To find the Frobenius series solution to the differential equation

$x^2 y'' + x y' - 9y = 0,$

we start by checking if $x = 0$ is a regular singular point. This is done by rewriting the equation in standard form:

$y'' + \frac{1}{x} y' - \frac{9}{x^2} y = 0.$

Here, both $\frac{1}{x}$ and $\frac{-9}{x^2}$ have singularities at $x = 0$, making it a regular singular point. Thus, we can use the Frobenius method, which assumes a solution of the form:

$y(x) = x^r \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+r}.$

Step 1: Substitute $y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}$

First, we calculate $y'$ and $y''$ in terms of the series:

1. First derivative $y'$: $y' = \sum_{n=0}^{\infty} a_n (n + r) x^{n + r - 1}.$

2. Second derivative $y''$: $y'' = \sum_{n=0}^{\infty} a_n (n + r)(n + r - 1) x^{n + r - 2}.$

Step 2: Substitute $y$, $y'$, and $y''$ into the differential equation

Substitute these into the original equation:

$x^2 \sum_{n=0}^{\infty} a_n (n + r)(n + r - 1) x^{n + r - 2} + x \sum_{n=0}^{\infty} a_n (n + r) x^{n + r - 1} - 9 \sum_{n=0}^{\infty} a_n x^{n + r} = 0.$

This simplifies to:

$\sum_{n=0}^{\infty} a_n (n + r)(n + r - 1) x^{n + r} + \sum_{n=0}^{\infty} a_n (n + r) x^{n + r} - 9 \sum_{n=0}^{\infty} a_n x^{n + r} = 0.$

Combine terms:

$\sum_{n=0}^{\infty} a_n \left[(n + r)(n + r - 1) + (n + r) - 9 \right] x^{n + r} = 0.$

Step 3: Form the Indicial Equation

For the series to satisfy the differential equation, the coefficient of each power of $x^{n+r}$ must be zero. Focusing on the lowest power of $x$, which is $x^r$, we obtain the indicial equation by setting $n = 0$:

$a_0 \left[ r(r - 1) + r - 9 \right] = 0.$

This simplifies to:

$a_0 \left( r^2 - 9 \right) = 0.$

Thus, the indicial equation is:

$r^2 - 9 = 0.$

Solving for $r$, we get:

$r = 3 \quad \text{or} \quad r = -3.$

Step 4: Solve for the Series Coefficients $a_n$

We now have two cases for $r$: $r = 3$ and $r = -3$. We proceed by substituting each value of $r$ separately to find the corresponding series solutions.

Case 1: $r = 3$

Substitute $r = 3$ into the recurrence relation obtained from setting the coefficients of $x^{n+r}$ to zero:

$a_n \left[ (n + 3)(n + 2) - 9 \right] = 0.$

Solving this recurrence will yield one solution.

Case 2: $r = -3$

Similarly, substituting $r = -3$ will yield the second independent solution.

The final solution will be a combination of these two independent solutions, forming the general solution:

$y(x) = C_1 y_1(x) + C_2 y_2(x),$

where $y_1(x)$ and $y_2(x)$ correspond to the series solutions obtained for $r = 3$ and $r = -3$, respectively.

Would you like me to go through the steps for calculating the explicit series terms for each case?

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Related Questions:

1. How do we handle regular singular points in other types of differential equations?
2. Why do we need two solutions for a second-order differential equation?
3. Can Frobenius series solutions be applied to non-homogeneous equations?
4. How can we verify the radius of convergence of a Frobenius series?
5. What are the alternative methods to solve second-order differential equations with variable coefficients?

Tip:

When solving differential equations with series, it’s often helpful to work carefully through the indicial equation, as the roots determine the nature of the solutions (e.g., two distinct solutions, repeated solutions).