(E2)O&W2.33(a-(i) (a)()We need to find the homogeneous and particular solutions with the final solution being the sum of the two. Let's start with the particular solution which we will denote yp(t). Let the particular solution take the form of the input for t> 0, (t)=Aet. Substituting into(P2 33-1)for t>0 we ha 3.Aet +2Ae Which a g noting the homogeneous solution as yn(t), we have yn(t)= Best where B and s are constants to be determined. Substituting into(P2 33-1)with the input set to o we have Bse+ 2Be 0 Thus the homogeneous solution takes the form Be-2t. The output is then given y(t)= yn(t)+ yp(t) We know that the system is initially at rest, so at t=0, the output has to be zero B Thus, the output, y(t) + Note that we had to solve for B with the additional information(in addition to the differential equation) that the system was at initial rest. This is because LCCDES are Not complete characterizations of systems. In general, we need more information to compute the output when given and input signal(E2) O&W 2.33 (a-(i)) (a) (i) We need to find the homogeneous and particular solutions with the final solution being the sum of the two. Let’s start with the particular solution which we will denote yp(t). Let the particular solution take the form of the input for t > 0, yp(t) = Ae3t . Substituting into (P2.33 − 1) for t > 0 we have, 3t 3t 3Ae3t + 2Ae = e 1 A = 5 Which 3t gives us the particular solution, yp(t) = 1 e . 5 For the homogeneous part, let’s try a general exponential, again for t > 0. De￾noting the homogeneous solution as yh(t), we have yh(t) = Best where B and s are constants to be determined. Substituting into (P2.33 − 1) with the input set to 0 we have, 3t Bse3t + 2Be = 0 s + 2 = 0 s = −2 Thus the homogeneous solution takes the form Be−2t . The output is then given by y(t) = yh(t) + yp(t) = Be−2t + 1 3t e 5 We know that the system is initially at rest, so at t = 0, the output has to be zero: y(t) = Be−2(0) + 1 3t e 5 1 B = −5 1 3t Thus, the output, y(t) = −1 e−2t + 5 e . 5 Note that we had to solve for B with the additional information (in addition to the differential equation) that the system was at initial rest. This is because LCCDEs are NOT complete characterizations of systems. In general, we need more information to compute the output when given and input signal. 3