Question

Transform the BVP u_t = u_{xx}; 0 < x < 1 u(0, t) = 2 u(1, t) = t into new one with homogenous boundary conditions, we get a. w_t = w_{xx} - x; 0 < x < 1 w(0, t) = 0 w(1, t) = 0 b. w_t = w_{xx} + 2x - 2; 0 < x < 1 w(0, t) = 0 w(1, t) = 0 c. w_t = w_{xx} + x - 1; 0 < x < 1 w(0, t) = 0 w(1, t) = 0 d. w_t = w_{xx} - x + 1; 0 < x < 1 w(0, t) = 0 w(1, t) = 0 e. w_t = w_{xx}; 0 < x < 1 w(0, t) = 0

          Transform the BVP
u_t = u_{xx}; 0 < x < 1
u(0, t) = 2
u(1, t) = t
into new one with homogenous boundary conditions, we get
a. w_t = w_{xx} - x; 0 < x < 1
w(0, t) = 0
w(1, t) = 0
b. w_t = w_{xx} + 2x - 2; 0 < x < 1
w(0, t) = 0
w(1, t) = 0
c. w_t = w_{xx} + x - 1; 0 < x < 1
w(0, t) = 0
w(1, t) = 0
d. w_t = w_{xx} - x + 1; 0 < x < 1
w(0, t) = 0
w(1, t) = 0
e. w_t = w_{xx}; 0 < x < 1
w(0, t) = 0

Transform the BVP
ut = uxx; 0 < x < 1
u(0, t) = 2
u(1, t) = t
into new one with homogenous boundary conditions, we get
a. wt = wxx - x; 0 < x < 1
w(0, t) = 0
w(1, t) = 0
b. wt = wxx + 2x - 2; 0 < x < 1
w(0, t) = 0
w(1, t) = 0
c. wt = wxx + x - 1; 0 < x < 1
w(0, t) = 0
w(1, t) = 0
d. wt = wxx - x + 1; 0 < x < 1
w(0, t) = 0
w(1, t) = 0
e. wt = wxx; 0 < x < 1
w(0, t) = 0

Added by Deborah M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

06/26/2023

Step 1

To do this, we can use the transformation $w(x, t) = u(x, t) - v(x)$, where $v(x)$ is a function that satisfies the boundary conditions of the original problem: Show more…

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Transform the BVP u_t = u_{xx}; 0 < x < 1 u(0, t) = 2 u(1, t) = t into new one with homogenous boundary conditions, we get a. w_t = w_{xx} - 2x; 0 < x < 1 w(0, t) = 0 w(1, t) = 0 b. w_t = w_{xx} + 2x - 2; 0 < x < 1 w(0, t) = 0 w(1, t) = 0 c. w_t = w_{xx} + x - 1; 0 < x < 1 w(0, t) = 0 w(1, t) = 0 d. w_t = w_{xx} - 2x + 1; 0 < x < 1 w(0, t) = 0 w(1, t) = 0 e. w_t = w_{xx}; 0 < x < 1 w(0, t) = 0