Consider the following boundary value problem (E) : u a2 0<2< 1, t>0.. (1) u(0, t) – 0, u(1, t) – 0, t > 0.. (2) u(x, 0) = f(x), 0 < r<1…(3), ue(1, 0) – 2, 0 4 b. A1 = 0,42 = -2, A3 = 3 and A, = 0, Vn > 4 c. An = 0, Vn 21 d. None of the above (3) By using the initial condition (4), we obtain that 2(–1)"+1 a. B Vn 21 %3D b. B Vn 21 с. В, — 0. d. None of the above d.

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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part 3 laplace pde
Consider the following boundary value problem (E) :
u
a2 0<2< 1, t>0.. (1)
u(0, t) – 0, u(1, t) – 0, t > 0.. (2)
u(x, 0) = f(x), 0 < r<1…(3),
ue(1, 0) – 2, 0 <r <1…(4),
where f(r) = 3 sin(wx) – 2 sin(3rz).
(1) By using the method of separation of variables, the family of solutions un(r, t) for equations
(1) and (2) is:
a. Un(x, t) = [An cos(nat) + B, sin(nxt)] cos (na2)
b. Un(r, t) – [An cos(nat) + Bn sin(nat)] sin(nar)
c. Un(2, t) = [A, cos(nax) + B, sin(naz)] sin(nat)
d. None of the above
+00
Let now u(r, t) = Un(x, t) be a solution of (E).
%3D
(2) By using the initial condition (3), we obtain that
a. A1 = 3,A2 = 0, A3 = -2 and An = 0, Vn > 4
b. A1 = 0,42 = -2, A3 = 3 and A, = 0, Vn > 4
c. An = 0, Vn 21
d. None of the above
(3) By using the initial condition (4), we obtain that
2(–1)"+1
a. B
Vn 21
%3D
b. B
Vn 21
с. В, — 0.
d. None of the above
d.
Transcribed Image Text:Consider the following boundary value problem (E) : u a2 0<2< 1, t>0.. (1) u(0, t) – 0, u(1, t) – 0, t > 0.. (2) u(x, 0) = f(x), 0 < r<1…(3), ue(1, 0) – 2, 0 <r <1…(4), where f(r) = 3 sin(wx) – 2 sin(3rz). (1) By using the method of separation of variables, the family of solutions un(r, t) for equations (1) and (2) is: a. Un(x, t) = [An cos(nat) + B, sin(nxt)] cos (na2) b. Un(r, t) – [An cos(nat) + Bn sin(nat)] sin(nar) c. Un(2, t) = [A, cos(nax) + B, sin(naz)] sin(nat) d. None of the above +00 Let now u(r, t) = Un(x, t) be a solution of (E). %3D (2) By using the initial condition (3), we obtain that a. A1 = 3,A2 = 0, A3 = -2 and An = 0, Vn > 4 b. A1 = 0,42 = -2, A3 = 3 and A, = 0, Vn > 4 c. An = 0, Vn 21 d. None of the above (3) By using the initial condition (4), we obtain that 2(–1)"+1 a. B Vn 21 %3D b. B Vn 21 с. В, — 0. d. None of the above d.
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