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