Final answer:
To solve the differential equation y'' - 2y' - 3y = 3t^2 + 4t - 5, assume a solution of the form y = At^2 + Bt + C. Take the derivatives of y and substitute them into the differential equation to solve for A, B, and C. Add the particular solution to the general solution of the homogeneous equation to obtain the complete solution.
Step-by-step explanation:
To solve the differential equation y'' - 2y' - 3y = 3t^2 + 4t - 5, we first assume a solution of the form y = At^2 + Bt + C, where A, B, and C are constants. Taking the derivatives of y, we find y' = 2At + B and y'' = 2A. Substituting these derivatives into the differential equation, we can equate the corresponding coefficients and solve for A, B, and C.
After finding the values of A, B, and C, we can substitute them back into the equation y = At^2 + Bt + C to obtain the particular solution.
Finally, we can add the particular solution to the general solution of the homogeneous equation (the solution when the right-hand side of the differential equation is zero) to get the complete solution.