Consider the scalar first-order state space system : x(t) = ax(t) + bu(t), x(0) = x0 where a and b are real constants and the solution reads x(t) : [0, T] R. The solution x(t) can be written as an explicit function of time as follows: x(t) = e atx0 + Z t 0 e a(t) bu( )d To show that this function is a solution to the system over the time interval [0, T], you may differentiate this function and plug in, but you will need the Leibnitz rule, which reads as follows: d dt Z b a f(x, t)dx = Z b a tf(x, t)dx + f(b, t) db dt f(a, t) da dt (a) Given u(t) = 0 for t [0, T], show that for every pair of real constants 1 and 2 the solution x3(t) = 1x1(t) + 2x2(t) is a solution that corresponds to the initial condition x30 = 1x10 + 2x20. (b) Use your result from the previous step to show that the set of solutions over the time interval [0, T] to with zero input but various initial conditions forms a vector space over the reals. (c) Given x0 = 0, show that for every pair of real constants 1 and 2 the solution x3(t) = 1x1(t) + 2x2(t) is a solution that corresponds to the input u3(t) = 1(t)u1(t) + 2(t)u2(t). (d) Use your result from the previous step to show that the set of solutions to over the time interval [0, T] to zero initial condition but various inputs u(t) forms a vector space over the reals. (e) Does the set of solutions to over the time interval [0, T] to various non-zero initial conditions and various non-zero inputs u(t) form a vector space over the reals?

Consider the scalar rst-order state space system 2 : $03) = arc(t) + bu(t), 33(0) = {So where a, and b are real constants and the solution reads 33(t) : [0, T] ) R. The solution $03) can be written as an explicit function of time as follows: t {303) = eatmo + / saltT)bu('r)d'r 0 To show that this function is a solution to the system 2 over the time interval [0, T], you may di'erentiate this function and plug in, but you will need the Leibnitz rule, which reads as follows: dt d(/b flmtdm)= fl); f(m,)tdx+ f(b t)::_ (\"30% (a) Given u(t) = 0 for t E [0,T], show that for every pair of real constants 0:1 and 012 the solution 333(t) = 011.1216) + agmgu) is a solution that correSponds to the initial condition $30 = 013310 + agmgo. (b) Use your result from the previous step to show that the set of solutions over the time interval [0,T] to E with zero input but various initial conditions forms a vector space over the reals. (c) Given $0 = 0, show that for every pair of real constants a1 and 022 the solution $302) = almt) + 022mm is a solution that correSponds to the input \"303) = a1(t)u1(t) + 012 (02592 (t) (d) Use your result from the previous step to show that the set of solutions to 2 over the time interval [0, T] to zero initial condition but various inputs 1505) forms a vector space over the reals. (e) Does the set of solutions to 2 over the time interval [0, T] to various non-zero initial conditions and various non-zero inputs 11,03) form a vector space over the reals