Math 555: Differential Equations

5.7 Bessel's Equation

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5.7.1 Bessel's Equation

Here we examine a very important differential equation named after Friedrich Wilhelm Bessel, Bessel's equation :

$$ x^2 y'' + xy' + (x^2 - \nu^2)y = 0, $$
where $\nu$ is a constant. When $x=0$, we have a regular singular point, since

$$ p_0 = \lim_{x\rightarrow 0} x\dfrac{Q(x)}{P(x)} = \lim_{x\rightarrow 0} x\dfrac{x}{x^2} = 1, $$
and
$$ q_0 = \lim_{x\rightarrow 0} x^2\dfrac{R(x)}{P(x)} = \lim_{x\rightarrow 0} x^2\dfrac{x^2-\nu^2}{x^2} = -\nu^2. $$
From here, we invoke Theorem 5.6.1 to get the indicial equation an find the values for $r$

$$ \begin{aligned} r(r-1) + p_0 r + q_0 &= 0 \\ r(r-1) + r - \nu^2 &= 0 \\ r &= \pm\nu. \end{aligned} $$
Clearly, the solution to the ODE changes based on the value of $\nu$. This value will be referred to as the order of the equation and the solutions are called Bessel functions of order $\nu$. We will look at 3 cases, $\nu = 0$, $\nu = \frac{1}{2}$, and $\nu = 1$ and the solutions for when $x \gt 0$.

5.7.2 Bessel Equation of Order Zero

For $\nu = 0$, the ODE reduces to

$$ x^2 y'' + xy' + x^2 y = 0,$$
with the indicial equation

$$ \begin{aligned} F(r) &= r(r - 1) + p_0 r + q_0 \\ &= r(r - 1) + r + 0\\ &= r^2 \end{aligned} $$
Both roots are zero, and we may represent a solution of the ODE as

$$ y_1(x) = 1 + \sum_{n=1}^\infty a_n(0)x^n,\quad x\gt 0 $$
where $a_n$ is given by the recurrence relation

$$ a_n (r) = \dfrac{-1}{F(r+n)}\sum_{k=0}^{n-1} a_k\Big((r+k)p_{n-k} + q_{n-k}\Big),\quad n\ge 1. $$

We express the coefficients here in the form

$$ a_n(r) $$
at some points. This is because the power series does in fact depend on the value for $r$. However, it is very inconvenient (and confusing) to write this for every step in the summation notation since it looks like multiplication. Therefore, it will be omitted unless a formula is being stated.

To use the techniques of the previous section, we need to understand that the power series of functions $xp(x)$ and $x^2 q(x)$ are interpreted as

$$ x p(x) = 1 = \sum_{n=0}^\infty p_n x^n,\quad x^2 q(x) = x^2 = \sum_{n=0}^\infty q_n x^n. $$
We see that $p_0 = 1$ and $q_2 = 1$, with all other $p_n,\ q_n = 0$. The coefficient $p_0$ never shows up in the recurrence relation, so we will only need to look for where $q_2$ shows up.

Now we are ready to proceed. Setting $n=1$, we can use the recurrence relation to find

$$ \begin{aligned} a_1(0) &= \dfrac{-1}{F(0+1)}\sum_{k=0}^{1-1} a_k\Big((0 + k)p_{1-k} + q_{1-k}\Big) \\ \\ &=\dfrac{-1}{F(0+1)} a_0\Big((0)p_{1} + q_{1}\Big) \\ \\ &= -\dfrac{a_0}{(0+1)^2} \cdot 0 \\ \\ &= 0 \end{aligned} $$
Setting $n=2$,
$$ \begin{aligned} a_2(0) &= \dfrac{-1}{F(2)}\sum_{k=0}^{1} a_k\Big((k)p_{2-k} + q_{2-k}\Big) \\ \\ &= \dfrac{-1}{2^2} \Big( a_0\left(0\cdot p_2 + q_2\right) + a_1\left(1\cdot p_1 + q_1\right)\Big) \\ \\ &= \dfrac{-1}{4} \Big( a_0\left(0\cdot 0 + 1\right) + 0\cdot\left(1\cdot 0 + 0\right)\Big) \\ \\ &= -\dfrac{a_0}{4} \end{aligned} $$
We can continue to use this formula to compute the $a_n$ terms, but we can shortcut the computations by looking at the general case

