Consider the following boundary value problem (E) : h + u = t2, r > 0, t>0 (1) (2) (3) u(r, 0) - cha, r>0 u(0, t) – 0, t>0 and suppose that ach(r) – sh(x) I = [*ch(2}dz = «ch{z) - sh(2), a +c. - 1 (1) By applying the Laplace transform to equation (1) of (E) (acting to the variable t) and by using equation (2) of (E), we obtain the following ODE: 3 a. Uz(r, s) + sU(r, s) = -cha – 2 b. U-(r, s) + sU(r, s) = -cha +- 2 c. Uz(r, s) + sU(r, s) = chx + d. None of the above (2) Using equation (3) of (E), the solution of the ODE obtained in part (2) is: shz s2 - 1 2 2 a. U(r, s) = chr- s2 - 1 s4 chr b. U(r, s) = shr 1 82 1 shr chr c. U(r,s) + 1 d. None of the above (3) The general solution of (E) is: (H(t – a) is the unit step function) a. u(z, t) = t³ + (t – x)®H(t – x) + cos(t – æ)H(t – x) b. u(r, t) = cha cost – shr sint + (t – r)*H(t – r) c. u(r, t) = cha cost – shr sint + t -(t- 2)*H(t – r) – cos(t – 2)H(t – r) d. None of the above
Consider the following boundary value problem (E) : h + u = t2, r > 0, t>0 (1) (2) (3) u(r, 0) - cha, r>0 u(0, t) – 0, t>0 and suppose that ach(r) – sh(x) I = [*ch(2}dz = «ch{z) - sh(2), a +c. - 1 (1) By applying the Laplace transform to equation (1) of (E) (acting to the variable t) and by using equation (2) of (E), we obtain the following ODE: 3 a. Uz(r, s) + sU(r, s) = -cha – 2 b. U-(r, s) + sU(r, s) = -cha +- 2 c. Uz(r, s) + sU(r, s) = chx + d. None of the above (2) Using equation (3) of (E), the solution of the ODE obtained in part (2) is: shz s2 - 1 2 2 a. U(r, s) = chr- s2 - 1 s4 chr b. U(r, s) = shr 1 82 1 shr chr c. U(r,s) + 1 d. None of the above (3) The general solution of (E) is: (H(t – a) is the unit step function) a. u(z, t) = t³ + (t – x)®H(t – x) + cos(t – æ)H(t – x) b. u(r, t) = cha cost – shr sint + (t – r)*H(t – r) c. u(r, t) = cha cost – shr sint + t -(t- 2)*H(t – r) – cos(t – 2)H(t – r) d. None of the above
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
laplace part 2
Transcribed Image Text:Consider the following boundary value problem (E) :
h + u = t2, r > 0, t>0 (1)
(2)
(3)
u(r, 0) - cha, r>0
u(0, t) – 0,
t>0
and suppose that
ach(r) – sh(x)
I = [*ch(2}dz = «ch{z) - sh(2),
a +c.
- 1
(1) By applying the Laplace transform to equation (1) of (E) (acting to the variable t) and by
using equation (2) of (E), we obtain the following ODE:
3
a. Uz(r, s) + sU(r, s) = -cha –
2
b. U-(r, s) + sU(r, s) = -cha +-
2
c. Uz(r, s) + sU(r, s) = chx +
d. None of the above
(2) Using equation (3) of (E), the solution of the ODE obtained in part (2) is:
shz
s2 - 1
2
2
a. U(r, s) = chr-
s2 - 1
s4
chr
b. U(r, s) =
shr
1
82
1
shr
chr
c. U(r,s)
+
1
d. None of the above
(3) The general solution of (E) is: (H(t – a) is the unit step function)
a. u(z, t) = t³ + (t – x)®H(t – x) + cos(t – æ)H(t – x)
b. u(r, t) = cha cost – shr sint + (t – r)*H(t – r)
c. u(r, t) = cha cost – shr sint + t -(t- 2)*H(t – r) – cos(t – 2)H(t – r)
d. None of the above
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