Solve the following ODE using the power series method
(1-x^2)y"-2xy'+2y=0
To solve the given ordinary differential equation (ODE) using the power series method, we will assume that the solution can be represented as a power series in terms of x. Let's assume the following power series solution:
y(x) = ∑[n=0 to ∞] a_n * x^n.
Then, we can find the derivatives of y(x) and substitute them into the ODE to determine the recurrence relation for the coefficients a_n.
First, let's find the derivatives of y(x) with respect to x:
y'(x) = ∑[n=0 to ∞] a_n * n * x^(n-1),
y''(x) = ∑[n=0 to ∞] a_n * n * (n-1) * x^(n-2).
Next, substitute these derivatives into the ODE:
(1 - x^2) * ∑[n=0 to ∞] a_n * n * (n-1) * x^(n-2) - 2x * ∑[n=0 to ∞] a_n * n * x^(n-1) + 2 * ∑[n=0 to ∞] a_n * x^n = 0.
Now, let's simplify this equation and collect the terms with the same power of x together:
∑[n=0 to ∞] a_n * [n * (n-1) * x^(n-2) - (n+2) * a_n * x^(n-1) + 2 * a_n * x^n] = 0.
Let's analyze each of the terms inside the summation separately:
Term with x^(n-2):
For n >= 2, the term becomes: a_n * n * (n-1) * x^(n-2).
Term with x^(n-1):
For n >= 1, the term becomes: -a_(n+2) * (n+2) * x^(n-1).
Term with x^n:
The term becomes: 2 * a_n * x^n.
Equating the above expressions to zero, we get the following recurrence relation for a_n:
a_n * [n * (n-1) - (n+2) * a_(n+2) + 2] = 0.
This equation holds for each value of n.
Now, we can solve this recurrence relation to find the values of a_n.
For n = 0:
a_0 * [0 * (-1) - (0+2) * a_2 + 2] = 0,
-2 * a_2 + 2 * a_0 = 0,
a_2 = a_0.
For n > 0:
a_n * [n * (n-1) - (n+2) * a_(n+2) + 2] = 0,
n * (n-1) - (n+2) * a_(n+2) + 2 = 0,
a_(n+2) = (n * (n-1) + 2) / (n+2).
Using this recurrence relation, we can calculate the values of a_n for different values of n.
Once we have the coefficients a_n, we can substitute them back into the power series solution y(x) = ∑[n=0 to ∞] a_n * x^n to obtain the final solution to the ODE.