Question 2. Find the general solution of the given ODE for t > 0, t²x" − t(t+2)x' + (t + 2)x = t³e¹. Note that x₁(t) = t solves the homogeneous ODE.
Question 2. Find the general solution of the given ODE for t > 0, t²x" − t(t+2)x' + (t + 2)x = t³e¹. Note that x₁(t) = t solves the homogeneous ODE.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Show that you have reasoned about the problem a way that is correct
![(a) i=√1 and eit = cost + i sin t.
(b) For all integer n ≥ 0 we have:
cos(NT) ) = (−1)”, sin(nx) = 0, sin ((2n + 1)) = (−1)", cos ((2n + 1)
-1))
= 0.
(c) Integration by Parts: fudv = uv-
Svdu.
(d) Quadratic formula: If ar² + br + c = 0 for a 0. Then,
L[sin at]
(e) Laplace Transform: F(s) = fest f(t) dt.
(f) Convolution: (f* g)(t) = f f (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:
and
=
(h) Properties of LT/ILT:
(1) LT/ILT are linear operators.
a
s² + a²
r =
u₁ (t) = -
(k) Change of variables:
-b ± √b² - 4ac
2a
L[cos at]
• Euler equation s = lnt or t=es,
Bernoulli equation: z = x-k
-k
• k-homogeneous equation: z =
(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)] = s¹F(s) — sn−¹ƒ(0) — sn−2 ƒ'(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)n d™ 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:
x₂(t) = x₁(t)
=
S
s² + a²
Te
e-Sp(t)dt
x²(t)
(j) Formula for Variation of Parameters: xp (t) = x₁(t)u₁(t) + x₂(t)u₂(t) when
x₂ (t)g(t)
- / [ ]
W(x1, x₂)(t)
or x = zt.
dt and
and L[t"]=
=
dt
u₂(t) = /[ W
•/
n!
gn+1
x₁ (t)g(t)
W(x1, x₂) (t)
dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd76a2d57-02e6-4734-ba8a-6fadc8c476a5%2F258aed8d-7fc5-4e82-8f8c-072e53643abb%2Fipo4nex_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) ) = (−1)”, sin(nx) = 0, sin ((2n + 1)) = (−1)", cos ((2n + 1)
-1))
= 0.
(c) Integration by Parts: fudv = uv-
Svdu.
(d) Quadratic formula: If ar² + br + c = 0 for a 0. Then,
L[sin at]
(e) Laplace Transform: F(s) = fest f(t) dt.
(f) Convolution: (f* g)(t) = f f (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:
and
=
(h) Properties of LT/ILT:
(1) LT/ILT are linear operators.
a
s² + a²
r =
u₁ (t) = -
(k) Change of variables:
-b ± √b² - 4ac
2a
L[cos at]
• Euler equation s = lnt or t=es,
Bernoulli equation: z = x-k
-k
• k-homogeneous equation: z =
(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)] = s¹F(s) — sn−¹ƒ(0) — sn−2 ƒ'(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)n d™ 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:
x₂(t) = x₁(t)
=
S
s² + a²
Te
e-Sp(t)dt
x²(t)
(j) Formula for Variation of Parameters: xp (t) = x₁(t)u₁(t) + x₂(t)u₂(t) when
x₂ (t)g(t)
- / [ ]
W(x1, x₂)(t)
or x = zt.
dt and
and L[t"]=
=
dt
u₂(t) = /[ W
•/
n!
gn+1
x₁ (t)g(t)
W(x1, x₂) (t)
dt
Transcribed Image Text:Question 2. Find the general solution of the given ODE for t > 0,
t²x" − t(t + 2)x' + (t + 2)x = t³et.
Note that ₁ (t) = t solves the homogeneous ODE.
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