## Solving the Differential Equation (x^2+1)y'' + xy' - y = 0 using Power Series

This article explores the solution of the second-order linear differential equation (x^2+1)y'' + xy' - y = 0 using the power series method. This technique allows us to find a solution in the form of an infinite series, which can be particularly useful when dealing with equations that lack elementary function solutions.

### The Power Series Approach

We assume a solution of the form:

y(x) = ∑_(n=0)^∞ a_n x^n = a_0 + a_1x + a_2x^2 + ...

where the coefficients a_n are constants to be determined.

**Step 1: Finding Derivatives**

Calculate the first and second derivatives of y(x):

y'(x) = ∑_(n=1)^∞ na_n x^(n-1) = a_1 + 2a_2x + 3a_3x^2 + ...

y''(x) = ∑_(n=2)^∞ n(n-1)a_n x^(n-2) = 2a_2 + 6a_3x + 12a_4x^2 + ...

**Step 2: Substituting into the Differential Equation**

Substitute y(x), y'(x), and y''(x) into the original differential equation:

(x^2+1) ∑*(n=2)^∞ n(n-1)a_n x^(n-2) + x ∑*(n=1)^∞ na_n x^(n-1) - ∑_(n=0)^∞ a_n x^n = 0

**Step 3: Simplifying and Shifting Indices**

Simplify the equation by multiplying and adjusting the indices to match the powers of x:

∑*(n=2)^∞ n(n-1)a_n x^n + ∑*(n=2)^∞ n(n-1)a_n x^(n-2) + ∑*(n=1)^∞ na_n x^n - ∑*(n=0)^∞ a_n x^n = 0

Now, shift the indices to make all powers of x match:

∑*(n=0)^∞ (n+2)(n+1)a*(n+2) x^n + ∑*(n=0)^∞ (n+2)(n+1)a*(n+2) x^n + ∑*(n=0)^∞ (n+1)a*(n+1) x^n - ∑_(n=0)^∞ a_n x^n = 0

**Step 4: Combining Terms and Recurrence Relation**

Combine the coefficients of the same powers of x:

∑*(n=0)^∞ [(n+2)(n+1)a*(n+2) + (n+2)(n+1)a_(n+2) + (n+1)a_(n+1) - a_n] x^n = 0

Since this equation must hold for all values of x, the coefficient of each power of x must be zero. This gives us the following recurrence relation:

(n+2)(n+1)a_(n+2) + (n+2)(n+1)a_(n+2) + (n+1)a_(n+1) - a_n = 0

Simplifying, we get:

a_(n+2) = (a_n - (n+1)a_(n+1)) / (2(n+2)(n+1))

**Step 5: Finding the Solution**

Using the recurrence relation and initial conditions (a_0 and a_1, which are arbitrary constants), we can calculate the coefficients a_n for all n. This allows us to construct the power series solution:

y(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + ...

**Important Note:** This power series solution may converge only for a specific range of x values, determined by the ratio test or other convergence tests.

### Example: Finding the First Few Terms

Let's assume a_0 = 1 and a_1 = 0 for simplicity. Using the recurrence relation, we can calculate:

- a_2 = (1 - 1
*0) / (2*2*1) = 1/4 - a_3 = (0 - 2
*1/4) / (2*3*2) = -1/24 - a_4 = (1/4 - 3*(-1/24)) / (2
*4*3) = 1/48 - ...

Therefore, the first few terms of the power series solution are:

y(x) = 1 + 0x + (1/4)x^2 - (1/24)x^3 + (1/48)x^4 + ...

This series represents a solution of the differential equation (x^2+1)y'' + xy' - y = 0.

### Conclusion

The power series method provides a powerful tool for solving a wide variety of differential equations, including those that do not have elementary function solutions. By finding a solution in the form of an infinite series, we can often obtain a representation of the solution that is valid over a specific range of values. This technique is particularly useful for analyzing the behavior of solutions near singular points or for obtaining approximations of solutions in specific regions.