Final Answer:
The general solution of the differential equation y'' + 3y' + 2y = e^(-2x) is y(x) = C₁e^(-x) + C₂e^(-2x) + (1/4)e^(-2x), where C₁ and C₂ are arbitrary constants.
Step-by-step explanation:
To find the general solution, let's solve the associated homogeneous equation first. The characteristic equation is obtained by setting the coefficients of the derivatives to zero:
r² + 3r + 2 = 0
Factoring this quadratic equation gives (r + 1)(r + 2) = 0, which yields two roots: r₁ = -1 and r₂ = -2. Therefore, the homogeneous solution is y_h(x) = C₁e^(-x) + C₂e^(-2x), where C₁ and C₂ are constants.
To find the particular solution, we use the method of undetermined coefficients, assuming a particular solution in the form of y_p(x) = A*e^(-2x). Substituting this into the original differential equation, we find A = 1/4.
Thus, the particular solution is y_p(x) = (1/4)e^(-2x). The general solution is the sum of the homogeneous and particular solutions: y(x) = y_h(x) + y_p(x) = C₁e^(-x) + C₂e^(-2x) + (1/4)e^(-2x).
In summary, the general solution to the given differential equation is y(x) = C₁e^(-x) + C₂e^(-2x) + (1/4)e^(-2x), where C₁ and C₂ are arbitrary constants.