Using Euler's method, what is f(1) estimated to be?

y'=y(2t-1)
y(0)=8
n=3
When I solve for h ((b-a)/n), I get 8/3, which is larger than 1, so I do not know how I could estimate at f(1) when t initial is ~2.67.

1 Answer
Apr 27, 2018

y(1)~~5.27

Explanation:

Here a=0 and b=1. For n=3, we get

h = (b-a)/n = 1/3

According to Euler method, we have

y^'(0) = y(0)(2times 0-1) =-8

y(1/3) ~~ y(0)+hy^'(0) = 8+1/3(-8) = 16/3

y^'(1/3) = y(1/3)(2times 1/3 -1) ~~16/3 times (-1/3)= -16/9

y(2/3) ~~ y(1/3)+hy^'(1/3) = 16/3+1/3(-16/9) = 128/27

y^'(2/3) = y(2/3)(2times 2/3 -1) ~~128/27 times (+1/3)=128/81

y(1) ~~ y(2/3)+hy^'(2/3) = 128/27+1/3(128/81) = 1280/243