$$ \begin{aligned} a_n(r) &= \dfrac{-1}{F(r+n)}\sum_{k=0}^{n-1} a_k\Big((r+k)p_{n-k} + q_{n-k}\Big) \\ \\ &= \dfrac{-1}{(r+n)^2} \Big( a_0\left((r+0)\cdot p_{n-1} + q_{n-1}\right) + \ldots + a_{n-1}\left((r+n-1)\cdot p_1 + q_1\right) \Big) \\ \\ &= \dfrac{-1}{(r+n)^2} \Big( a_{n-2}q_2 \Big) \\ \\ &= -\dfrac{a_{n-2}}{(r+n)^2},\qquad n\ge 2. \end{aligned} $$
Using this formula, we see that since $a_1=0$, every odd indexed coefficient $a_3,\ a_5,\ldots = 0$. We can also relabel the index $n = 2m$ and look to establish a general term for the series

$$ \begin{aligned} a_2 &= a_{2\cdot 1} = -\dfrac{a_0}{2^2} \\ \\ a_4 &= a_{2\cdot 2} = -\dfrac{a_2}{4^2} = \dfrac{a_0}{2^6} \\ \\ a_6 &= a_{2\cdot 3} = -\dfrac{a_4}{6^2} = -\dfrac{a_0}{2^6(3\cdot 2)^2} \\ &\ \ \vdots \\ a_{2m}(0) &= \dfrac{(-1)^m a_0}{2^{2m}(m!)^2},\qquad m = 1,\ 2,\ 3,\ldots\, . \end{aligned} $$
We now have an expression for the Bessel function of the first kind of order zero , denoted as

$$ J_0 (x) = y_1(x) = 1 + \sum_{m=1}^\infty \dfrac{(-1)^m x^{2m}}{2^{2m}(m!)^2}. $$

To find a second solution to form the fundamental set, we look to case 2 of Theorem 5.6.1, since both of our exponents at the singularity are 0. Hence,

$$ y_2(x) = y_1(x)\ln\vert x\vert + \sum_{n=1}^\infty a_n'(0)x^n. $$

To determine the values of $a_n'(0)$, we utilize the expression for $a_n(r)$, differentiate it with respect to $r$, and employ the recurrence relation. First, every odd coefficient $a_{2m-1}$ will be equal to zero. This is because our computation for $a_1(r)$ gives that

$$ a_1(r) = \dfrac{a_0}{(r+1)^2}\cdot 0, $$
which isn't just zero when $r=0$, but is zero in a neighborhood of $0$. This means that $a_1'(0) = 0$ must be zero as well. Furthermore, the recurrence relation gives that every odd $a_{2m-1}'(0) = 0$ by the same reasoning.

We now move to the even coefficients, beginning with the recurrence relation

$$ a_{2m}(r) = \dfrac{(-1)^m a_0}{(r+2)^2\ldots(r+2m)^2} = (-1)^m a_0 (r+2)^{-2}\ldots \ldots(r+2m)^{-2}. $$
To find $a_{2m}'(r)$, we take advantage of a clever trick for functions of the form

$$ f(x) = (x-\alpha_1)^{\beta_1}(x-\alpha_2)^{\beta_2}\ldots (x-\alpha_n)^{\beta_n}. $$
If $x$ is not equal to any of the $\alpha_j$ values, then the logarithmic derivative of $f(x)$ is

$$ \dfrac{f'(x)}{f(x)} = \dfrac{\beta_1}{x-\alpha_1} + \dfrac{\beta_2}{x-\alpha_2} + \ldots + \dfrac{\beta_n}{x-\alpha_n}. $$
So,
$$ \dfrac{a_{2m}'(r)}{a_{2m}(r)} = -2\left(\dfrac{1}{r+2} + \dfrac{1}{r+4} + \ldots + \dfrac{1}{r+2m}\right). $$
Which allows us to set $r = 0$ to find

