Final answer:
To find the general solution to the differential equation y'' − 3y' − 4y = 6e^(2t), solve for the complementary solution of the homogeneous equation and find a particular solution for the non-homogeneous part, then combine them.
Step-by-step explanation:
The question asks us to find the general solution of the differential equation y'' − 3y' − 4y = 6e^(2t). To find the general solution, we first find the complementary solution (y_c) of the associated homogeneous equation y'' − 3y' − 4y = 0, then find a particular solution (y_p) for the non-homogeneous part.
For the complementary solution, we solve the characteristic equation r^2 − 3r − 4 = 0, which factors to (r − 4)(r + 1) = 0, giving us r = 4 and r = −1. Therefore, the complementary solution is y_c = C_1e^(4t) + C_2e^(-t), where C_1 and C_2 are constants.
Finding a particular solution involves assuming a form based on the non-homogeneous part. Since the non-homogeneous part is 6e^(2t), and 2 is not a root of the characteristic equation, we assume a particular solution of the form y_p = Ae^(2t), where A is a constant to be determined. We find A by substituting y_p into the original differential equation and solving for A.
Once we have both y_c and y_p, the general solution is y(t) = y_c + y_p = C_1e^(4t) + C_2e^(-t) + Ae^(2t), where C_1, C_2, and A are constants determined by initial conditions or additional constraints.