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Q2) Solve the following partial differential equation: де a²0 =h². Ət Əx² Given the following boundary conditions: 0(0, t) = 60°C, 8 (20,t) = 60°C, 0(x,0) = 130 °C Prove that the temperature at the center of the wall and for the value of h²=0.04 cm²/s can be calculated by the following equation: ∞ 0 (t) = 60 + 280 (-1)k π2k +1° -e-((2k+1)²), 1014 k=0

Q2) Solve the following partial differential equation: де a²0 =h². Ət Əx² Given the following boundary conditions: 0(0, t) = 60°C, 8 (20,t) = 60°C, 0(x,0) = 130 °C Prove that the temperature at the center of the wall and for the value of h²=0.04 cm²/s can be calculated by the following equation: ∞ 0 (t) = 60 + 280 (-1)k π2k +1° -e-((2k+1)²), 1014 k=0

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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Q2) Solve the following partial differential equation: 20 =h². a²0 əx² at Given the following boundary conditions: 0(0, t) = 60°C, 0 (20, t) = 60°C, 0(x,0) = 130 °C Prove that the temperature at the center of the wall and for the value of h²=0.04 cm²/s can be calculated by the following equation: ∞ 280 0 (t) = 60 + Σ (−1)k -((2k+1)²) e 1014 TT 2k + 1 k=0
Transcribed Image Text:Q2) Solve the following partial differential equation: 20 =h². a²0 əx² at Given the following boundary conditions: 0(0, t) = 60°C, 0 (20, t) = 60°C, 0(x,0) = 130 °C Prove that the temperature at the center of the wall and for the value of h²=0.04 cm²/s can be calculated by the following equation: ∞ 280 0 (t) = 60 + Σ (−1)k -((2k+1)²) e 1014 TT 2k + 1 k=0
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