$$ a_{2m}'(0) = -2\left(\dfrac{1}{2} + \dfrac{1}{4} + \ldots + \dfrac{1}{2m}\right)a_{2m}(0). $$
If we define

$$ H_m = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \ldots + \dfrac{1}{m} $$
to be the m-th partial sum of the harmonic series , then

$$ a_{2m}'(0) = -H_m \dfrac{(-1)^m a_0}{2^{2m}(m!)^2}. $$
At last, this yields the formula for $y_2$:

$$ y_2(x) = J_0(x)\ln x + \sum_{m=1}^\infty \dfrac{(-1)^{m+1} H_m}{2^{2m}(m!)^2}x^{2m},\quad x \gt 0. $$
However, we are not quite finished yet. The canonical Bessel function of the second kind of order zero is typically written in a slightly different form, a linear combination of $J_0$ and $y_2$:

$$ Y_0(x) = \dfrac{2}{\pi}\left[ \left(\gamma + \ln\frac{x}{2}\right)J_0(x) + \sum_{m=1}^\infty \dfrac{(-1)^{m+1} H_m}{2^{2m}(m!)^2}x^{2m}\right],\quad x \gt 0, $$ where

$$\gamma = \lim_{n\rightarrow\infty} H_n - \ln n \cong 0.5772 $$
is the Euler-Mascheroni constant.

A general solution to the Bessel equation of order zero is a linear combination of these two functions

$$ y(x) = c_1 J_0 (x) + c_2 Y_0(x). $$
Here are the two functions plotted together:

5.7.3 Bessel Equation of Order One-Half

The computations to find the Bessel functions of order one-half and one are similar to the computations for order zero. They will be included in part, showing where relevant changes have occurred.

Set $\nu = \frac{1}{2}$ in Bessel's equation

$$ x^2 y'' + xy' + \left(x^2 - \frac{1}{4}\right) y = 0,$$
then the indicial equation is

$$\begin{aligned} F(r) &= r(r-1) + p_0 r + q_0 \\ &= r(r-1) + r - \frac{1}{4} \\ &= r^2 - \frac{1}{4}. \end{aligned} $$
We obtain the recurrence relation using the method above, noting that for the functions $xp(x)$ and $x^2 q(x)$ that

$$ x p(x) = 1 = \sum_{n=0}^\infty p_n x^n,\quad x^2 q(x) = x^2 - \frac{1}{4} = \sum_{n=0}^\infty q_n x^n. $$
Only $p_0 = 1$, $q_0 = -\frac{1}{4}$, and $q_2 = 1$ are nonzero, and only $q_2$ will appear in the recurrence relation

$$ \begin{aligned} a_n (r) &= \dfrac{-1}{F(r+n)}\sum_{k=0}^{n-1} a_k\Big((r+k)p_{n-k} + q_{n-k}\Big) \\ \\ &= \dfrac{-1}{(r+n)^2-\frac{1}{4}} \Big(a_0( (r)p_n + q_n) ) + a_1( (r+1)p_{n-1} + q_{n-1} ) + \ldots \\ \\ &\quad + a_{n-2}( (r + n-2 )p_2 + q_2) + a_{n-1}( (r + n - 1)p_1 + q_1 ) \Big) \\ \\ &= \dfrac{-1}{(r+n)^2-\frac{1}{4}} \cdot \Big( a_{n-2}q_n \Big) \\ \\ &= -\dfrac{a_{n-2}}{(r+n)^2-\frac{1}{4}} \end{aligned} $$
Setting $r = \frac{1}{2}$, this expression becomes

$$ a_n\left(\frac{1}{2}\right) = -\dfrac{a_{n-2}}{n(n+1)}. $$
The value $a_0$ is arbitrary, and $a_1 = 0$ (computed directly as in the previous example). By the recurrence relation, this implies that all odd $a_{2m+1} = 0$.

