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
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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pde using laplace transformation part 1 2 3
Consider the following boundary value problem (K) :
u(r, 0) = shr,
u(0, г) — 0,
- e-", a>0, t>0 (1)
* > 0
t >0
(2)
(3)
(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(r, s) + U (x, s) =
s(s+1)
b. Uz(r, s)
s(s+1)
1
c. Uz(x, s) :
%3D
8+1
d. None of the above
(2) Using equation (3) of (K), the solution of the ODE obtained in part (2) is:
1
+e*+ shr-
1
a. U(r, s) = r-
s(s+1)
1.
b. U(r,s) =
1.
+ shr
%3D
s(s+1)
s(s+ 1)
c. U(r,s) =
1
1
+ e"= + shr-+(-
(s+1)
+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 + e* + shx + sin(t – x)H(t – x)
(1 ++),.
b. u(x, t) = x(1 - e-t+ te²) + shr
c. u(x,t) = e-(z+1) – e-² – e-t + shr + 1
d. None of the above
Transcribed Image Text:Consider the following boundary value problem (K) : u(r, 0) = shr, u(0, г) — 0, - e-", a>0, t>0 (1) * > 0 t >0 (2) (3) (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(r, s) + U (x, s) = s(s+1) b. Uz(r, s) s(s+1) 1 c. Uz(x, s) : %3D 8+1 d. None of the above (2) Using equation (3) of (K), the solution of the ODE obtained in part (2) is: 1 +e*+ shr- 1 a. U(r, s) = r- s(s+1) 1. b. U(r,s) = 1. + shr %3D s(s+1) s(s+ 1) c. U(r,s) = 1 1 + e"= + shr-+(- (s+1) +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 + e* + shx + sin(t – x)H(t – x) (1 ++),. b. u(x, t) = x(1 - e-t+ te²) + shr c. u(x,t) = e-(z+1) – e-² – e-t + shr + 1 d. None of the above
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