Frobenius级数求解疑问：二阶微分方程幂级数解推导遇阻

Hey there! Let's break down this problem step by step to figure out where you might have gotten stuck, and share some practical tips for handling Frobenius series solutions with repeated roots.

First, Let's Verify the Index Equation & First Solution Recurrence #

You correctly identified the regular singularity at (x=0) and found the repeated root (r = \frac{1}{2}) from the index equation (4r(r-1) + 1 = 0). Let's rederive the recurrence relation for (y_1(x) = \sqrt{x}\sum_{k=0}^{\infty}c_kxk = \sum_{k=0}^{\infty}c_kx{k+\frac{1}{2}}) to confirm it's straightforward (you might have missed a simplification step earlier):

1. Substitute (y_1 = \sum c_kx^{k+\frac{1}{2}}), (y_1'' = \sum (k+\frac{1}{2})(k-\frac{1}{2})c_kx^{k-\frac{3}{2}}) into the original equation (4x^2y'' + (3x+1)y = 0):

   • (4x^2y'' = 4\sum (k+\frac{1}{2})(k-\frac{1}{2})c_kx^{k+\frac{1}{2}} = 4\sum (k^2 - \frac{1}{4})c_kx^{k+\frac{1}{2}})
   • ((3x+1)y_1 = 3\sum c_kx^{k+\frac{3}{2}} + \sum c_kx^{k+\frac{1}{2}} = 3\sum c_{k-1}x^{k+\frac{1}{2}} + \sum c_kx^{k+\frac{1}{2}}) (shift index for the (3x) term)

2. Combine like powers of (x):

   • For (k=0): (4(-\frac{1}{4})c_0 + c_0 = 0) (consistent with the index equation)
   • For (k \geq 1): (4(k^2 - \frac{1}{4})c_k + c_k + 3c_{k-1} = 0)

3. Simplify the recurrence:
   The (4(-\frac{1}{4})c_k + c_k) terms cancel out, leaving:
   [4k^2c_k + 3c_{k-1} = 0 \implies c_k = -\frac{3c_{k-1}}{4k^2}]

With (c_0 = 1) (arbitrary non-zero constant), the first few terms are:
[y_1(x) = \sqrt{x}\left(1 - \frac{3}{4}x + \frac{9}{64}x^2 - \frac{3}{256}x^3 + \dots\right)]

Key Tips to Avoid Sticking Points #

If you couldn't get this recurrence, the most likely issues are index-shifting errors or skipping simplification steps. Here are actionable tricks:

• Slow down index shifts: When reindexing terms (like converting (x^{k+\frac{3}{2}}) to (x^{m+\frac{1}{2}})), explicitly write out the new index (e.g., (m = k+1), so (k = m-1)) to avoid off-by-one mistakes.
• Simplify coefficients early: The ((k+\frac{1}{2})(k-\frac{1}{2})) term simplifies to (k^2 - \frac{1}{4}) via the difference of squares—this quick simplification eliminates messy fractions later.
• Validate with the first few terms: Plug the first 2-3 terms of (y_1) back into the original equation to check if they satisfy it (low-order terms should cancel out perfectly).

Handling the Second Solution (Repeated Root Case) #

Since (r_1 = r_2 = \frac{1}{2}), the second linearly independent solution requires a logarithmic term:
[y_2(x) = y_1(x)\ln x + \sqrt{x}\sum_{k=0}^{\infty}d_kxk]

This part is computationally tedious, but here's a streamlined approach:

1. Substitute (y_2) into the original equation. The (\ln x) terms will cancel out (since (y_1) is a solution), leaving an equation for the series (\sum d_kx^{k+\frac{1}{2}}).
2. Derive the recurrence for (d_k) by matching coefficients:
   [d_k = -\frac{2c_k}{k} - \frac{3d_{k-1}}{4k^2}]
   (Set (d_0 = 0) to avoid duplicating (y_1) in the solution)

For example, with (c_1 = -\frac{3}{4}):
[d_1 = -\frac{2(-\frac{3}{4})}{1} = \frac{3}{2}]
And so on for subsequent terms.

内容的提问来源于stack exchange，提问作者benny