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

Question

#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#.

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Answer 1 · 2018-04-27T00:19:56

#y(1)~~5.27#

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#