(**Note: This homework will not be collected, and will not be graded. Think of this as a Practise Midterm (midterm
will be shorter, but this is the difficulty you should expect!**)

**Problem 1**

show that the Region of Absolute Stability for the trapezoidal method is (-infinity,0).

**Problem 2**:

Consider an initial value problem of the form

dy/dt = F(y)

y(t_{0}) = y_{0}

This is called an autonomous differential equation, because F(y) does not depend explicitly on t. Give an expression
for the leading term of the truncation error for the Trapezoidal method

**Problem 3**:

Consider the initial value problem

dy/dt = t*sin(y)*cos(2*y)

0 <= t <= 2.51

y(0) = 1

Estimate a restriction on the timestep you can use applying a Runge Kutta scheme of order 3, such that the solution
exhibits qualitatively the correct behavior.

The region of absolute stability for a third order Runge Kutta scheme is (-2.51,0).

**Problem 4**:

Consider applying the Trapezoidal method

w_{k+1} = w_{k} + h/2*f( w_{k }, t_{k}) + h/2*f(w_{k+1 }t_{k+1})

To the differential equation dy/dt = 4t^{2}cos(y).

(a) Write out the equation that must be solved at each timestep to advance the solution from w_{k }to w_{k+1}.

(b) Write out the Newton's method iteration that could be used to solve the equation in (a).

(c) Newton's method requires an initial guess; what would you choose as an initial guess for the iteration in (b)
?

**Problem 5**:

Consider the third order differential equation

d^{3}y/dt^{3} - 3 dy/dt = 6

y(0) = 1

dy/dt(0) = 2

d^{2}y/dt^{2}(0) = 3

(a) Put this equation into an equivalent first order system form.

(b) Express the system in (a) using matrix/vector notation i.e. give an expression for d**u**/dt if your unknowns
are the components of a vector** u**.

**Problem 6**:

Assume that the numerical approximation of a certain method produces the results given below at a time t=2.0. The
methods was used with different timesteps, as indicated. Can you estimate the order of the method ? (Hint: Assume
that the result for the small timestep h=0.0001 "is almost exact").

h |
Approximation |

0.0001 |
1.00000 |

.1 |
0.999439 |

.05 |
0.999849 |

.025 |
0.9999605 |

.0125 |
0.99999 |