Consider the following boundary value problem (K) : e-t, æ > 0, t >0 (1) Əxət at u(r, 0) = shæ, u(0, t) = 0, r > 0 (2) (3) t>0 (1) By applying the Laplace transform to equation (1) of (K) (acting to the variable t) and by using equation (2) of (K), we obtain the following ODE: 1 a. Uz(x, s) + U (x, s) = %3D s(s+1) b. Uz(r, s) = %3D s(s+1) 1 c. Uz(r, s) = + e %3D s+1 d. None of the above (2) Using equation (3) of (K), the solution of the ODE obtained in part (2) is: 1 a. U(r, s) = r- +e s(8+ 1) + shr- 1. b. U(x, s) = 1. + shr s(s+1) s(s+1) 1 1 1 c. U(x, s) = x- + e* - + shr- +(- -)e- (s+1) 's2 +1 d. None of the above (3) The general solution of (K) is: (H(t – a) is the unit step function) a. u(r, t) = xe¬t +e² + shx + sin(t – x)H(t – r) b. u(x, t) = x(1 - e-t+te²) + shr c. u(x, t) = e-(z+t) – e¬¤ – e-t + shæ +1 d. None of the above
Consider the following boundary value problem (K) : e-t, æ > 0, t >0 (1) Əxət at u(r, 0) = shæ, u(0, t) = 0, r > 0 (2) (3) t>0 (1) By applying the Laplace transform to equation (1) of (K) (acting to the variable t) and by using equation (2) of (K), we obtain the following ODE: 1 a. Uz(x, s) + U (x, s) = %3D s(s+1) b. Uz(r, s) = %3D s(s+1) 1 c. Uz(r, s) = + e %3D s+1 d. None of the above (2) Using equation (3) of (K), the solution of the ODE obtained in part (2) is: 1 a. U(r, s) = r- +e s(8+ 1) + shr- 1. b. U(x, s) = 1. + shr s(s+1) s(s+1) 1 1 1 c. U(x, s) = x- + e* - + shr- +(- -)e- (s+1) 's2 +1 d. None of the above (3) The general solution of (K) is: (H(t – a) is the unit step function) a. u(r, t) = xe¬t +e² + shx + sin(t – x)H(t – r) b. u(x, t) = x(1 - e-t+te²) + shr c. u(x, t) = e-(z+t) – e¬¤ – e-t + shæ +1 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
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