Answer:
To solve the given ordinary differential equation (ODE): y‴ - 3y″ + 3y′ - y = 2ex(x - 2), where x > 0, we can use the method of undetermined coefficients.
Homogeneous Solution:
First, let's find the homogeneous solution by assuming y = e^(rx), where r is a constant.
Substituting this into the ODE, we get the characteristic equation:
r³ - 3r² + 3r - 1 = 0
Factoring the equation, we have:
(r - 1)³ = 0
This gives us a triple root r = 1.
Therefore, the homogeneous solution is:
y_h = (C₁ + C₂x + C₃x²)e^x, where C₁, C₂, and C₃ are constants.
Particular Solution:
Next, we find a particular solution for the non-homogeneous term 2ex(x - 2). Since this term is a polynomial multiplied by an exponential function, we assume the particular solution has the form:
y_p = Ax²ex + Bxex
Now, we can calculate the derivatives of y_p:
y_p' = (2Ax + Bx²)ex + (Ax² + Bx)ex
y_p'' = (2A + 2Bx + 2Ax² + Bx²)ex + (2Ax + Bx²)ex
y_p''' = (2B + 6Ax + 6Bx² + 6Ax²)ex + (2A + 2Bx + 2Ax² + Bx²)ex
Substituting these derivatives into the ODE, we have:
(2B + 6Ax + 6Bx² + 6Ax²)ex + (2A + 2Bx + 2Ax² + Bx²)ex - 3[(2A + 2Bx + 2Ax² + Bx²)ex + (Ax² + Bx)ex] + (Ax² + Bx)ex - (Ax² + Bxex) = 2ex(x - 2)
Simplifying and grouping like terms, we get:
(6B - 3B - 2A + A)x²ex + (6A - 3A + B - B)xex + (2B + 2A)ex = 2ex(x - 2)
Comparing the coefficients of like terms, we have the following equations:
6B - 3B - 2A + A = 0 --> 3B - A = 0
6A - 3A + B - B = 0 --> 3A = 0
2B + 2A = 2 --> B + A = 1
From the second equation, we have A = 0.
Substituting this into the first and third equations, we get B = 0 and B + A = 1, respectively.
Therefore, we find that A = 0 and B = 1.
Hence, the particular solution is:
y_p = xex
General Solution:
The general solution is the sum of the homogeneous and particular solutions:
y = y_h + y_p = (C₁ + C₂x + C₃x²)e^x + xex
So, the general solution to the given ODE is:
y = (C₁ + C₂x + C₃x²)e^x + xex, where C₁, C₂, and C₃ are arbitrary constants.
Explanation: