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Solving the Differential Equation (1-x^2)y'' - 2xy' + 2y = 0 using Power Series
This article will explore the solution of the second-order linear differential equation (1-x^2)y'' - 2xy' + 2y = 0 using the power series method. This method involves assuming the solution can be expressed as an infinite series and then finding the coefficients of that series.
1. Power Series Representation
Let's assume the solution to the differential equation can be expressed as a power series:
y(x) = ∑_(n=0)^∞ a_n x^n
where a_n are the coefficients to be determined.
2. Finding the Derivatives
We need the first and second derivatives of y(x):
y'(x) = ∑_(n=1)^∞ n a_n x^(n-1)
y''(x) = ∑_(n=2)^∞ n(n-1) a_n x^(n-2)
3. Substituting into the Differential Equation
Substituting these expressions into the original differential equation gives:
(1-x^2) ∑(n=2)^∞ n(n-1) a_n x^(n-2) - 2x ∑(n=1)^∞ n a_n x^(n-1) + 2 ∑_(n=0)^∞ a_n x^n = 0
4. Manipulating the Series
We need to manipulate the series so that they all have the same power of x. To achieve this, we reindex the series:
- For the first term: Replace n with n+2, so the power of x becomes x^n.
- For the second term: Replace n with n+1, so the power of x becomes x^n.
The equation now becomes:
∑(n=0)^∞ (n+2)(n+1) a(n+2) x^n - ∑(n=0)^∞ (n+1) a(n+1) x^(n+1) - ∑(n=0)^∞ 2n a_n x^(n+1) + ∑(n=0)^∞ 2 a_n x^n = 0
5. Combining Terms and Solving the Recurrence Relation
Combine the terms with the same power of x and set the coefficient of each power of x to zero. This gives us the following recurrence relation:
(n+2)(n+1) a_(n+2) - (n+1) a_(n+1) - 2n a_n + 2 a_n = 0
Simplifying the equation, we get:
(n+2)(n+1) a_(n+2) - (n+1) a_(n+1) - 2(n-1) a_n = 0
This equation allows us to find the coefficients a_(n+2) in terms of a_(n+1) and a_n.
6. Solving for the Coefficients
To find the explicit form of the solution, we need to solve the recurrence relation and determine the initial coefficients a_0 and a_1. The recurrence relation can be rewritten as:
a_(n+2) = ( (n+1) a_(n+1) + 2(n-1) a_n ) / ((n+2)(n+1))
Starting with n=0, we can find a_2, a_4, a_6,... in terms of a_0. Similarly, starting with n=1, we can find a_3, a_5, a_7,... in terms of a_1.
7. General Solution
The general solution of the differential equation is then expressed as a linear combination of two linearly independent solutions:
y(x) = a_0 (1 + a_2 x^2 + a_4 x^4 + ...) + a_1 (x + a_3 x^3 + a_5 x^5 + ...)
where a_0 and a_1 are arbitrary constants and the coefficients a_2, a_3, a_4,... are determined by the recurrence relation.
Conclusion
The power series method provides a way to solve the given differential equation and express the solution as an infinite series. This method gives us a general solution where the coefficients of the series are determined by a recurrence relation. The power series method can be applied to a wide range of differential equations, offering a powerful tool for finding solutions.