Question 2: The function y(t) satisfies the ordinary differential equation dy dt = f(t, y) for t> 0. (4) The variable t takes discrete values, tη with a constant step-size h = tn+1 − tn, for all n € N. The approximation of y(tn) is denoted yn. (a) Derive an implicit scheme by integrating (4) over the interval [tn, tn+1], using the trapezium quadrature rule. (b) Consider the case f(y) = -λy where > 0 is a constant. (i) Find the exact solution of equation (4), subject to the initial condition y(0) = yo, and identify the behaviour of the solution as t→ +∞. (ii) Write down the difference equation corresponding to the implicit scheme derived in part (a) and show that Yn+1 = 1 - Xh/2 1+ Xh/2 Yn. (iii) Solve the difference equation (i.e. write down yn in terms of yo) and show that this implicit scheme is unconditionally stable. = (c) Verify that y(t) tlnt2t is a solution of equation (4) for f(t, y) initial condition y(1) = 2. = 1+y/t, with Write down the implicit scheme obtained in part (a) in an explicit form (i.e. express Yn+1 in terms of yn). Then, starting at t = 1 with h = 1 and yo = 2, approximate y at t = 2 and calculate the error of the approximation. (d) The global error of this numerical scheme can be analysed for f(y) = -λy. The global error is the difference between the exact solution y(t), obtained in part ((b)i), and the approximate solution, obtained in ((b)iii), when n = t/h. Show, by expanding the global error as a Taylor series in powers of h, that the method is of second order.