We start by substituting t = 0 {\textstyle t=0} into the equation we found for y → ( t ) {\textstyle {\vec {y}}(t)} in part b),
Set this equal to [ 1 − 1 ] T {\textstyle [1\ \ -1]^{T}} and solve for c 1 {\textstyle c_{1}} and c 2 {\textstyle c_{2}}
We can see that c 1 = − 1 {\textstyle c_{1}=-1} and c 2 = 0 {\textstyle c_{2}=0} solve the system. Thus, we have