Run the program with a Δt corresponding to 20 time steps per period of the oscillations (see Sect. 8.4.3 for how to find such a Δt).

1.Consider (8.43)–(8.44) modeling an oscillating engineering system. This 2×2 ODE system can be solved by the Backward Euler scheme, which is based on discretizing derivatives by collecting information backward in time. More specifically, u (t) is approximated as u (t) ≈ u(t) − u(t − Δt) Δt . A general vector ODE u
= f (u, t), where u and f are vectors, can use this approximation as follows: un − un−1 Δt = f (un, tn), which leads to an equation for the new value un: un − Δtf (un, tn) = un−1 . For a general f , this is a system of nonlinear algebraic equations. However, the ODE (8.43)–(8.44) is linear, so a Backward Euler scheme leads to a system of two algebraic equations for two unknowns: un − Δtvn = un−1, (8.89) vn + Δtω2un = vn−1 . (8.90) a) Solve the system for un and vn. b) Implement the found formulas for un and vn in a program for computing the entire numerical solution of (8.43)–(8.44). c) Run the program with a Δt corresponding to 20 time steps per period of the oscillations (see Sect. 8.4.3 for how to find such a Δt). What do you observe? Increase to 2000 time steps per period. How much does this improve the solution?

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