Using Taylor\'s Series method taking algorithm of order 3, solve the initial value problem y\' = 1 + y; y(0) = 0 with h = 0.5 At x = 1 Solution %% % taylororder 3 rule %% clear all fork=1:2 h=0.5/k ; tmax=1; n=tmax /h; x(1)= 0; y(1)= 0; tout=x(1); yout=y(1).\'; true(1) = 0; ero(1) =0; trout=true(1).\'; erout=ero(1); ero(1) =0; disp(\' x actual Taylor(order3)error \') disp(\'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\') fori=1:(n + 1 ) x(i+1) = x(1)+ i*h; T3 = (1+y(i)+(h/2)*(1+y(i)+x(i) )+h^2*(1+x(i)+y(i)) ); y(i+1) = y(i) + h*T3; % true(i+1) = exp(x(i)) -1; ero(i+1) =abs( y(i) -true(i+1) ); % tout=[tout;x(i)]; yout=[yout,y(i).\']; trout=[trout,true(i).\']; erout=[erout,ero(i).\']; % ifrem(i-1 ,k) ==0% x(i)<= 0.1 %rem(i-1,10) == 0 fprintf(\'%5.2f%12.7f%12.7f%15.7f\ \' ,x(i),true(i+1),y(i), ero(i+1) ); end% if end% i.