Use Euler's method with step size 0.2 to estimate y(1.4), where y(x) is the solution of the initial-value problem below. Give your answer correct to 4 decimal places.

y' = x - xy
y(1) = 0
h = 0.2

Since I am at y(1) = 0 and not y(0) = 0 would I just do this twice? As in:

y1 = 0 + 0.2*(1.2 - (1.2)(0))
y1 = 0.24

y2 = 0.24 + 0.2*(1.4 - (1.4)(0.24))
y2 = 0.4528?

Am I starting this out or setting it up wrong? Please help. Thx

y(x=1.2) = y(x=1) + 0.2* y'(1)
= 0 + 0.2* (1 - 0) = 0.2

y(x=1.4) = y(x=1.2) + 0.2 (y'@x=1.2)
= 0.2 + 0.2(1.2 - 1.2*0.2)
= 0.2 + 0.2(0.96) = 0.3920

2 answers

Yes that is the right answer
give me an anser