Find the general solution of the differential equation: d^2x/dt^2 + 5dx/dt + 6x = 2cos(t) - sin(t)

First solve complementary function, i.e. d²x/dt² + 5dx/dt + 6x = 0. To do so, let x = e^mt, where m = arbitrary constant. Differentiating gives dx/dt = m e^mt and d²x/dt² = m² e^mt. Substituting into the complementary function and cancelling e^mt gives us a quadratic equation in m: m² + 5m + 5 = 0. Factorising: (m + 3)(m + 2) = 0 -> m = -3 and m =-2. Hence the general solution to the complementary function is: x(t) = Ae^-3t + Be^-2t. Now we have to solve the particular integral of 2cos(t) - sin(t). To do so, we guess the solution is of the form x = C cos(t) + D sin(t). Differentiating gives dx/dt = -C sin(t) + D cos(t) and d²x/dt² = -C cos(t) - D sin(t). Substituting into the particular integral and cancelling cos(t) gives 5C + 5D = 2. Substituting into the particular integral and cancelling sin(t) gives 5C - 5D = 1. Solving these two simultaneous equations, we find that C = 0.3 and D = 0.1. Putting it all together, the general solution to the ODE is x(t) = Ae^-3t + Be^-2t + 0.3 cos(t) + 0.1 sin(t).