Question

5. Consider the IVP r = {2+x?, *(0) = 1. Complete the following table for the numerical solutions of given IVP with step-size h 0.05. by Euler's Method by Improved Euler's Method 0 0.05 0.1 1

thanks for all help

Answer 1

c' = t? + x2, 210) = 1, h = 0.05 1 The iteration formula for Euler method is Ynts = Ynth to for the differential equation dy = fit, y). dt SO, t2 + 2?, x (0) = 1 dx at xo = 1 Xn+1 = 2n + h (th? + Xp? ), where, tn to + nh. For n = 0, 1, 2, 3, ... Therefore, for n = 0 x %o + h I to? + Xo) to = 0, 20 = 1, h = 0.05 ... 2y = 1 + 0.05 02 + 1² ) or, X = 1.05 Now, for n = 1 2 X2 x, thiti + X,?) te = toth or, te =h (to = 0 ) X2 = 1.05 + 0.05 ((0.05)? + 12.05)2] له Or, *2 = 1.10525

Hence, rt X by Euler's Method 1 0 1.05 0.05 1.10525 0.1 Further, the iterative formula for Improved Euler's Method is 2 Kim = thị 1, kan = (toth)² + Inthkan Ants = Yn + balkon tkan), where tn = t nh , For n = 0, 1, 2, SO, FOI n = 0 10 to = 0, X=1 SO, Kypr kzo = (to + h 1? + Io + hkzo h=0.05 kzo = 10 +0.05)2 + 1 + 0.05 x 1 So, kao = 1.0525 h (kio tko ) = 20 + 2 = 1+0.05 ( 1 + 1.0525) or, Xo = 1.0513125

Now, for n = 1 kis = t² + x 2 t; = toth oy, ty = 0.05 1 So, kjs = (0.05)2 + (1.0513125)2 i.e., key = 1.10735797265625 K21 = (teth)? + Zithkat 10.05+0:05)2 + 1.0513125 + 0.05X 1.00775797265625 Or, ks = 1.116 70039863281 X2 = x + h (kutkas) 2 1.10775797265625 + = 1.0513125 + 0.05 2 1.11670039863281 81) oy, xy = 1.106 923 959 28223 Hence, t x by Improved Euler Method 1 o 1.0513125 0.05 1.106923959 28223 0.1