Final Answer:
The complementary solution (yc) to the given second-order linear homogeneous differential equation y'' + 3y' + 2y = 0 is yc = C₁e^(-2x) + C₂e^(-x), where C₁ and C₂ are arbitrary constants.
Step-by-step explanation:
To find the complementary solution (yc), we first solve the characteristic equation associated with the homogeneous part of the given differential equation. The characteristic equation is obtained by replacing y'' with λ², y' with λ, and y with 1 in the differential equation, resulting in the equation λ² + 3λ + 2 = 0. Factoring this quadratic equation yields (λ + 1)(λ + 2) = 0, giving us two distinct roots λ₁ = -1 and λ₂ = -2.
The complementary solution is then expressed as yc = C₁e^(λ₁x) + C₂e^(λ₂x), where C₁ and C₂ are arbitrary constants. Substituting the values of λ₁ and λ₂ into the equation, we get yc = C₁e^(-x) + C₂e^(-2x). This is the final expression for the complementary solution (yc).
In the given problem, the differential equation also includes a particular solution (yp) on the right side. However, since we are only asked to find the complementary solution in this part, the final answer for yc is C₁e^(-x) + C₂e^(-2x). If the particular solution (yp) is required, it would be added to yc to obtain the complete solution to the non-homogeneous differential equation.