Question

Consider the IVPs: 
        (A)  y'+2y = 1,   0<t<1 ,  y(0)=2.
        (B)  y' = y(1-y), 0<t<1 ,  y(0)=1/2.

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  Δt=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 + Δt 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  Δt=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 Δt 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  Δt=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 ?

Answer 1