91Ó°ÊÓ

To find the homogeneous solution, we solve the homogeneous equation without the forcing term, \[y'' - 3y' + 2y = 0.\] This is a second-order linear ODE with constant coefficients. The characteristic equation corresponding to the homogeneous equation is given by: \[r^2 - 3r + 2 = 0.\] Solve it to get the roots \(r_1 = 1\) and \(r_2 = 2\). Therefore, the general solution for the homogeneous equation is: \[y_h(t) = C_1 e^{t} + C_2 e^{2t},\] where \(C_1\) and \(C_2\) are constants to be determined by the initial conditions.

To find the particular solution, we consider the forced equation: \[y'' - 3y' + 2y = \delta(t-1).\] We are going to find a Green's function, \(G(t,\tau)\), such that: \[y_p(t) = \int_{}^{} G(t,\tau) \delta(\tau-1) d\tau.\] We start by finding the general Green's function for the linear ODE problem. For this, consider \(y'' - 3y' + 2y = \delta(t-\tau)\). This ODE can be split into two equations for \(t<\tau\) and \(t>\tau\): 1. \(t < \tau: \quad y'' - 3y' + 2y = 0\) 2. \(t > \tau: \quad y'' - 3y' + 2y = 0\) The general solutions for both equations will be different than our homogeneous solution \(y_h(t)\): 1. \(t < \tau: \quad G(t,\tau) = A_1(\tau) e^{t} + B_1(\tau) e^{2t}\) 2. \(t > \tau: \quad G(t,\tau) = A_2(\tau) e^{t} + B_2(\tau) e^{2t}\) Now, we have to impose jump conditions at \(t=\tau\): 1. \(G(\tau,\tau^{-}) = G(\tau,\tau^{+})\) 2. \(G'(\tau,\tau^{-}) - G'(\tau,\tau^{+}) = 1\) Solve these equations to find: 1. \(A_1(\tau)=\dfrac{e^{2\tau}-e^{\tau}}{3}\); \(B_1(\tau)=e^{\tau}\) 2. \(A_2(\tau)=0\), \(B_2(\tau)=e^{2\tau}\) So, the Green's function is given by: \[G(t,\tau) = \begin{cases} \frac{e^{2\tau}-e^\tau}{3} e^{t} + e^\tau e^{2t} & t < \tau \\ e^{2t+\tau} & t > \tau \end{cases}\] Therefore, the particular solution is given by: \[y_p(t) = \int_{}^{} G(t,\tau) \delta(\tau-1) d\tau = G(t,1).\]

The complete solution is the sum of the homogeneous and particular solutions, \[y(t) = y_h(t) + y_p(t) = C_1 e^{t} + C_2 e^{2t} + G(t,1).\] Apply the initial conditions \(y(0) = 1\) and \(y'(0) = 0\) to find: \[1 = C_1 + C_2 + G(0,1), \quad 0 = C_1 + 2C_2 + G'(0,1).\] Solving for \(C_1\) and \(C_2\), we get: \[C_1 = \dfrac{1}{3}\] and \[C_2 = \dfrac{2}{3}.\] Substitute the values of \(C_1\) and \(C_2\) to get the solution of the given initial-value problem: \[y(t) = \boxed{\frac{1}{3} e^{t} + \frac{2}{3} e^{2t} + G(t,1)}.\]