sin at]
(e) Laplace Transform: F(s) =
e st f(t) dt.
(f) Convolution: (f * g)(t) = ff 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) = tn for n ≥ 0:
=
(h) Properties of LT/ILT:
) = 0, sin ((2n + 1) 7) = (−1)”, cos (
; ((2n +1) 7/2) =
= 0.
a
s² + a²
(1) LT/ILT are linear operators.
u₁(t): ==
(k) Change of variables:
T =
and L[cos at]
(j) Formula for Variation of Parameters: xp(t) =
- / [w
x2 (t)g(t)
W(x₁, x₂)(t)]
dt
(2) Shift/Scaling: L[eat f(t)] = F(s — a) for constant a R and L[f(at)] = F(²) for constant a > 0.
(3) LT of derivatives: L[f(n) (t)] = s¹ F(s) — sn−¹ƒ(0) — sn-2 f'(0) — ... — fn−¹(0) for n ≥ 1.
• LT of integrals: L [ f(0)d0] = F(s) when ƒ(0) = 0.
(4) Derivative of LT or LT of multiplication by polynomials: L[t" f(t)] = (−1)nd F(s) for n ≥ 1.
(5) Unit Step Function: L[H₂(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:
x₂(t) = x₁(t)
• Euler equation s = lnt or t = es,
• Bernoulli equation: z = x-k,
• k-homogeneous equation: z =
=
S
s² + a²
√ [²
x₁(t)u₁(t) + x2(t)u₂(t) when
x₁ (t)g(t)
or x = zt.
e-S p(t)dt
and L[t"]
x² (t)
and u₂(t) =
n!
sn+1
dt
- / [W ()] de
dt
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