For the even coefficients $a_{2m}$,

$$ a_{2m} = -\dfrac{a_{2m-2}}{2m(2m+1)},\quad m\ge 1 $$
which has an easy to compute general term

$$ a_{2m} = \dfrac{(-1)^m a_0}{(2m+1)!},\quad m\ge 1. $$
Applying our theorem and setting $a_0 = 1$, we have that the solution $y_1$ has the form

$$ y_1(x) = x^{1/2}\left(1 + \sum_{n=1}^\infty \dfrac{(-1)^m x^{2m}}{(2m+1)!} \right) = x^{-1/2}\left(\sum_{n=0}^\infty \dfrac{(-1)^m x^{2m+1}}{(2m+1)!} \right). $$
The series in this expression is the series for $\sin x$, and this allows us to define the Bessel function of the first kind of order one-half

$$ J_{1/2}(x) = \sqrt{\dfrac{2}{\pi x}}\sin x. $$
To find the Bessel function of the second kind of order one-half, we note that the roots $\pm \frac{1}{2}$ differ by $1 = N$, a positive integer. This places us in Case 3 of Theorem 5.6.1, where

$$ y_2(x) = a J_{r_2}(x)\ln x + x^{r_2}\left(1 + \sum_{n=1}^\infty c_n\left(r_2\right) x^n\right), $$
for constants $a$ and $c_n(r_2)$ which are given by

$$ \begin{aligned} a & = \lim_{r \rightarrow r_2} \left(r-r_2\right)a_N\left(r_2\right) \\ \\ &= \lim_{r \rightarrow -1/2} \left(r+\frac{1}{2}\right)a_1\left(-\frac{1}{2}\right) \\ \\ c_n\left(r_2\right) &= \dfrac{d}{dr}\left. \left[ \left(r-r_2\right) a_n \left(r\right) \right] \right\vert_{\ r = r_2} \\ \\ c_n\left(-\frac{1}{2}\right) &= \dfrac{d}{dr}\left. \left[ \left(r+\frac{1}{2}\right) a_n \left(r\right) \right] \right\vert_{\ r = -1/2} \end{aligned} $$

To show that $a = 0$, we need only to show that $a_1(-1/2)$ is finite. Using the direct formula:

$$ \begin{aligned} a_1 (r) &= \dfrac{-1}{F(r+1)}\sum_{k=0}^{1-1} a_k\Big((r+k)p_{1-k} + q_{1-k}\Big) \\ \\ &= \dfrac{-a_0}{F(r+1)} \Big( (r)p_1 + q_1 \Big) \\ \\ &= 0, \end{aligned} $$
since $p_1 = q_1 = 0$.

We now focus on finding the values of the coefficients $c_n(r_2)$,

$$ \begin{aligned} c_n\left(r_2\right) &= \dfrac{d}{dr}\left. \left[ \left(r-r_2\right) a_n \left(r\right) \right] \right\vert_{\ r = r_2} \\ \\ &= \left.\left[ a_n(r) + (r-r_2)a_n'(r) \right] \right\vert_{\ r = r_2} \end{aligned} $$
We see that we are going to need the values of $a_n'(r)$ to compute them. To find those, we use the recurrence relations for $a_n$ to separate the coefficients into the even and odd sets of terms. Due to $a_1$ being zero, the odd terms are all likewise zero by the recurrence relation. Examining the even terms:

$$\begin{aligned} a_{2m}(r) &= \dfrac{(-1)^m a_0}{\left[ (r+2)^2 - \frac{1}{4} \right]\left[ (r+4)^2 - \frac{1}{4} \right]\ldots\left[ (r+2m)^2 - \frac{1}{4} \right]} \\ \\ &= \dfrac{(-1)^m a_0}{\left[ (r+2\cdot 1)^2 - \frac{1}{4} \right]\left[ (r+2\cdot 2)^2 - \frac{1}{4} \right]\ldots\left[ (r+2m)^2 - \frac{1}{4} \right]} \end{aligned}$$
If we look at each factor in the denominator as $(r+2k)^2 - \frac{1}{4}$ for $k = 1,\ 2\,\ldots,\ m$, it is possible to rewrite each of these as

