Since ∆y/∆t is the slope of the line approximating the curve,
the ∆y is just the slope times ∆t
So, if dy/dt = f(t)
yk+1 = yk + f(t)*∆t
dy/dt = 2y^(1/2) y(0)= 2
What is Euler's method for computing yk+1 from yk with step size delta t when yo= 10
1 answer