Consider the nonlinear system with a cubic term = (3) with initial condition (0) = 1....

 

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Consider the nonlinear system with a cubic term = (3) with initial condition (0) = 1. Using DSolve (In Mathematica) I was able to find that the solution is +1 x(t)= = 2t-c where c is a constant that determines the initial condition. (4) (a) Demonstrate that (4) is really the solution to (3). (b) If we did not know the solution, we can consider the iteration = 2k+(t)=1+ With xo(t) 1, carry out the calculation for k = 1,2,3. You may do this either by hand (somewhat tedious) or use a computer algebra system (CAS). (c) Plot xo(t), x1(t), x2(t), and x3(t) along with x*(t). If you are using a CAS, you can add in x(t) and x(t). You should observe how the x(t) are converging to the solution x*(t). Consider the nonlinear system with a cubic term = (3) with initial condition (0) = 1. Using DSolve (In Mathematica) I was able to find that the solution is +1 x(t)= = 2t-c where c is a constant that determines the initial condition. (4) (a) Demonstrate that (4) is really the solution to (3). (b) If we did not know the solution, we can consider the iteration = 2k+(t)=1+ With xo(t) 1, carry out the calculation for k = 1,2,3. You may do this either by hand (somewhat tedious) or use a computer algebra system (CAS). (c) Plot xo(t), x1(t), x2(t), and x3(t) along with x*(t). If you are using a CAS, you can add in x(t) and x(t). You should observe how the x(t) are converging to the solution x*(t).