$$ (r+2k)^2 - \frac{1}{4} = \left(r + 2k - \frac{1}{2}\right)\left(r+2k + \frac{1}{2}\right) $$
using the difference of squares. This is useful because it allows us to note that none of the roots of these quadratics will ever be $-\frac{1}{2}$ for a positive integer $k$ and sets up the logarithmic derivative trick from earlier. Hence,

$$ \dfrac{a_{2m}'(r)}{a_{2m}(r)} = -\dfrac{1}{r + \frac{3}{2}} -\dfrac{1}{r + \frac{5}{2}} -\dfrac{1}{r + \frac{7}{2}} -\dfrac{1}{r + \frac{9}{2}}- \ldots -\dfrac{1}{r + 2m - \frac{1}{2}} -\dfrac{1}{r + 2m + \frac{1}{2}}, $$
which provides an explicit formula for $a_n'(r)$. Note however, that $(r-r_2)a_n'(r)$ will be zero if $r = r_2$, since there are no factors in $a_n'(r)$'s denominator to eliminate the $r-r_2$ factor. Thus $c_n(r_2) = a_n(r_2)$.

Recalling that we set $a_0 = 1$ earlier, we have the general term

$$ c_n\left(-\frac{1}{2}\right) = \dfrac{(-1)^m }{(2m)!}. $$
Substituting this into the general form

$$ \begin{aligned} y_2(x) &= x^{-1/2}\left(1 + \sum_{n = 1}^\infty \dfrac{(-1)^m x^{2m}}{(2m)!}\right) \\ \\ &= x^{-1/2}\cos x. \end{aligned} $$
This is the basic form of the second solution. It is typical to choose $a_0 = \sqrt{\frac{2}{\pi}}$ to obtain the Bessel function of the second kind of order one-half

$$ J_{-1/2}(x) = \sqrt{\dfrac{2}{\pi x}}\cos x. $$
The general solution to the Bessel equation of order one-half is

$$ y(x) = c_1 J_{1/2}(x) + c_2 J_{-1/2}(x), $$
and their plots look like:

5.7.4 Bessel Equation of Order One

One more time with $\nu = 1$. Our equation becomes

$$ x^2 y'' + xy' + (x^2 - 1)y - 0. $$
We note that $xp(x) = 1$ and $x^2q(x) = x^2 - 1$, so $p_0 = 1$, $q_0 = -1$, and $q_2 = 1$ with all other $p_k, q_k = 0$. Therefore, the indicial equation is

$$ \begin{align*} F(r) &= r(r-1) + p_0 r + q_0 \\ &= r(r-1) + r - 1 \\ &= r^2 - 1. \end{align*} $$
The exponents at the singularity are $r_1 = 1$ and $r_2 = -1$, which differ by a positive integer. To find the recurrence relation, we start with the general form

$$ a_n(r) = \dfrac{-1}{F(r+n)} \sum_{k=0}^{n-1} a_k\left(p_{n-k}(r+k) + q_{n-k}\right) $$
and note that of our non-zero values for $p_k$ and $q_k$ only $q_2 = 1$ will appear in the recurrence relation. Therefore the only non-zero element of the sum will occur when $n-k = 2$ or when $k = n-2$. Hence, our recurrence relation simplifies greatly to

$$ a_n(r) = -\dfrac{a_{n-2}}{(r+n)^2 - 1}, \quad n\geq 2 $$
and reduces even more when setting $r = r_1 = 1$

$$ a_n = -\dfrac{a_{n-2}}{(n+2)n},\quad n\geq 2. $$
Before moving on, we note from direct computation using the recurrence equation when $n = 1$ that

