Final answer:
To solve the given initial-value problem y'' + 4y' + 4y = (5 + x)e⁻²ˣ, we find the particular solution and solve for the constants using the initial conditions.
Step-by-step explanation:
To solve the given initial-value problem y'' + 4y' + 4y = (5 + x)e⁻²ˣ, we need to find the particular solution that satisfies the initial conditions y(0) = 4 and y'(0) = 6.
We can solve this using the method of undetermined coefficients. First, we find the complementary solution by solving the associated homogeneous equation y'' + 4y' + 4y = 0. The characteristic equation is r² + 4r + 4 = 0, which has a double root of -2.
The complementary solution is y_c(x) = c₁e^(-2x) + c₂xe^(-2x), where c₁ and c₂ are constants. To find the particular solution, we guess the form of the solution as (Ax + B)e^(-2x), where A and B are constants.
Plugging this guess into the differential equation, we equate coefficients of like terms. We obtain A = -1 and B = 1/4. Therefore, the particular solution is y_p(x) = (-x + 1/4)e^(-2x).
The general solution is the sum of the complementary solution and the particular solution: y(x) = y_c(x) + y_p(x). Finally, substituting the initial conditions, we can determine the values of c₁ and c₂. Therefore, the solution to the given initial-value problem is y(x) = (-x + 1/4)e^(-2x) + 4e^(-2x).