91Ó°ÊÓ

When solving a second order linear non-homogeneous differential equation like the one given, we start by finding the complementary solution. This involves solving the associated homogeneous equation. For our equation, the associated homogeneous version is: \[y'' - 3y' = 0\]Here, we ignore the non-homogeneous part (i.e., the right-hand side). The goal is to solve the homogeneous differential equation first.

To find the complementary solution, we use the characteristic equation. This equation comes from assuming a solution of the form \[y = e^{rt}\] for the homogeneous part. Substituting this into the homogeneous differential equation, we get the characteristic or auxiliary equation. In this exercise, the characteristic equation is:\[r^2 - 3r = 0\]Factoring this, we find:\[r(r - 3) = 0\]The solutions, or roots, of this characteristic equation are \[r = 0\] and \[r = 3\]. Therefore, the complementary solution is:\[y_h = C_1 + C_2 e^{3t}\]

The next step is to find the particular solution to account for the non-homogeneous part of the equation, which is \[t^2 - e^{3t}\]. To find the particular solution, we guess a form that reflects the terms on the right-hand side. Using the method of undetermined coefficients, we assume the particular solution has a form that would include both polynomial and exponential functions, such as:\[y_p = At^2 + Bt + C + De^{3t}\].This form covers the polynomial term \[t^2\] and the exponential term \[e^{3t}\].

Once we assume a form for the particular solution, we need to determine the coefficients. First, we find the first and second derivatives of the assumed particular solution:\[y_p' = 2At + B + 3De^{3t}\]\[y_p'' = 2A + 9De^{3t}\]Next, we substitute these back into the original differential equation:\[2A + 9De^{3t} - 3(2At + B + 3De^{3t}) = t^2 - e^{3t}\]Simplifying and collecting like terms, we compare coefficients on both sides of the equation to solve for \[A, B,\] and \[D\]:\[-6A = 1,\quad 3B = 2A,\quad 9D = -1\]This leads to \[A = -\frac{1}{6},\quad B = -\frac{1}{9},\quad D = -\frac{1}{9}\]. Thus, our particular solution is:\[y_p = - \frac{1}{6} t^2 - \frac{1}{9} t - \frac{1}{9} e^{3t}\]Combining this with the complementary solution gives the general solution:\[y = y_h + y_p = C_1 + C_2 e^{3t} - \frac{1}{6} t^2 - \frac{1}{9} t - \frac{1}{9} e^{3t}\]