Approximate the solution to $y' = te^{3t} - 2y$ using Adams-Bashforth Three-step method QUESTION
Consider the IVP 
$y' = te^{3t} - 2y$ for $0 \le t \le 1$ with $y(0) = 0$
and actual solution 
$$y(t) = \frac{1}{5}te^{3t} - \frac{1}{25}e^{3t} + \frac{1}{25}e^{-2t}$$
(a) Use the Adams-Bashforth Three step method to approximate the solution, h=0.2, to the IVP. Use exact starting values.
(b) Use the Adams-Bashforth Three step method to approximate the solution, with $h=0.2$, to the IVP. 
ATTEMPT
The Adams-Bashforth method looks as follows:
$$y_{i+1} = y_i + \frac{h}{12} (23f(t_i, y_i) - 16f(t_{i-1}, y_{i-1}) + 5f(t_{i-2},y_{i-2}))$$
I know we're supposed to use the Runge-Kutta method to find out what $$(23f(t_i, y_i) - 16f(t_{i-1}, y_{i-1}) + 5f(t_{i-2},y_{i-2}))$$ is. I don't know how. I only know the Runge-Kutta steps to find out $y_{i+r}$, $r\in N$ 
I'm not sure how to use it to find out $y_{i-r}$, where $r \in N$
PLEASE HELP
(Not homeweork. Studying for a test)
 A: We are given the IVP:
$\tag 1 y' = te^{3t} - 2y$ for $0 \le t \le 1$ with $y(0) = 0$
and actual solution 
$\tag 2 y(t) = \frac{1}{5}te^{3t} - \frac{1}{25}e^{3t} + \frac{1}{25}e^{-2t}$
(a) Use the Adams-Bashforth Three step method to approximate the solution, h=0.2, to the IVP. Use exact starting values.
Exact (Using the exact solution and time steps given by h.)


*

*$y(0.0) = 0$

*$y(0.2) = 0.0268128$

*$y(0.4) = 0.150778$

*$y(0.6) = 0.49602$

*$y(0.8) = 1.33086$

*$y(1.0) = 3.2191$


Using the Adams-Bashforth three-step method, we have:
$w_0 = 0$
$w_1 = 0.0.0268128$
$w_2 = 0.150778$
$\displaystyle w_{i+1} = w_i + \frac{h}{12} (23f(t_i, w_i) - 16f(t_{i-1}, w_{i-1}) + 5f(t_{i-2},w_{i-2}))$
Can you take it from here for the exact method?
(b) Use the Adams-Bashforth Three step method to approximate the solution, with $h=0.2$, to the IVP. 
For this case, we need to use the Runge-Kutta to find the starting values. 
We would use the Third Order Runge - Kutta
Using the 3rd-order RK For the starting values, we would set:


*

*$t = a = 0$, $w_0 = \alpha = 0$, and then for $i= 1, 2, \ldots, N$ do: 

*$\displaystyle k_1 = f(t, w)$

*$\displaystyle k_2 = f(t + \frac{1}{2} h, w + \frac{1}{2}k_1 h)$

*$\displaystyle k_3 = f(t + h, w -k_1 h + 2k_2 h)$

*$\displaystyle w_i = w_{i-1} + \frac{1}{6}\left(k1 + 4 k_2 + k_3\right)h$

*$\displaystyle t = a + i h$


After we have those starting values, we would proceed as before with the Adams-Bashforth (just have starting values that used Runge-Kutta as opposed to exact).
Does that all make sense?
