Runge-Kutta method: K11=x2(k) K21=-3x1(k)x2(k)+kh K12=x2(k)+0.5hK21 K22=-3[x1(k)+0.5hK11] [x2(k)+0.5hK21]+kh+0.5h K13=x2(k)+0.5hK22 K23=-3[x1(k)+0.5hK12] [x2(k)+0.5hK22]+kh+0.5h K14=x2(k)+0.5hK23 K24=-3[x1(k)+0.5hK13] [x2(k)+0.5hK23]+kh+h tk+1=tk+h Hence, we have ?1(?+1)=?1(?)+1/6ℎ(?11 +2?12 +2?13 +?14) ?2(?+1)=?2(?)+1/6ℎ(?21 +2?22 +2?23 +?24) t=0.1 (k=0): x1(1)=2.0754 x2(1)=0.5542 t=0.2 (k=1): x1(2)=2.1175 x2(2)=0.3146  whenthe step length of h=0.1, find out the time response of this dynamic system at t = 0.1, and t t = 0.2,how can we get x1(1),x2(1)

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