12. u(0, t) 100, u(1, t) = 100, f(x) = 50 x(1x), L = 1, c = 1. An = C, and u(x, t) = u₁(x) + u₂(x, t). |u2(x,t) = Ebne n=1 (6) (7) (8) -X²/t u1 (X) sin bn=f(f(x)-( -x+ Ti)) sin x dx. T₂ - T1 L Nonzero Boundary Conditions Consider the heat boundary value problem du = c² ², u(0, t) = T₁ and NT L u(x, 0)= f(x), 0 < x 0, u(L, t) = T2, t> 0, 0

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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heat boundry value PDE

12. u(0, t) = 100, u(1, t) = 100, f(x) = 50 x(1 − x), L = 1, c = 1.
An = C, and
e
u(x, t) = u₁(x) + u₂(x, t).
(6)
(7)
(8)
u2(x,t) = Σ bne-at
n=1
u₁(x)
bn = ² √ ¹ (f(x) – T2 - T
(
L
L
Nπ
sin -X,
L
Nonzero Boundary Conditions
Consider the heat boundary value problem
du = c²3242,
a²u
It
Nπ
-x+ T₁)) sin x dx.
u(0, t) T₁ and
=
0 < x <L, t> 0,
u(L, t) = T2, t> 0,
u(x,0) = f(x), 0<x<L.
The problem is nonhomogeneous when T₁ and T₂ are not both zer
Transcribed Image Text:12. u(0, t) = 100, u(1, t) = 100, f(x) = 50 x(1 − x), L = 1, c = 1. An = C, and e u(x, t) = u₁(x) + u₂(x, t). (6) (7) (8) u2(x,t) = Σ bne-at n=1 u₁(x) bn = ² √ ¹ (f(x) – T2 - T ( L L Nπ sin -X, L Nonzero Boundary Conditions Consider the heat boundary value problem du = c²3242, a²u It Nπ -x+ T₁)) sin x dx. u(0, t) T₁ and = 0 < x <L, t> 0, u(L, t) = T2, t> 0, u(x,0) = f(x), 0<x<L. The problem is nonhomogeneous when T₁ and T₂ are not both zer
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