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Consider the following boundary value problem (F) : Pu Ər2. I> 0, t>0… (1) du u(x,0) = 0, (r, 0) = 1, x >0-- (2) %3D at u(0, t) = sin t, t>0. (3), Let U(1, s) be the Laplace transform of u(r,t) acting to the variable t i.e: U (r, s) = L(u(x, t), and suppose that U (x, s) is bounded as s→ +o. (1) By apply the Laplace transform to the equation(1) of (F) and by using the equation (2) of (F), we obtain the following ODE: a. Urz(r, s) – s®U (r, s) = –1 b. Uz(r, s) – sºU(r, s) = –1 c. Uzr(r, s) – s²U«(r, s) = –1 d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r, s) = A(s)e-sz + B(s)e®z + ! b. U(r, s) = A(s)e¬sz + B(s)e* + c. U(r, s) = A(s)e¬sz + B(s)e*¤ d. None of the above (3) Using the fact that, the Laplace transform of u(x, t) is bounded as s → +∞ and using the equation(3) of (F), we obtain: a. U(r, s) = (T - )e-sz +! b. U(r, s) = (1,a -)e-sz c. U(r, s) = ( -)e=sz + d. None of the above (4) The general solution of (F) is: (H(t – a) is the unit step function) a. u(x,t) = t – (t – r)H(t – x) + sin(t – x)H(t – x) b. u(x,t) = sin(t –- 1)H(t – x) c. u(x,t) = t – (t – 2)H(t – z) d. None of the above %3D

Consider the following boundary value problem (F) : Pu Ər2. I> 0, t>0… (1) du u(x,0) = 0, (r, 0) = 1, x >0-- (2) %3D at u(0, t) = sin t, t>0. (3), Let U(1, s) be the Laplace transform of u(r,t) acting to the variable t i.e: U (r, s) = L(u(x, t), and suppose that U (x, s) is bounded as s→ +o. (1) By apply the Laplace transform to the equation(1) of (F) and by using the equation (2) of (F), we obtain the following ODE: a. Urz(r, s) – s®U (r, s) = –1 b. Uz(r, s) – sºU(r, s) = –1 c. Uzr(r, s) – s²U«(r, s) = –1 d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r, s) = A(s)e-sz + B(s)e®z + ! b. U(r, s) = A(s)e¬sz + B(s)e* + c. U(r, s) = A(s)e¬sz + B(s)e*¤ d. None of the above (3) Using the fact that, the Laplace transform of u(x, t) is bounded as s → +∞ and using the equation(3) of (F), we obtain: a. U(r, s) = (T - )e-sz +! b. U(r, s) = (1,a -)e-sz c. U(r, s) = ( -)e=sz + d. None of the above (4) The general solution of (F) is: (H(t – a) is the unit step function) a. u(x,t) = t – (t – r)H(t – x) + sin(t – x)H(t – x) b. u(x,t) = sin(t –- 1)H(t – x) c. u(x,t) = t – (t – 2)H(t – z) d. None of the above %3D

Advanced Engineering Mathematics
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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laplace transform non homo part 23
Consider the following boundary value problem (F) : Pu r> 0, t>0. (1) H2 du u(r,0) = 0, a (r, 0) = 1, r>0.. (2) %3D u(0, t) = sin t, t>0• (3), Let U(r, s) be the Laplace transform of u(x,t) acting to the variable t i.e: U (r, s) = L(u(x, t), and suppose that U(r, s) is bounded as s→ +o. (1) By apply the Laplace transform to the equation(1) of (F) and by using the equation (2) of (F), we obtain the following ODE: a. Urz(x, s) – s®U (1, s) = –1 b. Uz(r, s) – s°U(x, s) = -1 c. Uzz(x, 8) – s²U#(r, s) = –1 d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r, s) = A(s)e-s" + B(s)e® +: b. U(r, s) = A(s)e-s + B(s)e + ; c. U(r, s) = A(s)e¬s" + B(s)e*¤ d. None of the above (3) Using the fact that, the Laplace transform of u(x, t) is bounded as s → +0 and using the equation(3) of (F), we obtain: a. U(r, 8) = ( - )e-sz + ! b. U(r, 8) = ( - )e c. U(r, 8) = ( -)e=s= + d. None of the above (4) The general solution of (F) is: (H(t – a) is the unit step function) a. u(r, t) = t – (t – x)H(t – x) + sin(t – x)H(t – x) b. u(x, t) = sin(t – x)H(t – x) c. u(x, t) =t – (t – 2)H(t – x) d. None of the above 1
Transcribed Image Text:Consider the following boundary value problem (F) : Pu r> 0, t>0. (1) H2 du u(r,0) = 0, a (r, 0) = 1, r>0.. (2) %3D u(0, t) = sin t, t>0• (3), Let U(r, s) be the Laplace transform of u(x,t) acting to the variable t i.e: U (r, s) = L(u(x, t), and suppose that U(r, s) is bounded as s→ +o. (1) By apply the Laplace transform to the equation(1) of (F) and by using the equation (2) of (F), we obtain the following ODE: a. Urz(x, s) – s®U (1, s) = –1 b. Uz(r, s) – s°U(x, s) = -1 c. Uzz(x, 8) – s²U#(r, s) = –1 d. None of the above (2) The solution of the ODE obtained in part (1) is: a. U(r, s) = A(s)e-s" + B(s)e® +: b. U(r, s) = A(s)e-s + B(s)e + ; c. U(r, s) = A(s)e¬s" + B(s)e*¤ d. None of the above (3) Using the fact that, the Laplace transform of u(x, t) is bounded as s → +0 and using the equation(3) of (F), we obtain: a. U(r, 8) = ( - )e-sz + ! b. U(r, 8) = ( - )e c. U(r, 8) = ( -)e=s= + d. None of the above (4) The general solution of (F) is: (H(t – a) is the unit step function) a. u(r, t) = t – (t – x)H(t – x) + sin(t – x)H(t – x) b. u(x, t) = sin(t – x)H(t – x) c. u(x, t) =t – (t – 2)H(t – x) d. None of the above 1
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