$$ \begin{align*} a_1(1) &= \dfrac{-1}{F(1+1)} \sum_{k=0}^{1-1} a_k\left(p_{n-k}(r+k) + q_{n-k}\right) \\ &= \dfrac{-1}{(1+1)^2 - 1} \left( a_0\Big[ (1+0)p_1 + q_1\Big] \right) \\ &= \dfrac{-a_0}{3}\Big[1(0) + 0\Big] \\ &= 0 \end{align*} $$
which based on the recurrence relation implies that all odd-indexed $a_{2m+1} = 0$. We focus on the even-indexed coefficients and label them $n = 2m$ for $m\in\mathbb{N}$. This lets us write

$$ a_{2m} = -\dfrac{a_{2m-2}}{(2m+2)2m} = -\dfrac{a_{2m-2}}{2^2(m+1)m},\quad m \geq 1.$$
If we start writing out terms, the pattern would be pretty clear. This sequence is alternating and in the denominator we will have $2m$ factors of $2$ and the $m+1$ and $m$ factors will collect as factorials giving us

$$ a_{2m} = \dfrac{(-1)^m a_0}{2^{2m}(m+1)! m!},\quad m\geq 1.$$
We now have

$$ y_1(x) = x^1\left(1 + \sum_{m=1}^\infty \dfrac{(-1)^m x^{2m}}{2^{2m}(m+1)! m!}\right) $$
which (after choosing $a_0 = \frac{1}{2}$) is the Bessel function of the first kind of order one

$$ J_1(x) = \dfrac{x}{2}\sum_{m=0}^\infty \dfrac{(-1)^m x^{2m}}{2^{2m}(m+1)! m!}. $$
A quick check with the Ratio Test confirms this series converges absolutely for all $x$.

Next, we move to find our second fundamental solution $y_2$. We will be using $r = r_2 = -1$ in our computations and looking for a solution in the form

$$ y_2(x) = aJ_1(x)\ln|x| + |x|^{r_2}\left(1 + \displaystyle\sum_{n=1}^{\infty} c_n(r_2)x^n\right) $$
where the additional coefficients are given by

$$ c_n(r_2) = a_n(r_2) + (r-r_2)a_n'(r_2). $$
Recalling our equation from above

$$ \begin{align*} a_n(r) &= -\dfrac{a_{n-2}}{(r+n)^2 - 1} \\ \\ a_n(-1)&= -\dfrac{a_{n-2}}{(n-1)^2 - 1} \\ \\ &= -\dfrac{a_{n-2}}{(n-2)n}, \quad n\geq 2. \end{align*} $$
Based on $y_1$, it is reasonable to suspect that the odd terms will once again all be zero. We confirm this by computing

$$ a_1(r) = \dfrac{-1}{(r+1)^2 - 1}\sum_{k=0}^{1-1}a_0\left[ (r+0)p_1 + q_1 \right] = \dfrac{-a_0}{(r+1)^2 - 1}\cdot \left[ (r)0 + 0\right] = 0. $$
Thus $a_1(-1) = 0$ and also $a_1'(-1) = 0$ since the derivative of zero is zero. It follows from the recurrence relation that all odd $a_{2m+1} = 0$ and hence each $c_{2m+1} = a_{2m+1}(-1) + (r+1)a_{2m+1}(-1) = 0$.

For even values of $n = 2m$, we look at our recurrence relation

$$ \begin{align*} a_{2m}(r) &= \dfrac{-a_{n-2}}{(r+2m)^2 -1} \\ \\ &= \dfrac{(-1)^m a_0}{\left[(r+2)^2 -1\right]\left[(r+4)^2 -1\right]\ldots\left[(r+2m)^2 -1\right]} \\ \\ &= \dfrac{(-1)^m a_0}{\left[(r+1)(r+3)\right]\left[(r+3)(r+5)\right]\ldots\left[(r+2m-1)(r + 2m+1)\right]} \end{align*} $$
Here we note that every factor except $r+1$ and $r+2m+1$ appears twice. This means that it is possible to reorder terms so that

$$ a_{2m}(r) = \dfrac{(-1)^m a_0}{\color{#ec008c}{(r+1)}(r+3)\ldots(r+2m-1)\cdot(r+3)(r+5)\ldots(r+2m+1)}. $$
The $r+1$ factor in the denominator is our enemy. When $r=-1$ bad things happen, so we must do some careful algebra to make this factor vanish. The next several steps are meant to show that this factor does indeed disappear.

