Use the Runge-Kutta method to find approximate values of the solution of the initial-value problem y'+3y=x^2-3xy+y^2, y(0)=2; h=0.05 at the points xi=x0+ih, where x0 is the point where the initial condition is imposed and I=1, 2.

Question

anonymous · Answer

Which order of R-K? If I remember correctly, first order is the same as the Euler method. I'd have to check my notes (which I don't have on me at the moment) for the higher order ones.

anonymous · Answer

The first page here describes the algorithm for the second order version: http://www.ee.nthu.edu.tw/bschen/files/c16-1.pdf

idealist10 · Answer

Well, my book didn't say which order so I don't know.

anonymous · Answer

I'm comfortable with using this if you are.
$$y_{n+1}=y_n+hf\left(x_n+\frac{1}{2}h,~y_n+\frac{1}{2}hf(x_n,~y_n)\right)$$
where $f(x,y)=y'=x^2-3xy+y^2-3y$. The step size is $h=0.05$.
I like setting up tables for these sorts of problems for organizational purposes.
$$\begin{array}{c|c|c|c|c|c}
\color{lightgray}{(1)}&\color{lightgray}{(2)}&\color{lightgray}{(3)}&\color{lightgray}{(4)}&\color{lightgray}{(5)}&\color{lightgray}{(6)}&\color{lightgray}{(7)}\
n&x_n&y_n&f(x_n,y_n)&x_n+\frac{1}{2}h&y_n+\frac{1}{2}h\color{lightgray}{(4)}&y_n+hf(\color{lightgray}{(5)}+\color{lightgray}{(6)})\
\hline
0&0&2&1&0.025&2.025&1.893\
\hline
1&0.05&1.893&\cdots&\cdots&\cdots&\cdots\
\hline
2&0.1&\cdots&\cdots&\cdots&\cdots&\cdots\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots
\end{array}$$

idealist10 · Answer

Go for it, I'm fine with that formula.

anonymous · Answer

It looks like you only have to do it for $i=1$ and $i=2$ ($i=n$ in the table above), so that's nice.

idealist10 · Answer

So how do I start the problem with I=1?

anonymous · Answer

Go from left to right. Plug $x=0.05$ and $y=1.893$ into the derivative function $f(x,y)$ - this gives the fourth column.

Fifth column is simply the $x_n$ of that row plus half the step size: column 2 + h/2

Sixth column is a given row's approximation for $y_n$ plus half the step size times the derivative: column 3 + h/2 (column 4)

anonymous · Answer

Notice that column 7, row 1 is the same as column 3, row 2.

idealist10 · Answer

Thank you so much! I've got the right answers!