Let me see if I remember this, hectictar.....
y' = f (x, y) = 2xy
The initial condition says that y(x0) = y0 ...so..... y(0) = 1 and (x0, y0) = (0, 1)
So f( x0, y0 ) = 2(x0)(y0) = 2(0)(1) = 0
And h = .1 [ this is known as the step size ]
So....we need to find (x1, y1)
x1 = x0 + h = 0 + .1 = .1
y1 = y0 + h f(x0, y0) = 1 + .1 ( 0) = 1
So......our second point (x1, y1) is (.1, 1) and f (x1, y1) = 2(.1)(1) = .2
Now.......find (x2, y2 )
x2 = x1 + h = .1 + .1 = .2
y2 = y1 + h f(x1, y1) = 1 + .1 ( .2) = 1.02
So.....our third point (x2, y2) is (.2, 1.02) and f (x2, y2 ) = 2(.2)(1.02) = .408
This iteration continues until we get to x10 = 1
I assume you have some program that you can use to do this???...it's very time consuming, by hand!!
Anyway ...... Here's a pretty good explanation of the method :
http://calculuslab.deltacollege.edu/ODE/7-C-1/7-C-1-h-c.html