If we assume that $r\neq -1$ for now and apply the logarithmic derivative trick from above,

$$ \dfrac{a_{2m}'(r)}{a_{2m}(r)} = -1 \left(\dfrac{1}{r+1} + \dfrac{2}{r+3} + \dfrac{2}{r+5} + \cdots + \dfrac{2}{r+2m-1} + \dfrac{1}{r+2m+1}\right). $$
This gives an explicit formula for $a_{2m}'(r)$ if we multiply by $a_{2m}(r)$ on both sides. Next we recall the definition of $c_{2m}(r)$ and work through some algebra (I plugged in $r_2 = -1$ for convenience, but the other $r$'s are still variables for now)

$$ \begin{align*} c_{2m}(r) &= a_{2m}(r) + (r+1)a_{2m}'(r) \\ \\ &= a_{2m}(r) - (r+1)a_{2m}(r)\left(\dfrac{1}{r+1} + \dfrac{2}{r+3} + \dfrac{2}{r+5} + \cdots + \dfrac{2}{r+2m-1} + \dfrac{1}{r+2m+1}\right) \\ \\ &= a_{2m}(r)\left[1 - \left(\dfrac{r+1}{r+1} + \dfrac{2(r+1)}{r+3} + \cdots + \dfrac{2(r+1)}{r+2m-1} + \dfrac{r+1}{r+2m+1}\right)\right] \\ \\ &= -a_{2m}(r)(r+1)\left(\dfrac{2}{r+3} + \dfrac{2}{r+5} + \cdots + \dfrac{2}{r+2m-1} + \dfrac{1}{r+2m+1}\right) \end{align*} $$
We are very happy to see $a_{2m}(r)(r+1)$ here because it handles our problem factor in the denominator of $a_{2m}(r)$. It is now time to plug in $r = -1$:

$$ \begin{align*} c_{2m}(-1) &= \dfrac{(-1)^{m+1} a_0}{2\cdot 4\ldots(2m-2)\cdot 2\cdot 4 \ldots (2m)}\cdot\left(\dfrac{2}{2} + \dfrac{2}{4} + \cdots + \dfrac{2}{2m-2} + \dfrac{1}{2m}\right) \\ \\ &= \dfrac{(-1)^{m+1} a_0}{2^{2m-1}m!(m-1)!}\dfrac{1}{2}\left(2 + \dfrac{2}{2} + \dfrac{2}{3} + \cdots + \dfrac{2}{m-1} + \dfrac{1}{m}\right) \\ \\ &= \dfrac{(-1)^{m+1} a_0}{2^{2m}m!(m-1)!}\left( \underbrace{1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{m-1} + \dfrac{1}{m}}_{H_m} + \underbrace{1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{m-1} }_{H_{m-1}}\right) \\ \\ &= \dfrac{(-1)^{m+1}\left(H_m + H_{m-1}\right)}{2^{2m}m!(m-1)!},\quad m\geq 1 \end{align*} $$
where $H_m$ is the m-th partial sum of the harmonic series ($H_0 = 0$) and we let $a_0 = 1$ for convenience.

Our arduous journey is finally complete, with our second fundamental solution

$$ y_2(x) = -J_1(x)\ln x + x^{-1}\left( 1 - \sum_{m=1}^\infty \dfrac{(-1)^m\left(H_m + H_{m-1}\right)}{2^{2m}m!(m-1)!}x^{2m}\right). $$
Following the convention from our text, we may now state the formula for the Bessel function of the second kind of order one

$$ Y_1(x) = \dfrac{2}{\pi}\Big(-y_2(x) + (\gamma - \ln 2)J_1(x)\Big). $$
where $\gamma$ is again the Euler-Mascheroni constant.

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