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Steps to Solve Differential Equations with Indicial Roots Not Differing by an Integer

Hey there! ๐Ÿ‘‹ Differential equations can be tricky, especially when those indicial roots don't play nice and differ by integers. It's like trying to fit puzzle pieces that almost, but don't quite, match. Let's break down the steps for the case where they *don't* differ by an integer. It's all about understanding the Frobenius method and a bit of careful algebra. Let's get started! ๐Ÿค“
๐Ÿงฎ Mathematics

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โœ… Best Answer

๐Ÿ“š Introduction to Solving Differential Equations with Indicial Roots Not Differing by an Integer

When faced with a second-order linear ordinary differential equation of the form:

$P(x)y'' + Q(x)y' + R(x)y = 0$

where $P(x)$, $Q(x)$, and $R(x)$ are polynomials, and particularly when $x=0$ is a regular singular point, the Frobenius method comes to the rescue. A regular singular point means that $(x-x_0)\frac{Q(x)}{P(x)}$ and $(x-x_0)^2\frac{R(x)}{P(x)}$ are analytic at $x=x_0$. When the indicial roots (roots of the indicial equation) do not differ by an integer, the process simplifies considerably. Let's explore the steps:

๐Ÿ“œ History and Background

The Frobenius method, named after Ferdinand Georg Frobenius, provides a way to find an infinite series solution to a second-order ordinary differential equation. This method is crucial when dealing with equations that do not have elementary solutions. The study of singular points and their impact on solutions dates back to the 19th century, laying the foundation for modern differential equation theory.

๐Ÿ”‘ Key Principles

  • ๐Ÿ” Identify a Regular Singular Point: Determine if the equation has a regular singular point, typically at $x=0$. This involves checking the analyticity of $(x-x_0)\frac{Q(x)}{P(x)}$ and $(x-x_0)^2\frac{R(x)}{P(x)}$ at $x=x_0$.
  • ๐Ÿงฉ Assume a Frobenius Series Solution: Postulate a solution of the form:

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

    where $r$ is a constant to be determined, and $a_n$ are coefficients.
  • โž• Compute Derivatives: Calculate the first and second derivatives of the assumed solution:

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

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

  • โœ๏ธ Substitute into the Differential Equation: Replace $y$, $y'$, and $y''$ in the original differential equation with their series representations.
  • ๐Ÿ”ข Determine the Indicial Equation: Collect terms with the lowest power of $x$, and set the coefficient of that term to zero. This will give you the indicial equation, a quadratic equation in $r$.
  • ๐ŸŒฑ Solve for Indicial Roots: Solve the indicial equation to find the roots $r_1$ and $r_2$. If $r_1 - r_2$ is NOT an integer, proceed with the next steps.
  • ๐Ÿ”„ Find Recurrence Relation: Equate the coefficients of other powers of $x$ to zero to find a recurrence relation for the coefficients $a_n$. This relation will express $a_n$ in terms of previous coefficients.
  • ๐Ÿ’ก Compute Coefficients: Use the recurrence relation to find the coefficients $a_n$ for each root $r_1$ and $r_2$. Since $r_1 - r_2$ is not an integer, you'll get two linearly independent solutions directly.
  • โœ… Write the General Solution: Construct the general solution by taking a linear combination of the two linearly independent solutions:

    $y(x) = c_1 y_1(x) + c_2 y_2(x)$

    where $y_1(x)$ corresponds to the solution found using $r_1$, and $y_2(x)$ corresponds to the solution found using $r_2$, and $c_1$ and $c_2$ are arbitrary constants.

๐Ÿงช Real-World Example

Consider the differential equation:

$2x^2y'' - xy' + (1+x)y = 0$

Here, $P(x) = 2x^2$, $Q(x) = -x$, and $R(x) = 1+x$. Then we assume a solution of the form $y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}$. After substituting and collecting terms, the indicial equation is found from the lowest power of $x$:

$2r(r-1) - r + 1 = 0 \implies 2r^2 - 3r + 1 = 0$

The roots are $r_1 = 1$ and $r_2 = \frac{1}{2}$. Since $1 - \frac{1}{2} = \frac{1}{2}$ is not an integer, we can find two linearly independent solutions directly using these roots and solving for the recurrence relation. The solutions are:

$y_1(x) = x + \frac{2}{5}x^2 + ...$

$y_2(x) = x^{\frac{1}{2}} + \frac{1}{3}x^{\frac{3}{2}} + ...$

The general solution is then:

$y(x) = c_1 y_1(x) + c_2 y_2(x)$

๐ŸŒ Conclusion

Solving differential equations when indicial roots do not differ by an integer involves a systematic application of the Frobenius method. By carefully identifying the regular singular point, assuming a series solution, finding the indicial roots, and computing the recurrence relation, you can construct the general solution. This approach provides a powerful tool for tackling a wide range of differential equations encountered in physics, engineering, and other scientific disciplines.

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