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Solving the Differential Equation: (1-x^2)y'' - 2xy' + 2y = 0
This article will discuss the solution to the second-order linear homogeneous differential equation:
(1-x^2)y'' - 2xy' + 2y = 0
This equation is a special case of Legendre's differential equation, which arises in various areas of physics and mathematics, including potential theory, quantum mechanics, and the study of special functions.
Recognizing the Form and Solution
The equation is a Legendre's equation because it can be rewritten in the form:
(1-x^2)y'' - 2xy' + n(n+1)y = 0
Where n = 1 in our case. The solutions to Legendre's equation are known as Legendre polynomials, which are a set of orthogonal polynomials defined by the following Rodrigues formula:
P_n(x) = (1/2^n n!) d^n/dx^n (x^2 - 1)^n
Therefore, the general solution to our differential equation is:
y(x) = c_1 P_1(x) + c_2 P_0(x)
where c_1 and c_2 are arbitrary constants.
Finding the Legendre Polynomials
We can find the specific Legendre polynomials needed for our solution by applying the Rodrigues formula:
- P_0(x) = 1
- P_1(x) = x
General Solution
Substituting these polynomials into our general solution, we get:
y(x) = c_1 x + c_2
This is the general solution to the given differential equation. It represents a family of linear functions, where the constants c_1 and c_2 determine the slope and intercept of each function.
Conclusion
This article demonstrates how to solve the differential equation (1-x^2)y'' - 2xy' + 2y = 0 by recognizing its connection to Legendre's equation and using the known solutions, Legendre polynomials. The general solution is a linear function, highlighting the relationship between differential equations and their corresponding families of solutions.