benjamin_allen
benjamin_allen Jan 16, 2026 โ€ข 0 views

What is Case 3 in Frobenius method for differential equations?

Okay, so I'm kinda stuck on Frobenius method, specifically Case 3. ๐Ÿ˜ซ I get the general idea, but when the roots are equal or differ by an integer, I always mess up the second solution. Can someone explain Case 3 in a way that actually makes sense? ๐Ÿ™
๐Ÿงฎ Mathematics

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derek_garcia Jan 7, 2026

๐Ÿ“š Understanding Case 3 of the Frobenius Method

The Frobenius method is a technique for finding series solutions to second-order linear ordinary differential equations of the form:

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

where $P(x)$, $Q(x)$, and $R(x)$ are analytic functions. The method is particularly useful when dealing with equations that have regular singular points. Case 3 arises when the roots of the indicial equation are equal or differ by an integer.

๐Ÿ“œ History and Background

The Frobenius method is named after Ferdinand Georg Frobenius. He developed this method in the late 19th century to solve differential equations with variable coefficients, expanding upon earlier work by Fuchs. The method provides a systematic way to find solutions near regular singular points, which are points where the coefficients of the differential equation become singular in a specific way.

๐Ÿ”‘ Key Principles of Case 3

  • ๐Ÿ” Indicial Equation: First, we assume a solution of the form $y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}$, where $r$ is a constant to be determined. Substituting this into the differential equation and solving for $r$ leads to the indicial equation.
  • ๐Ÿ”ข Roots of the Indicial Equation: Case 3 occurs when the roots $r_1$ and $r_2$ of the indicial equation are equal ($r_1 = r_2$) or differ by an integer ($r_1 - r_2 = N$, where $N$ is an integer).
  • โœจ First Solution: The first solution, $y_1(x)$, is found as usual using the larger root (or the only root if they are equal).
  • ๐Ÿ’ก Second Solution: Finding the second linearly independent solution, $y_2(x)$, is more complicated. The form of $y_2(x)$ depends on whether the roots are equal or differ by an integer.

โž— Equal Roots ($r_1 = r_2 = r$)

If the roots are equal, the second solution has the form:

$y_2(x) = y_1(x) \ln(x) + \sum_{n=1}^{\infty} b_n x^{n+r}$

  • ๐Ÿ“ Steps:
    • ๐Ÿงช Substitute $y_2(x)$ into the original differential equation.
    • โž— Solve for the coefficients $b_n$. This often involves recurrence relations.

โž– Roots Differ by an Integer ($r_1 - r_2 = N$)

If the roots differ by an integer, the second solution has the form:

$y_2(x) = C y_1(x) \ln(x) + \sum_{n=0}^{\infty} b_n x^{n+r_2}$

where $C$ is a constant that may be zero.

  • ๐Ÿ“ Steps:
    • ๐Ÿงช Substitute $y_2(x)$ into the original differential equation.
    • โž— Solve for the coefficients $b_n$ and the constant $C$. Sometimes, $C$ turns out to be zero, simplifying the solution.

๐ŸŒ Real-world Examples

Consider the differential equation:

$x^2 y'' + x y' + x^2 y = 0$

This is a Bessel equation of order zero. The indicial equation is $r^2 = 0$, so $r_1 = r_2 = 0$. The two linearly independent solutions are $J_0(x)$ and $Y_0(x)$, where $Y_0(x)$ involves a logarithmic term.

๐Ÿ“ Example when roots differ by an integer

Consider the differential equation:

$x(1-x)y'' + (1-5x)y' - 4y = 0$

The indicial equation is $r^2 - 4r + 4 = 0$. Then $r = 2$. This corresponds to the case where the roots differ by an integer.

โœ… Conclusion

Case 3 of the Frobenius method requires careful consideration when the roots of the indicial equation are equal or differ by an integer. The second solution involves a logarithmic term, which can make the calculations more complex. Understanding the underlying principles and practicing with various examples is key to mastering this technique.

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