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As shown in lectures, the derivative of a function y(x) can be calculated using the second order central, and first order forward and backward difference schemes 1 αι 0 0 0 0 0 dy (xN-1) ~ (YN-1 — YN-2)/A. dx The double derivative can be approximated as 0 0 B₁ Y1 a2 B2 0 α3 dy -(xi) ≈ (Yi+1 − Yi−1)/(2A) dx 0 0 72 B3 d'y dx² 0 0 0 0 0 0 0 dy Show that if you discretise the equation shown in Q1 using the differentiation schemes above and implementing the boundary conditions, you will get a set of equations that can be expressed as - (xo) ≈ (y₁ − Yo)/A - dx d²y dr² (™N−1) ≈ (YN-1 — 2YN−2+ YN-3)/A². 0 0 0 Y3 ; (xi) ≈ (Yi–1 d²y dx² (ªo) ≈ (yo − 2y₁ +Y2)/A² ... - - 2yi + Yi+1)/A² 0 0 0 0 aN-3 BN-3 YN-3 0 aN-2 0 0 0 0 0 0 BN-2 YN-2 0 1 Yo Y1 Y2 Y3 YN-3 YN-2 YN-1 a Q₁ Q2 Q3 : QN-3 QN-2 b

As shown in lectures, the derivative of a function y(x) can be calculated using the second order central, and first order forward and backward difference schemes 1 αι 0 0 0 0 0 dy (xN-1) ~ (YN-1 — YN-2)/A. dx The double derivative can be approximated as 0 0 B₁ Y1 a2 B2 0 α3 dy -(xi) ≈ (Yi+1 − Yi−1)/(2A) dx 0 0 72 B3 d'y dx² 0 0 0 0 0 0 0 dy Show that if you discretise the equation shown in Q1 using the differentiation schemes above and implementing the boundary conditions, you will get a set of equations that can be expressed as - (xo) ≈ (y₁ − Yo)/A - dx d²y dr² (™N−1) ≈ (YN-1 — 2YN−2+ YN-3)/A². 0 0 0 Y3 ; (xi) ≈ (Yi–1 d²y dx² (ªo) ≈ (yo − 2y₁ +Y2)/A² ... - - 2yi + Yi+1)/A² 0 0 0 0 aN-3 BN-3 YN-3 0 aN-2 0 0 0 0 0 0 BN-2 YN-2 0 1 Yo Y1 Y2 Y3 YN-3 YN-2 YN-1 a Q₁ Q2 Q3 : QN-3 QN-2 b

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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You are required to find the numerical solution to the following ordinary differential equation dy dx d²y dx² 6y = 15 sin(12x) in the domain 0 ≤ x ≤ 1. You are required to solve this equation using numerical methods with the boundary conditions y(0) = − 1 and y(1) = 2. Note that the input for the sin term should be in radians (not degrees).
Transcribed Image Text:You are required to find the numerical solution to the following ordinary differential equation dy dx d²y dx² 6y = 15 sin(12x) in the domain 0 ≤ x ≤ 1. You are required to solve this equation using numerical methods with the boundary conditions y(0) = − 1 and y(1) = 2. Note that the input for the sin term should be in radians (not degrees).
Q1 As shown in lectures, the derivative of a function y(x) can be calculated using the second order central, and first order forward and backward difference schemes 1 0 HOO...... dy ·(XN−1) ≈ (YN-1 — YN-2)/A. dx The double derivative can be approximated as αι B₁ 0 X2 0 0 0 0 0 0 0 0 0 Y1 B₂ d²y dx² Show that if you discretise the equation shown in Q1 using the differentiation schemes above and implementing the boundary conditions, you will get a set of equations that can be expressed as α3 0 0 dy (xi) ~ (Yi+1 — Yi-1)/(2A) dx 72 B3 0 0 0 0 d'y dx² dy · (xo) ≈ (Y₁ — Yo)/A 0 0 0 Y3 dx ; (x₁) ≈ (Yi−1 − 2Yi + Yi+1)/A² d²y dx² (xN−1) ≈ (YN-1 – 2yn-2 + YN-3)/▲². aN-3 0 -(xo) ≈ (yo − 2y₁ + Y2) /▲² 2 - 0 0 0 0 0 0 ...O BN-3 YN-3 aN-2 BN-2 YN-2 0 0 1 0 Yo Y1 Y2 Y3 : YN-3 YN-2 YN-1 a Q₁ Q2 Q3 - QN-3 QN-2 b
Transcribed Image Text:Q1 As shown in lectures, the derivative of a function y(x) can be calculated using the second order central, and first order forward and backward difference schemes 1 0 HOO...... dy ·(XN−1) ≈ (YN-1 — YN-2)/A. dx The double derivative can be approximated as αι B₁ 0 X2 0 0 0 0 0 0 0 0 0 Y1 B₂ d²y dx² Show that if you discretise the equation shown in Q1 using the differentiation schemes above and implementing the boundary conditions, you will get a set of equations that can be expressed as α3 0 0 dy (xi) ~ (Yi+1 — Yi-1)/(2A) dx 72 B3 0 0 0 0 d'y dx² dy · (xo) ≈ (Y₁ — Yo)/A 0 0 0 Y3 dx ; (x₁) ≈ (Yi−1 − 2Yi + Yi+1)/A² d²y dx² (xN−1) ≈ (YN-1 – 2yn-2 + YN-3)/▲². aN-3 0 -(xo) ≈ (yo − 2y₁ + Y2) /▲² 2 - 0 0 0 0 0 0 ...O BN-3 YN-3 aN-2 BN-2 YN-2 0 0 1 0 Yo Y1 Y2 Y3 : YN-3 YN-2 YN-1 a Q₁ Q2 Q3 - QN-3 QN-2 b
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