Answer:
The value of the differential equation is y(x) = (C₁ + C₂x)e^(-x).
Explanation:
To solve the given differential equation using the method of undetermined coefficients, we assume that the particular solution has the form:
y_p(x) = (Ax^2 + Bx + C)(x - 1)^2
where A, B, and C are constants to be determined.
First, let's find the derivatives of y_p(x):
y_p'(x) = 2Ax(x - 1)^2 + (Ax^2 + Bx + C)(2(x - 1))
y_p''(x) = 2A(x - 1)^2 + 4A(x - 1)(x) + 2(Ax^2 + Bx + C)
y_p'''(x) = -2A(x - 1) + 4A(x) + 4Ax - 4A + 2(Ax^2 + Bx + C)
y_p⁽⁴⁾(x) = 4A + 4A + 2Ax + 4A + 2A = 10A + 2Ax
Now, substitute these derivatives into the differential equation:
10A + 2Ax + 2(2A(x - 1)^2 + 4A(x - 1)(x) + 2(Ax^2 + Bx + C)) + (Ax^2 + Bx + C) = (x - 1)^2
Simplifying this equation, we get:
10A + 2Ax + 4A(x - 1)^2 + 8A(x - 1)(x) + 4(Ax^2 + Bx + C) + Ax^2 + Bx + C = (x - 1)^2
Now, equating the coefficients of like powers of x, we get the following system of equations:
10A + 4A = 0 (coefficient of x⁰)
2A - 8A + B = 0 (coefficient of x¹)
4A + B + C = 0 (coefficient of x²)
Solving this system of equations, we find:
A = 0
B = -2A = 0
C = -4A = 0
Therefore, the particular solution y_p(x) = 0.
To find the general solution, we need to find the complementary solution by solving the homogeneous equation:
y'' + 2y' + y = 0
The characteristic equation is:
r^2 + 2r + 1 = 0
The roots of the characteristic equation are r = -1 (double root).
Therefore, the complementary solution is:
y_c(x) = (C₁ + C₂x)e^(-x)
where C₁ and C₂ are constants.
The general solution is the sum of the particular and complementary solutions:
y(x) = y_c(x) + y_p(x)
= (C₁ + C₂x)e^(-x)
Therefore, y(x) = (C₁ + C₂x)e^(-x).