"solution \\space \\\\\ngiven\\\\\nfind \\space y(0.1)\\\\\nfor \\space y' \\space =(y-x)(y+x), \\space y(0)=1, \\space with \\space step \\space lenght \\space 0.02 \\space using \\space euler \\space method\\\\\nEulerr \\space method \\space \\\\\nwhere \\space we \\space take \\space y' \\space as \\space f(x,y)\\\\\n[1] \\space\ny_1(0.02) \\space = \\space y_0 \\space +h. \\space f(x_0,y_0)=1+(0.02).f(0,1)=1+(0.02).(1)=1+(0.02)=1.02\\\\\n[2] \\space y_2(0.04) \\space = \\space y_1 \\space +h. \\space f(x_1,y_1)=1.02+(0.02)f(0.02,1.02)=1.02+(0.02)\u22c5(1.04)=1.02+(0.0208)=1.0408\\\\\n[3] \\space y_3(0.06) \\space = \\space y_2 \\space +h. \\space f(x_2,y_2)=1.0408+(0.02)f(0.04,1.0408)=1.0408+(0.02)\u22c5(1.0817)=1.0408+(0.0216)=1.0624\\\\\n\n[4] \\space y_4(0.08) \\space = \\space y_3 \\space +h. \\space f(x_3,y_3)=1.0624+(0.02)f(0.06,1.0624)=1.0624+(0.02)\u22c5(1.1252)=1.0624+(0.0225)=1.0849\\\\\n[5] \\space y_5(0.1) \\space = \\space y_4+h. \\space f(x_4,y_4)=1.0849+(0.02)f(0.08,1.0849)=1.0849+(0.02)\u22c5(1.1707)=1.0849+(0.0234)=1.1084\\\\\nhence \\space \\\\\ny(0.1)=1.1084\\\\"
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