Find the power series solution of the differential equation:

(x² -1)d²y/dx²+3xdy/dx+xy=0

in powers of x

To find the power series solution of the differential equation \((x^2 - 1) \frac{d^2y}{dx^2} + 3x \frac{dy}{dx} + xy = 0\), we will assume a solution of the form \(y(x) = \sum_{n=0}^{\infty} a_n x^n\). This approach allows us to express the solution as a power series centered at \(x = 0\). Let's break down the steps involved in solving this equation.

Step 1: Set Up the Power Series

We start by substituting the power series into the differential equation. First, we need to compute the derivatives of \(y(x)\):

• The first derivative is given by:

  \(\frac{dy}{dx} = \sum_{n=1}^{\infty} n a_n x^{n-1}\)

• The second derivative is:

  \(\frac{d^2y}{dx^2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\)

Step 2: Substitute into the Differential Equation

Next, we substitute these derivatives back into the original equation:

\((x^2 - 1) \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} + 3x \sum_{n=1}^{\infty} n a_n x^{n-1} + x \sum_{n=0}^{\infty} a_n x^n = 0\)

Now, we can expand each term:

• For \((x^2 - 1) \frac{d^2y}{dx^2}\):

  \((x^2 - 1) \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty} n(n-1) a_n x^n - \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\)

• For \(3x \frac{dy}{dx}\):

  \(3x \sum_{n=1}^{\infty} n a_n x^{n-1} = 3 \sum_{n=1}^{\infty} n a_n x^n\)

• For \(xy\):

  \(x \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+1} = \sum_{n=1}^{\infty} a_{n-1} x^n\)

Step 3: Combine and Rearrange

Now we combine all these series into a single series:

\(\sum_{n=2}^{\infty} n(n-1) a_n x^n - \sum_{n=0}^{\infty} n(n-1) a_n x^{n-2} + 3 \sum_{n=1}^{\infty} n a_n x^n + \sum_{n=1}^{\infty} a_{n-1} x^n = 0\)

To align the indices, we can shift the second sum:

\(- \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = - \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n\)

Step 4: Form the Recurrence Relation

Now we can combine all the series to form a single series in terms of \(x^n\):

\(\sum_{n=0}^{\infty} \left[n(n-1) a_n + 3n a_n + a_{n-1} - (n+2)(n+1) a_{n+2}\right] x^n = 0\)

For this equation to hold for all \(x\), the coefficients of each power of \(x^n\) must equal zero:

\(n(n-1) a_n + 3n a_n + a_{n-1} - (n+2)(n+1) a_{n+2} = 0\)

Step 5: Solve the Recurrence Relation

This gives us a recurrence relation that we can use to find the coefficients \(a_n\). Rearranging, we have:

\(a_{n+2} = \frac{n(n+3) a_n + a_{n-1}}{(n+2)(n+1)}\)

Starting with initial conditions, typically \(a_0\) and \(a_1\), we can compute subsequent coefficients. For instance:

• If \(a_0 = 1\) and \(a_1 = 0\), we can find \(a_2\), \(a_3\), and so on.
• Continuing this process will yield the coefficients for the power series solution.

Final Thoughts

By following these steps, you can systematically derive the power series solution to the given differential equation. The key is to carefully manage the series expansions and ensure that all terms are correctly aligned to form a valid recurrence relation. This method not only provides a solution but also deepens your understanding of how power series can be applied to solve differential equations.