5.2 — Frobenius Method (Regular Singular Points)

Solution

Step 1: Check that $x = 0$ is a regular singular point

Divide by $2x$:

$$y'' + \frac{1}{2x}y' + \frac{1}{2x}y = 0 \quad\Longrightarrow\quad p(x) = \frac{1}{2x},\qquad q(x) = \frac{1}{2x}$$ $$xp(x) = \frac{1}{2}, \qquad x^2q(x) = \frac{x}{2}$$

Both are analytic at $x = 0$, so $x = 0$ is a regular singular point with $p_0 = \tfrac{1}{2}$ and $q_0 = 0$.

Step 2: Indicial equation

$$r(r-1) + \tfrac{1}{2}r + 0 = 0 \quad\Longrightarrow\quad r\left(r - \tfrac{1}{2}\right) = 0$$ $$r_1 = \frac{1}{2}, \qquad r_2 = 0, \qquad r_1 - r_2 = \frac{1}{2} \notin \mathbb{Z}$$

This is Case 1: two independent Frobenius series, no logarithm.

Step 3: Substitute $y = \sum_{n\ge 0} a_n x^{n+r}$ and build the recurrence

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

Shift the index in the last sum so every term carries $x^{n+r-1}$, then equate coefficients:

$$a_n\Big[2(n+r)(n+r-1) + (n+r)\Big] + a_{n-1} = 0$$ $$\boxed{\;a_n\,(n+r)\big(2n+2r-1\big) = -a_{n-1},\qquad n \ge 1\;}$$

The $n = 0$ term reproduces the indicial equation $a_0\,r(2r-1) = 0$, confirming $r = \tfrac{1}{2},\,0$.

Step 4: First solution, $r_1 = \tfrac{1}{2}$

Here $(n+\tfrac{1}{2})(2n) = n(2n+1)$, so $a_n = -\dfrac{a_{n-1}}{n(2n+1)}$. With $a_0 = 1$:

$$a_1 = -\frac{1}{3},\qquad a_2 = \frac{1}{30},\qquad a_3 = -\frac{1}{630}$$ $$y_1(x) = x^{1/2}\left(1 - \frac{x}{3} + \frac{x^2}{30} - \frac{x^3}{630} + \cdots\right)$$

Step 5: Second solution, $r_2 = 0$

Now the bracket is $n(2n-1)$, so $a_n = -\dfrac{a_{n-1}}{n(2n-1)}$. With $a_0 = 1$:

$$a_1 = -1,\qquad a_2 = \frac{1}{6},\qquad a_3 = -\frac{1}{90}$$ $$y_2(x) = 1 - x + \frac{x^2}{6} - \frac{x^3}{90} + \cdots$$

General solution:

$$\boxed{y(x) = c_1\,x^{1/2}\left(1 - \frac{x}{3} + \frac{x^2}{30} - \cdots\right) + c_2\left(1 - x + \frac{x^2}{6} - \cdots\right)}$$

Solution

Step 1: Indicial equation

Dividing by $x$ gives $p(x) = 1/x$ and $q(x) = -1/x$, so $xp = 1$ and $x^2q = -x$ are analytic: $x = 0$ is regular singular with $p_0 = 1$, $q_0 = 0$.

$$r(r-1) + r = r^2 = 0 \quad\Longrightarrow\quad r_1 = r_2 = 0$$

A repeated root — Case 2 — so the second solution must contain $\ln x$.

Step 2: Recurrence and first solution

With $y = \sum a_n x^{n+r}$ the coefficient of $x^{n+r-1}$ gives

$$a_n(n+r)^2 = a_{n-1} \quad\xrightarrow{\;r\,=\,0\;}\quad a_n = \frac{a_{n-1}}{n^2}$$

Taking $a_0 = 1$ gives $a_n = \dfrac{1}{(n!)^2}$, so

$$y_1(x) = \sum_{n=0}^{\infty}\frac{x^n}{(n!)^2} = 1 + x + \frac{x^2}{4} + \frac{x^3}{36} + \frac{x^4}{576} + \cdots$$

Step 3: Second solution with the logarithm

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

Substituting, the $\ln x$ terms cancel automatically (because $y_1$ solves the equation) and what remains is

$$2y_1'(x) + \sum_{n\ge 1}\Big[n^2 b_n - b_{n-1}\Big]x^{n-1} = 0 \quad\Longrightarrow\quad \boxed{\;n^2 b_n = b_{n-1} - 2n\,a_n\;}$$

with $b_0 = 0$. Running it:

$$b_1 = -2a_1 = -2,\qquad 4b_2 = b_1 - 4a_2 = -3 \Rightarrow b_2 = -\frac{3}{4},$$ $$9b_3 = b_2 - 6a_3 = -\frac{11}{12} \Rightarrow b_3 = -\frac{11}{108},\qquad b_4 = -\frac{25}{3456}$$

In closed form $b_n = -\dfrac{2H_n}{(n!)^2}$, where $H_n = 1 + \tfrac{1}{2} + \cdots + \tfrac{1}{n}$ is the $n$-th harmonic number:

$$\boxed{y_2(x) = y_1(x)\ln x - 2\sum_{n=1}^{\infty}\frac{H_n}{(n!)^2}x^n = y_1(x)\ln x - \left(2x + \frac{3x^2}{4} + \frac{11x^3}{108} + \cdots\right)}$$

Solution

Step 1: Indicial equation

Here $p(x) = 0$ and $q(x) = -1/x$, so $xp = 0$ and $x^2q = -x$: regular singular with $p_0 = q_0 = 0$.

$$r(r-1) = 0 \quad\Longrightarrow\quad r_1 = 1,\quad r_2 = 0,\qquad r_1 - r_2 = 1 \in \mathbb{Z}^+$$

This is Case 3. Always start from the larger root.

