Final answer:
The general solution to the differential equation y'' - 5y' + 6y = 2eᵗ is y = c₁e^(2t) + c₂e^(3t) + (2/3)e^t.
Step-by-step explanation:
To find the general solution to the differential equation y'' - 5y' + 6y = 2eᵗ, we first solve the associated homogeneous equation y'' - 5y' + 6y = 0. The characteristic equation is r² - 5r + 6 = 0, which factors as (r - 2)(r - 3) = 0. Therefore, the homogeneous solution is y_h = c₁e^(2t) + c₂e^(3t), where c₁ and c₂ are constants.
Next, we find a particular solution to the non-homogeneous equation. A particular solution can be guessed as y_p = Ae^t, where A is a constant. Substituting this into the non-homogeneous equation, we get Ae^t - 5Ae^t + 6Ae^t = 2e^t. Simplifying, we find that A = 2/3. Therefore, the particular solution is y_p = (2/3)e^t.
The general solution to the non-homogeneous equation is the sum of the homogeneous solution and the particular solution: y = y_h + y_p = c₁e^(2t) + c₂e^(3t) + (2/3)e^t.