Therefore, from the above solution we can conclude that : x(t) = 10 He³ 3t e³t - 1.6 [៦] e -t -8.4e2t + -2e2t Use the method of variation of parameters to solve the initial value problem x' = Ax+f(t), x(a)=x using the following values. 4 -1 8 A= 16 |, f(t) = 7 1 + 4t x(0) = eAt -t = - 4 16t 1-4t x(t)= = ...