Answer to Quiz 10 Fri 21 Nov 2003

Apply two steps of the modified Euler method to the initial value problem

         dy
        ---- = 2 x y, y(0) = 1.
         dx

Use h = 0.2. In the table below, write first the formula above the number (with six decimal places in scientific format).

+---------++------+--------------------+--------------------------+
|    n    || x(n) |       yy(n)        |         y(n)             |
+---------++------+--------------------+--------------------------+
| formula ||      |                    |                          |
|    0    ||  0.0 |        --          |     1.00000E+0           |
|  value  ||      |                    |                          |
+---------++------+--------------------+--------------------------+
| formula ||      | y(0)+h*2*x(0)*y(0) | y(0)+h/2*(2*x(0)*y(0)    |
|    1    ||  0.2 |                    |           +2*x(1)*yy(1)) |
|  value  ||      |   = 1.00000E+0     |   = 1.04000E+0           |
+---------++------+--------------------+--------------------------+
| formula ||      | y(1)+h*2*x(1)*y(1) | y(1)+h/2*(2*x(1)*y(1)    |
|    2    ||  0.4 |                    |           +2*x(2)*yy(2)) |
|  value  ||      |   = 1.12320E+0     |   = 1.17146E+0           |
+---------++------+--------------------+--------------------------+

Use the method of undetermined coefficients to derive an Adams-Moulton formula which uses two function evaluations.
Answer:
```
   y(n+1) = y(n) + c(1)*f(n+1) + c(2)*f(n)
```
where
```
           x(n+1)
         /
	 |
         | f(x,y(x)) dx = c(1)*f(n+1) + c(2)*f(n)
         |
	 / 
          x(n)
```
has to agree for f = 1 and f = x, so c(1) and c(2) satisfy a linear system. Without loss of generality, we take x(n) = 0 and x(n+1) = h.
```
   f = 1 :  c(1)*1 + c(2)*1 = h

   f = x :  c(1)*h + c(2)*0 = h^2/2
```
So the Adams-Moulton formula we just found
```
   y(n+1) = y(n) + h/2*(f(n+1) + f(n))
```
is our familiar modified Euler method.