I am studying ODEs and came across this exam question:
I have the solution here also:
I have been working on this exam question all day and have been stuck for hours. What I don't understand is how the expansion has come about and why (i.e. hxT and also (xn+h)). Please could someone explain this to me as this would really really help!
Thank You
Ok, let me try to explain and guess what was really meant.
If you have an ODE $$ y\prime = f(x, y)$$
and want to solve it numerically, you must have a way to approximate derivatives numerically (by finite amount of data) and the interval of interest for a solution of ODE. This interval is represented by some grid; solution is known only at vertices of grid. Usually grid consists of equally spaced points. That's there $x_{n+1} = x_n + h$ and $x_{n-1} = x_n - h$ formulas came from.
So, it seems that ODE is approximated with this difference scheme:
$$ \frac{y_{n+1}-(1+\alpha)y_n + \alpha y_{n-1}}{h} =\frac{1}{2} \Bigl ((3-\alpha)f_n -(1+\alpha)f_{n-1} \Bigr ) $$
or, as it was written in your task
$$ y_{n+1}-(1+\alpha)y_n + \alpha y_{n-1} =\frac{h}{2} \left ((3-\alpha)f_n -(1+\alpha)f_{n-1} \right ). $$
So, we approximate the derivative at point $x_n$ with a $\frac{y_{n+1}-(1+\alpha)y_n + \alpha y_{n-1}}{h}$, using the information at $x_n$ and neighbour point $x_{n-1}$ and $x_{n+1}$. R.H.S. represents the value of $f(x_n, y_n)$.
For me, it seems that $T$ is an error that we get when we plug exact solution into difference operator. So, $h\cdot T$ appeared just for the sake of simplicity.
Basically all these methods assume that you have a smooth enough RHS of ODE. So, if it is smooth enough, you can use Taylor expansion of this function until this already mentioned "enough" terms. And that's the method: you plug exact solution into numerical method (symbolically, of course) and see how numerical result differs from exact value. Hope this explains, why expansion came.