y%26%2339%3B%26%2339%3B-2xy%26%2339%3B%2B2y%3D0-power-series.jpeg "(1-x^2)y''-2xy'+2y=0 Power Series")
Solving the Differential Equation (1-x^2)y''-2xy'+2y=0 with Power Series
This article explores the solution of the differential equation (1-x^2)y''-2xy'+2y=0 using the power series method. This method is particularly useful for solving differential equations where finding an explicit solution using standard techniques might be challenging.
Understanding the Power Series Method
The power series method involves expressing the solution y(x) as an infinite series:
y(x) = a<sub>0</sub> + a<sub>1</sub>x + a<sub>2</sub>x<sup>2</sup> + a<sub>3</sub>x<sup>3</sup> + ... = ∑<sub>n=0</sub><sup>∞</sup> a<sub>n</sub>x<sup>n</sup>
where a<sub>n</sub> are constants to be determined.
We then differentiate y(x) twice to obtain y' and y'' and substitute these expressions back into the original differential equation. This will result in an equation involving the coefficients a<sub>n</sub>. By solving for these coefficients, we obtain the power series solution of the differential equation.
Applying the Method to the Differential Equation
Let's apply this method to our equation:
-
Find y' and y'':
y' = a<sub>1</sub> + 2a<sub>2</sub>x + 3a<sub>3</sub>x<sup>2</sup> + ... = ∑<sub>n=1</sub><sup>∞</sup> na<sub>n</sub>x<sup>n-1</sup>
y'' = 2a<sub>2</sub> + 6a<sub>3</sub>x + 12a<sub>4</sub>x<sup>2</sup> + ... = ∑<sub>n=2</sub><sup>∞</sup> n(n-1)a<sub>n</sub>x<sup>n-2</sup>
-
Substitute y, y', and y'' into the equation:
(1-x<sup>2</sup>)∑<sub>n=2</sub><sup>∞</sup> n(n-1)a<sub>n</sub>x<sup>n-2</sup> - 2x∑<sub>n=1</sub><sup>∞</sup> na<sub>n</sub>x<sup>n-1</sup> + 2∑<sub>n=0</sub><sup>∞</sup> a<sub>n</sub>x<sup>n</sup> = 0
-
Simplify the equation:
To simplify, we need to manipulate the indices of the summations to ensure they all have the same power of x. This involves shifting the indices and adjusting the starting values accordingly:
∑<sub>n=0</sub><sup>∞</sup> (n+2)(n+1)a<sub>n+2</sub>x<sup>n</sup> - ∑<sub>n=2</sub><sup>∞</sup> n(n-1)a<sub>n</sub>x<sup>n</sup> - ∑<sub>n=1</sub><sup>∞</sup> 2na<sub>n</sub>x<sup>n</sup> + 2∑<sub>n=0</sub><sup>∞</sup> a<sub>n</sub>x<sup>n</sup> = 0
-
Combine terms with the same power of x:
(2a<sub>2</sub> + 2a<sub>0</sub>) + (6a<sub>3</sub> - 2a<sub>1</sub>)x + ∑<sub>n=2</sub><sup>∞</sup> [(n+2)(n+1)a<sub>n+2</sub> - (n(n-1) + 2n - 2)a<sub>n</sub>]x<sup>n</sup> = 0
-
Set coefficients to zero:
For the equation to hold for all values of x, each coefficient must be equal to zero:
- 2a<sub>2</sub> + 2a<sub>0</sub> = 0
- 6a<sub>3</sub> - 2a<sub>1</sub> = 0
- (n+2)(n+1)a<sub>n+2</sub> - (n(n-1) + 2n - 2)a<sub>n</sub> = 0, for n ≥ 2
-
Solve for the recurrence relation:
The last equation gives us a recurrence relation for the coefficients:
a<sub>n+2</sub> = [(n<sup>2</sup> + n - 2) / (n+2)(n+1)]a<sub>n</sub> = [(n-1)(n+2) / (n+2)(n+1)]a<sub>n</sub> = [(n-1)/(n+1)]a<sub>n</sub>
-
Determine the first few coefficients:
Using the initial conditions (a<sub>0</sub> and a<sub>1</sub>), and the recurrence relation, we can calculate the values of the coefficients. For example:
- a<sub>2</sub> = a<sub>0</sub>
- a<sub>3</sub> = (1/3)a<sub>1</sub>
- a<sub>4</sub> = (1/5)a<sub>2</sub> = (1/5)a<sub>0</sub>
- a<sub>5</sub> = (2/7)a<sub>3</sub> = (2/21)a<sub>1</sub>
-
Write the general solution:
Substitute the obtained coefficients back into the power series expression for y(x) to obtain the general solution:
y(x) = a<sub>0</sub>(1 + x<sup>2</sup> + (1/5)x<sup>4</sup> + ...) + a<sub>1</sub>(x + (1/3)x<sup>3</sup> + (2/21)x<sup>5</sup> + ...)
Conclusion
The power series method provides a powerful tool for solving differential equations that may not have readily available analytical solutions. By expressing the solution as an infinite series and following the steps outlined above, we can derive a general solution in the form of a power series. This solution can then be used to analyze the behavior of the system and understand its properties.