y%3De%5E-x-log-x.jpeg "(d^2+2d+1)y=e^-x Log X")
Solving the Differential Equation: (d^2+2d+1)y=e^-x log x
In this article, we will solve the differential equation (d^2+2d+1)y=e^-x log x. This differential equation is a second-order linear ordinary differential equation, and we will use a combination of techniques to find its general solution.
Step 1: Find the Homogeneous Solution
The homogeneous part of the differential equation is (d^2+2d+1)y=0. This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is:
r^2 + 2r + 1 = 0
Factoring the quadratic equation, we get:
(r+1)^2 = 0
which has a repeated root r=-1. Therefore, the homogeneous solution is:
y_h = c1e^-x + c2xe^-x
where c1 and c2 are arbitrary constants.
Step 2: Find the Particular Solution
To find the particular solution, we will use the method of undetermined coefficients. Since the right-hand side of the equation is e^-x log x, we assume that the particular solution has the form:
y_p = Ae^-x log x + Be^-x
where A and B are undetermined coefficients.
Substituting y_p into the differential equation, we get:
(d^2+2d+1)(Ae^-x log x + Be^-x) = e^-x log x
Expanding and simplifying, we get:
(-Ae^-x log x - Ae^-x - Be^-x + 2Ae^-x log x + 2Be^-x + Ae^-x) = e^-x log x
Collecting like terms, we get:
(A - A - B + 2A + 2B) e^-x log x + (-A + 2B) e^-x = e^-x log x
Equating the coefficients of e^-x log x and e^-x, we get:
A - A - B + 2A + 2B = 1 -A + 2B = 0
Solving these equations, we get:
A = 1 B = 1/2
Therefore, the particular solution is:
y_p = e^-x log x + (1/2)e^-x
Step 3: Find the General Solution
The general solution is the sum of the homogeneous solution and the particular solution:
y = y_h + y_p = c1e^-x + c2xe^-x + e^-x log x + (1/2)e^-x
Simplifying, we get:
y = (c1 + 1/2)e^-x + c2xe^-x + e^-x log x
This is the general solution to the differential equation (d^2+2d+1)y=e^-x log x.
Conclusion
In this article, we have solved the differential equation (d^2+2d+1)y=e^-x log x using the method of undetermined coefficients. The general solution involves the homogeneous solution and the particular solution, and it provides the complete description of the behavior of the system modeled by this differential equation.