Step 2: Larger root $r_1 = 1$

The coefficient of $x^{n+r-1}$ gives $a_n(n+r)(n+r-1) = a_{n-1}$, so with $r = 1$:

$$a_n = \frac{a_{n-1}}{n(n+1)} \quad\Longrightarrow\quad a_n = \frac{1}{n!\,(n+1)!}$$ $$y_1(x) = x + \frac{x^2}{2} + \frac{x^3}{12} + \frac{x^4}{144} + \cdots = \sum_{n=0}^{\infty}\frac{x^{n+1}}{n!\,(n+1)!}$$

Step 3: Try the smaller root $r_2 = 0$ — and watch it fail

With $r = 0$ the recurrence reads $a_n\,n(n-1) = a_{n-1}$. At the critical index $n = r_1 - r_2 = 1$:

$$a_1\cdot 1\cdot 0 = a_0 \quad\Longrightarrow\quad 0 = a_0,$$

contradicting $a_0 \ne 0$. No pure Frobenius series exists at $r_2$, so the logarithm is forced.

Step 4: Logarithmic second solution

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

Substituting and matching the coefficient of $x^k$ gives

$$\boxed{\;k(k+1)\,b_{k+1} = b_k - (2k+1)a_k\;}$$

At $k = 0$ this forces $b_0 = a_0 = 1$; $b_1$ is free (it only adds a multiple of $y_1$), so take $b_1 = 0$. Then

$$2b_2 = b_1 - 3a_1 = -\frac{3}{2} \Rightarrow b_2 = -\frac{3}{4},\qquad 6b_3 = b_2 - 5a_2 = -\frac{7}{6} \Rightarrow b_3 = -\frac{7}{36},\qquad b_4 = -\frac{35}{1728}$$ $$\boxed{y_2(x) = y_1(x)\ln x + 1 - \frac{3}{4}x^2 - \frac{7}{36}x^3 - \frac{35}{1728}x^4 - \cdots}$$

Solution

Step 1: Put the equation in standard form

$$y'' + \underbrace{\frac{x-2}{(x-2)^2(x+2)^2}}_{p(x)}y' + \underbrace{\frac{1}{(x-2)^2(x+2)^2}}_{q(x)}y = 0, \qquad p(x) = \frac{1}{(x-2)(x+2)^2}$$

Both $x = 2$ and $x = -2$ are singular points.

Step 2: Test $x = 2$

$$(x-2)p(x) = \frac{1}{(x+2)^2}, \qquad (x-2)^2q(x) = \frac{1}{(x+2)^2}$$

Both are analytic at $x = 2$ (value $\tfrac{1}{16}$), so $x = 2$ is a regular singular point, with $p_0 = q_0 = \tfrac{1}{16}$.

Step 3: Test $x = -2$

$$(x+2)p(x) = \frac{1}{(x-2)(x+2)}$$

This blows up as $x \to -2$, so it is not analytic there.

$$\boxed{x = 2 \text{ is regular singular};\qquad x = -2 \text{ is irregular singular}}$$

Solution

Step 1: Verify the singular point and read off $p_0, q_0$

Dividing by $3x$: $p(x) = \dfrac{1}{3x}$, $q(x) = -\dfrac{1}{3x}$, so $xp = \tfrac{1}{3}$ and $x^2q = -\tfrac{x}{3}$ are analytic: regular singular, $p_0 = \tfrac{1}{3}$, $q_0 = 0$.

Step 2: Indicial equation

$$r(r-1) + \tfrac{1}{3}r = 0 \quad\Longrightarrow\quad r\left(r - \tfrac{2}{3}\right) = 0 \quad\Longrightarrow\quad r_1 = \frac{2}{3},\; r_2 = 0$$

Since $r_1 - r_2 = \tfrac{2}{3} \notin \mathbb{Z}$, this is Case 1: two Frobenius series, no logarithm.

Step 3: Recurrence

Substituting $y = \sum a_n x^{n+r}$ and collecting the coefficient of $x^{n+r-1}$:

$$a_n\Big[3(n+r)(n+r-1) + (n+r)\Big] = a_{n-1} \quad\Longrightarrow\quad a_n(n+r)(3n+3r-2) = a_{n-1}$$

Step 4: Both solutions

For $r_1 = \tfrac{2}{3}$: $a_n = \dfrac{a_{n-1}}{n(3n+2)}$, giving $a_1 = \tfrac{1}{5}$, $a_2 = \tfrac{1}{80}$, $a_3 = \tfrac{1}{2640}$.

