Question 4. Apply Laplace Transform to solve the following Integro-Differential Equation when x(0) = 1 and - 1/2 √² (1 - 0) ² x (0) do = t 0 x' (t) -
Question 4. Apply Laplace Transform to solve the following Integro-Differential Equation when x(0) = 1 and - 1/2 √² (1 - 0) ² x (0) do = t 0 x' (t) -
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 4. Apply Laplace Transform to solve the following Integro-Differential Equation when x(0) = 1 and
1
x' (t) — ½ √ √ ² (t – 0)²³x(0) d0 = −t
2
![Some (not necessarily all) of the following formulas and definitions might be helpful for the final.
(a) i=√1 and eit = cost + i sin t.
(b) For all integer n ≥ 0 we have:
cos(NT) = (-1)", sin(n) = 0, sin ((2n +1)
(c) Integration by Parts: fudv=uv - fvdu.
(d) Quadratic formula: If ar² + br + c = 0 for a 0. Then,
a
s² + a²
(1) LT/ILT are linear operators.
(e) Laplace Transform: F(s) = f est f(t)dt.
(f) Convolution: (f * g) (t) = f f(0)g(t – 0)d0. Note (f * g)(t) = (g* f) (t).
(g) LT of f(t) = sin at, f(t) = cos at and f(t) = tn for n ≥ 0:
L[sin at]
and L[cos at]
(h) Properties of LT/ILT:
r =
1)) = (-1)", cos((2n + 1)
==
-S[W
-b ± √b² - 4ac
2a
x₂ (t) = x₁(t)
1/[²
S
s² + a²
(2) Shift/Scaling: L[eat f(t)] = F(s − a) for constant a R and L[ƒ(at)] = ¹ F(2) for constant a > 0.
(3) LT of derivatives: L[f(n) (t)] = snF(s) — sn-¹f(0) — sn-2 f'(0)
fn-1 (0) for n ≥ 1.
● LT of integrals: L [ f(0)d0] = F(s) when f(0) = 0.
S
(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[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:
• Euler equation s = lnt or t= es,
• Bernoulli equation: z = x-k,
• k-homogeneous equation: z = ½ or x = zt.
and L[t"]
e-S p(t)dt
x²(t)
dt
- ¹977) · = 0.
--
=
(j) Formula for Variation of Parameters: xp(t) = x₁(t)u₁(t) + x₂(t)u₂ (t) when
x₂(t)g(t)
x₁ (t)g(t)
u₁(t):
dt and
W(x₁, x₂)(t)]
*
u₂(t) = =
7(x1, x2)(t)]
(k) Change of variables:
n!
gn+1
dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd76a2d57-02e6-4734-ba8a-6fadc8c476a5%2Fee747c4b-4266-4402-88f0-3f905f8c3054%2Fo80dah_processed.png&w=3840&q=75)
Transcribed Image Text:Some (not necessarily all) of the following formulas and definitions might be helpful for the final.
(a) i=√1 and eit = cost + i sin t.
(b) For all integer n ≥ 0 we have:
cos(NT) = (-1)", sin(n) = 0, sin ((2n +1)
(c) Integration by Parts: fudv=uv - fvdu.
(d) Quadratic formula: If ar² + br + c = 0 for a 0. Then,
a
s² + a²
(1) LT/ILT are linear operators.
(e) Laplace Transform: F(s) = f est f(t)dt.
(f) Convolution: (f * g) (t) = f f(0)g(t – 0)d0. Note (f * g)(t) = (g* f) (t).
(g) LT of f(t) = sin at, f(t) = cos at and f(t) = tn for n ≥ 0:
L[sin at]
and L[cos at]
(h) Properties of LT/ILT:
r =
1)) = (-1)", cos((2n + 1)
==
-S[W
-b ± √b² - 4ac
2a
x₂ (t) = x₁(t)
1/[²
S
s² + a²
(2) Shift/Scaling: L[eat f(t)] = F(s − a) for constant a R and L[ƒ(at)] = ¹ F(2) for constant a > 0.
(3) LT of derivatives: L[f(n) (t)] = snF(s) — sn-¹f(0) — sn-2 f'(0)
fn-1 (0) for n ≥ 1.
● LT of integrals: L [ f(0)d0] = F(s) when f(0) = 0.
S
(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[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:
• Euler equation s = lnt or t= es,
• Bernoulli equation: z = x-k,
• k-homogeneous equation: z = ½ or x = zt.
and L[t"]
e-S p(t)dt
x²(t)
dt
- ¹977) · = 0.
--
=
(j) Formula for Variation of Parameters: xp(t) = x₁(t)u₁(t) + x₂(t)u₂ (t) when
x₂(t)g(t)
x₁ (t)g(t)
u₁(t):
dt and
W(x₁, x₂)(t)]
*
u₂(t) = =
7(x1, x2)(t)]
(k) Change of variables:
n!
gn+1
dt
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