Consider the IVPs: (A) y'+2y = 1, 0. (B) y' = y(1-y), 0.1. For each one, do the following: a. Find the exact solution y(t) and evaluate it at t=1. b. Apply Euler's method with ?t=1/4 to find Y4 ? y(1). Make a table of tn, Yn for n=0,1,2,3,4. c. Find the error at t=1.2. Euler's method is obtained by approximating y'(tn) by a forward finite difference. Use the backward difference approximation to y'(tn+1) to derive the Backward Euler Method: Yn+1 = Yn + ?t f(tn+1, Yn+1) , n=0,1,2,... Note that now the unknown Yn+1 appears inside f(.,.), so this equation needs to be solved for Yn+1 at each time-step!!! whence it is also called Implicit Euler Method. For the simple ODEs (A), (B) above, the updating equation can be solved by hand, but in general a root-finder (like Newton-Raphson) is needed. This scheme is also 1st order, but it has better stability properties than Explicit Euler. For each IVP problem (A), (B), do the following: a. Apply the Backward Euler Method with ?t=1/4 to find Y4 ? y(1). b. Find the error at t=1 and compare with Explicit Euler.3. Use the centered difference approximation to y'(tn) to derive the so called Midpoint Method: Yn+1 = Yn-1 + 2 ?t f(tn, Yn) , n=1,2,... Note that this requires both Yn-1 and Yn to produce Yn+1. It is a 2-step method, hence not self-starting (need Y0 and Y1 before it can be applied), so some single-step method (like Euler) must be used to start it off. However, it has the advantage of being a 2nd order method, and explicit. For each IVP problem (A), (B), do the following: a. Apply the Midpoint Method with ?t=1/4 to find Y4 ? y(1). b. Find the error at t=1 and compare with Explicit Euler and with Implicit Euler. Which method seems to be doing better in this case ?