For $r_2 = 0$: $a_n = \dfrac{a_{n-1}}{n(3n-2)}$, giving $a_1 = 1$, $a_2 = \tfrac{1}{8}$, $a_3 = \tfrac{1}{168}$.

$$\boxed{y = c_1x^{2/3}\left(1 + \frac{x}{5} + \frac{x^2}{80} + \frac{x^3}{2640} + \cdots\right) + c_2\left(1 + x + \frac{x^2}{8} + \frac{x^3}{168} + \cdots\right)}$$

Solution

Step 1: Indicial equation

$p(x) = \dfrac{1-x}{x}$ and $q(x) = -\dfrac{1}{x}$, so $xp = 1-x$ and $x^2q = -x$ are analytic: $p_0 = 1$, $q_0 = 0$.

$$r(r-1) + r = r^2 = 0 \quad\Longrightarrow\quad r_1 = r_2 = 0 \quad(\text{Case 2})$$

Step 2: Recurrence for the first solution

Substituting $y = \sum a_n x^{n+r}$, the coefficient of $x^{n+r-1}$ gives

$$(n+r)^2a_n = (n+r-1)a_{n-1} + a_{n-1} = (n+r)a_{n-1} \quad\Longrightarrow\quad a_n = \frac{a_{n-1}}{n+r}$$

At $r = 0$: $a_n = \dfrac{a_{n-1}}{n} = \dfrac{1}{n!}$, so

$$y_1(x) = \sum_{n=0}^{\infty}\frac{x^n}{n!} = e^{x}$$

Check directly: $xe^x + (1-x)e^x - e^x = 0$. ✓

Step 3: Second solution by reduction of order

With $p(x) = \dfrac{1-x}{x}$ we get $\displaystyle\int p\,dx = \ln x - x$, so $e^{-\int p\,dx} = \dfrac{e^{x}}{x}$ and

$$y_2 = y_1\int \frac{e^{-\int p\,dx}}{y_1^2}dx = e^{x}\int \frac{e^{x}}{x\,e^{2x}}dx = e^{x}\int\frac{e^{-x}}{x}\,dx$$

Expanding $\dfrac{e^{-x}}{x} = \dfrac{1}{x} - 1 + \dfrac{x}{2} - \dfrac{x^2}{6} + \cdots$ and integrating term by term:

$$\int\frac{e^{-x}}{x}dx = \ln x + \sum_{n=1}^{\infty}\frac{(-1)^n x^n}{n\,n!} = \ln x - x + \frac{x^2}{4} - \frac{x^3}{18} + \cdots$$ $$\boxed{y_2(x) = e^{x}\ln x - x - \frac{3x^2}{4} - \frac{11x^3}{36} - \frac{25x^4}{288} - \cdots}$$

exactly the Case 2 form $y_2 = y_1\ln x + \sum_{n\ge1} b_n x^n$ with $b_1 = -1$, $b_2 = -\tfrac{3}{4}$, $b_3 = -\tfrac{11}{36}$, $b_4 = -\tfrac{25}{288}$.

Solution

Step 1: Indicial equation

$xp = 2$ and $x^2q = x^2$ are analytic, so $p_0 = 2$, $q_0 = 0$:

$$r(r-1) + 2r = r(r+1) = 0 \quad\Longrightarrow\quad r_1 = 0,\; r_2 = -1,\qquad r_1 - r_2 = 1 \in \mathbb{Z}^+$$

Case 3 — the logarithm may be needed. We must check the critical index.

Step 2: Recurrence

Collecting the coefficient of $x^{n+r-1}$ in $xy'' + 2y' + xy$:

$$a_n(n+r)(n+r+1) + a_{n-2} = 0 \quad\Longrightarrow\quad a_n = \frac{-a_{n-2}}{(n+r)(n+r+1)}$$

Step 3: The smaller root $r_2 = -1$ — the critical test

At $r = -1$ the factor is $(n-1)n$, which vanishes at $n = 1$. The equation at $n = 1$ reads

$$a_1\cdot 0\cdot 1 + a_{-1} = 0 \quad\Longrightarrow\quad 0 = 0,$$

which is automatically satisfied. So $a_1$ is free, no logarithm is needed, and the single Frobenius series at $r_2 = -1$ already carries both solutions.

Step 4: Sum the series

With $a_0 = 1$, $a_1 = 0$: $a_2 = -\tfrac{1}{2}$, $a_4 = \tfrac{1}{24}$, giving

$$y = x^{-1}\left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\right) = \frac{\cos x}{x}$$

With $a_0 = 0$, $a_1 = 1$ the odd branch gives $x^{-1}\left(x - \tfrac{x^3}{6} + \tfrac{x^5}{120} - \cdots\right) = \dfrac{\sin x}{x}$.

$$\boxed{y(x) = c_1\frac{\cos x}{x} + c_2\frac{\sin x}{x}}$$