- 2y²,y(0) =0. 1+x² 4) Consider the IVP y'= х a) Verify that y= is the...

(4) Here, y(0)=0 Cit?),1- x(22) So dya (Hx2)2 CH2)2 (-2?+2x2 Or, y'+ ay? +2 (Hje CH2)Z -ay2 (+22 We also have y(0)=0 -ay2 I+22 Hence y: is a solution of the differential equation Ya Solution of the above Differential equation by Euler's Method by taking h=.1 by Euler's Method We have the Iteration Scheme yn+1 = yn + hf (xm , Yn) (b) Yn+1 = Yn + hf (xm ,Yn) with h=.1 actual y error(%) 0.2 0.19320316 0.19230769 0.465640623 0.4 0.34828856 0.34482759 1.003683012 0.6 0.44490933 0.44117647 0.846115408 0.8 0.48877129 0.48780488 0.198114476 1 0.49698952 0.5 0.602095918 1.2 0.48524429 0.49180328 1.333661512 1.4 0.46392331 0.47297297 1.91335677 1.6 0.43894036 0.4494382 2.335770672 1.8 0.41338392 0.4245283 2.625120872 2 0.3887547 0.4 2.811325643 Table-1

(c) Solution of the above Differential equation by Euler's Method by taking h= .05 by Euler's Method We have the Iteration Scheme yn+1 = yn + hf (xn , Yn) Yn+1 = Yn + hf (xn , Yn) with h=05 actual y | x error(%) 0.2 0.1929632 0.1923077 0.3408394 04 0.3467563 0.3448276 0.5593209 0.6 0.4431224 0.4411765 0.4410795 0.488315 0.4878049 0.1045714 0.8 0.4985338 0.5 0.2932485 0.4885843 0.4918033 0.6545347 1.2 0.4685154 0.472973 0.942452 1. 1.6 0.4442494 0.4494382 1.1544989 1.8 0.4190022 0.4245283 1.3017084 2 0.3944079 0.4 1.3980358 Table-2 (d) We observe that as we move further from the initial point the error decreases gradually in table -2 as compare to Table-1. (e) Certainly the 2nd Iteration is better as the error at the same point is lesser with the same number of iteration. This is due to the smaller value of step or mesh length.