91Ó°ÊÓ

First, find the complementary function by solving the homogeneous equation, i.e. the equation $$ 2 y^{\prime \prime} +3 y^{\prime} + y = 0. $$ To do this, let \(y = e^{rt}\) and substitute into the homogeneous equation to get the auxiliary equation: $$ 2r^2 + 3r + 1 = 0. $$ Apply the quadratic formula to solve for \(r\): $$ r = \frac{-3 \pm \sqrt{3^2 - 4(2)(1)}}{4} = \frac{-3 \pm \sqrt{1}}{4}. $$ The solutions are \(r_1 = -\frac{1}{2}\) and \(r_2 = -1\). Hence, the complementary function can be represented as: $$ y_c(t) = C_1 e^{\frac{-1}{2}t} + C_2 e^{-t}. $$

To find the particular solution, let's guess a solution and adjust for the appropriate order. Given that the right side of the equation shows a quadratic function in \(t\) and a trigonometric function, we can make the following guesses: $$ y_p(t) = At^2 + Bt + C + D\sin t + E\cos t. $$ Now differentiate \(y_p(t)\) twice: $$ y_p'(t) = 2At + B + D\cos t - E\sin t, $$ and $$ y_p''(t) = 2A - D\sin t - E\cos t. $$ Substitute \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into the given differential equation and simplify: $$ 2(2A - D\sin t - E\cos t) + 3(2At + B + D\cos t - E\sin t) + (At^2 + Bt + C + D\sin t + E\cos t) = t^2 + 3\sin t. $$ This leads to the following system of equations: $$ \begin{cases} A = \frac12, \\ 6A + B = 0, \\ 2C + 3D + E = 3, \\ 2A + 3E = 0, \\ 3B - 2D + D = 0 \end{cases}. $$ Solving this system, we find the coefficients: $$ A = \frac12, \quad B = -3, \quad C = \frac{5}{2}, \quad D = 1, \quad E = -\frac23. $$ Thus, the particular solution has the form: $$ y_p(t) = \frac{1}{2}t^2 - 3t + \frac{5}{2} + \sin t - \frac{2}{3}\cos t. $$

Finally, add both the complementary function and the particular solution to obtain the general solution: $$ y(t) = y_c(t) + y_p(t) = C_1 e^{\frac{-1}{2}t} + C_2 e^{-t} + \frac{1}{2}t^2 - 3t + \frac{5}{2} + \sin t - \frac{2}{3}\cos t. $$