Question

Given (dy/dx)=(3x^3+6xy^2-x)/(2y) with y=0.707 at x= 0, h=0.1 obtain a solution by the fourth order Runge-Kutta method for a range x=0 to 0.5

Given

(dy/dx)=(3x^3+6xy^2-x)/(2y)

with y=0.707 at x= 0, h=0.1 obtain a solution by the fourth order Runge-Kutta method for a range x=0 to 0.5

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Answer #1

hiven: dng . a v o 707) (0.1 又 S547) to) o003547 (0+ , 0. 10 7+ 0.00 8547 = (0,1)れ0.0s, 0.70877) 21Kg= 0.00358 0.707十은卡 0.003547 toico 358 1 o.0o 132 yl = 0.710596 о . 007318 к:@u) t (o.ISノ0.714255)133((os, 0 716297) ya = o. 72.2165 kl = 0.01624K2=o.oa g K3 .0320) k4# 0.02336 уз = 0.743365 733 toO.3 h-(01) t ( 0-3, o. 743365) ka = 0.0313kq= 0.049125 6 s4 = 0.751466 ki = 0.04914 K2 (o (0-4s, 0.80604) ka= 0,063579 k3 = 0.06543gs = 0.89696 kl = 0.08584 , 0 533) k2 = 0.11 g k3 = (01)れo.ss, 0.103965. K4 O. (64 849g6 = 0.84696 + 0.08589 = 0.9666 A t A t X:0,3,4(0-2)ニ 0.122165

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Given (dy/dx)=(3x^3+6xy^2-x)/(2y) with y=0.707 at x= 0, h=0.1 obtain a solution by the fourth order Runge-Kutta method for a range x=0 to 0.5
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