Example 4:
Using Euler's method find the approximate solution of y' = (y -
x)/(y + x), y(0) = 1.0 at x = 0.1 by taking h = 0.02
Solution:
f(x, y) = (y - x)/(y + x),
y1 = y0 + h f(x0, y0) =
1.0 + 0.02* ( (1.0 - 0.0)/(1.0 + 0.0) ) = 1.02
y2 = y1 + h f(x1, y1) =
1.02 + 0.02* ( (1.02 - .02)/(1.02 + .02) ) = 1.0392
y3 = y2 + h f(x2, y2) =
1.0392 + 0.02* ( (1.0392 - .04)/(1.0392 + .04) ) = 1.0577
y4 = y3 + h f(x3, y3) =
1.0577 + 0.02* ( (1.0577 - .06)/(1.0577 + .06) ) = 1.0756
y5 = y4 + h f(x4, y4) =
1.0756 + 0.02* ( (1.0756 - .08)/(1.0756 + .08) ) = 1.0928
Similarly
y[0.12] = 1.1094785695147549
y[0.14] = 1.1255744751575762
y[0.16] = 1.1411496068948328
y[0.18] = 1.156230879425498
y[0.20] = 1.1708425888179266
y[0.22] = 1.185006761753221
y[0.24] = 1.1987434465115816
y[0.26] = 1.2120709572133086
y[0.28] = 1.225006080222667
y[0.30] = 1.237564249677352
y[0.32] = 1.2497596976361904
y[0.34] = 1.2616055832152824
y[0.36] = 1.273114104217541
y[0.38] = 1.2842965940876045
y[0.40] = 1.2951636064962846
y[0.42] = 1.3057249894414982
y[0.44] = 1.3159899504204386
y[0.46] = 1.3259671139614
y[0.48] = 1.335664572588753
y[0.50] = 1.3450899321200915
y[0.52] = 1.3542503520520834
y[0.54] = 1.3631525816745895
y[0.56] = 1.3718029924560713
y[0.58] = 1.3802076071632443
y[0.60] = 1.3883721261112296
y[0.62] = 1.3963019508846237
y[0.64] = 1.404002205822987
y[0.66] = 1.4114777575246609
y[0.68] = 1.41873323258928
y[0.70] = 1.4257730337908368
y[0.72] = 1.4326013548488228
y[0.74] = 1.4392221939441487
y[0.76] = 1.4456393661086555
y[0.78] = 1.451856514601621
y[0.80] = 1.4578771213733517
y[0.82] = 1.4637045167044138
y[0.84] = 1.4693418880990288
y[0.86] = 1.4747922885024298
y[0.88] = 1.480058643904342
y[0.90] = 1.485143760384074
y[0.92] = 1.490050330646846
y[0.94] = 1.4947809400958176
y[0.96] = 1.4993380724797414
y[0.98] = 1.5037241151521437
y[1.0] = y[0.98] + h
f = 1.5037 + 0.02* ( (1.5037 - .98)/(1.5037
+ .98) )
=1.5079413639743826
