Thread: eulers method...

i'm trying to read and understand from my textbook how to do this method and i am lost. can anyone explain it to me?
my problem is...
dy/dx= x-y^2
y(0)=1
x is less than or equal to 2 and greater than or equal to 0
change in x= .25


use eulers method to approximate the solution of the initial-value problems over the stated interval. present the answer as a table and a graph.

thank you so much i appreciate it

You could be over-thinking it. We're jsut using the first derivative t approximate the curve over a small displacement.

Your first point (0,1)
Your second point (0+0.25, 1+What?)
We have dy/dx = x - y^2 ==> 0 - 1^2 = -1
y-displacement is -1*0.25 = -0.25
Your second point (0+0.25, 1 - 0.25) = (0.25, 0.75)

Now you have a second point. Calculate the slope there and find the third point.