4.4.2: Roots of Indicial Equation

IN THIS SECTION WE WILL CONSIDER the types of solutions one can obtain of the differential equation,

\[y^{\prime \prime}(x)+a(x) y^{\prime}(x)+b(x) y(x)=0 \nonumber \]

when \(x=0\) is a regular singular point. In this case, we assume that \(x a(x)\) and \(x^{2} b(x)\) have the convergent Maclaurin series expansions

\[ \begin{aligned} x a(x) &=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \\ x^{2} b(x) &=b_{0}+b_{1} x+b_{2} x^{2}+\ldots \end{aligned} \label{4.40} \]

Using the Frobenius Method, we assume \(y(x)\) and its derivatives are of the form

\[\begin{aligned} y(x) &=\sum_{n=0}^{\infty} c_{n} x^{n+r} \\ y^{\prime}(x) &=\sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r-1} \\ y^{\prime \prime}(x) &=\sum_{n=0}^{\infty} c_{n}(n+r)(n+r-1) x^{n+r-2} \end{aligned} \label{4.41} \]

Inserting these series into the differential equation, we obtain

\[\sum_{n=0}^{\infty} c_{n}\left[(n+r)(n+r-1)+(n+r) x a(x)+x^{2} b(x)\right] x^{n+r-2}=0 \nonumber \]

Using the expansions for \(x a(x)\) and \(x^{2} b(x)\), we note that the lowest power of \(x\) is \(n+r-2\) when \(n=0\). The coefficient for the \(n=0\) term must vanish:

\[c_{0}\left[r(r-1)+a_{0} r+b_{0}\right]=0\nonumber \]

Assuming that \(c_{0} \neq 0\), we have the indicial equation

\[r(r-1)+a_{0} r+b_{0}=0\nonumber \]

The roots of the indicial equation determine the type of behavior of the solution. This amounts to considering three different cases. Let the roots of the equation be \(r_{1}\) and \(r_{2}\). Then,

i. Distinct roots with \(r_{1}-r_{2} \neq\) integer.

In this case we have two linearly independent solutions,

\[\begin{aligned} &y_{1}(x)=|x|^{r_{1}} \sum_{n=0}^{\infty} c_{n} x^{n}, \quad c_{0}=1 \\ &y_{2}(x)=|x|^{r_{2}} \sum_{n=0}^{\infty} d_{n} x^{n}, \quad d_{0}=1 \end{aligned} \nonumber \]

ii. Equal roots: \(r_{1}=r_{2}=r\).

The form for \(y_{1}(x)\) is the same, but one needs to use the Method of Reduction of Order to seek the second linearly independent solution.

\[\begin{aligned} &y_{1}(x)=|x|^{r} \sum_{n=0}^{\infty} c_{n} x^{n}, \quad c_{0}=1 \\ &y_{2}(x)=|x|^{r} \sum_{n=0}^{\infty} d_{n} x^{n}+y_{1}(x) \ln |x|, \quad d_{0}=1 \end{aligned} \nonumber \]

iii. Distinct roots with \(r_{1}-r_{2}=\) positive integer.

Just as in the last case, one needs to find a second linearly independent solution.

\[\begin{aligned} &y_{1}(x)=|x|^{r_{1}} \sum_{n=0}^{\infty} c_{n} x^{n}, \quad c_{0}=1 \\ &y_{2}(x)=|x|^{r_{2}} \sum_{n=0}^{\infty} d_{n} x^{n}+\alpha y_{1}(x) \ln |x|, \quad d_{0}=1 \end{aligned} \nonumber \]

The constant \(\alpha\) can be subsequently determined and in some cases might vanish.

For solutions near regular singular points, \(x=x_{0}\), one has a similar set of cases but for expansions of the form \(y_{1}(x)=\left|x-x_{0}\right|^{r_{1}} \sum_{n=0}^{\infty} c_{n}\left(x-x_{0}\right)^{n} .\)

Example \(\PageIndex{1}\)

\(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\dfrac{1}{4}\right) y=0\).

In this example \(x=0\) is a singular point. We have \(a(x)=1 / x\) and \(b(x)=\left(x^{2}-1 / 4\right) / x^{2}\). Thus,

\[\begin{aligned} x a(x) &=1 \\ x^{2} b(x) &=x^{2}-\dfrac{1}{4} \end{aligned} \nonumber \]

So, \(a_{0}=1\) and \(b_{0}=-\dfrac{1}{4} .\) The indicial equation becomes

\[r(r-1)+r-\dfrac{1}{4}=0 \nonumber \]

Simplifying, we have \(0=r^{2}-\dfrac{1}{4}\), or \(r=\pm \dfrac{1}{2}\).

For \(r=+\dfrac{1}{2}\), we insert the series \(y(x)=\sqrt{x} \sum_{n=0}^{\infty} c_{n} x^{n}\) into the differential equation, collect like terms by re-indexing, and find

\[ \begin{aligned} 0 &=\sum_{n=0}^{\infty} c_{n}\left[\left(n+\dfrac{1}{2}\right)\left(n-\dfrac{1}{2}\right)+\left(n+\dfrac{1}{2}\right)+x^{2}-\dfrac{1}{4}\right] x^{n-\dfrac{3}{2}} \\ &=\sum_{n=0}^{\infty} c_{n}\left[n^{2}+n+x^{2}\right] x^{n-\dfrac{3}{2}} \\ &=\left[\sum_{n=0}^{\infty} n(n+1) c_{n} x^{n}+\sum_{n=0}^{\infty} c_{n} x^{n+2}\right] x^{-\dfrac{3}{2}} \\ &=\left[\sum_{k=0}^{\infty} k(k+1) c_{k} x^{k}+\sum_{n=2}^{\infty} c_{k-2} x^{k}\right] x^{-\dfrac{3}{2}} \\ &=2 c_{1} x+\sum_{n=2}^{\infty}\left[k(k+1) c_{k}+c_{k-2}\right] x^{k-\dfrac{3}{2}} \end{aligned} \label{4.42} \]

This gives \(c_{1}=0\) and

\[c_{k}=-\dfrac{1}{k(k+1)} c_{k-2}, \quad k \geq 2 \nonumber \]

Iterating, we have \(c_{k}=0\) for \(k\) odd and

\[\begin{array}{ll} k=2: & c_{2}=-\dfrac{1}{3 !} c_{0} . \\ k=4: & c_{4}=-\dfrac{1}{20} c_{2}=\dfrac{1}{5 !} c_{0} . \\ k= 6: & c_{6}=-\dfrac{1}{42} c_{4}=\dfrac{1}{7 !} c_{0} . \end{array} \nonumber \]

This gives

\[\begin{aligned} y_{1}(x) &=\sqrt{x} \sum_{n=0}^{\infty} c_{n} x^{n} \\ &=\sqrt{x}\left(c_{0}-\dfrac{1}{3 !} c_{0} x^{2}+\dfrac{1}{5 !} c_{0} x^{4}-\ldots\right) \\ &=\dfrac{c_{0}}{\sqrt{x}}\left(x-\dfrac{1}{3 !} x^{3}+\dfrac{1}{5 !} x^{5}-\ldots\right)=\dfrac{c_{0}}{\sqrt{x}} \sin x \end{aligned} \nonumber \]

Similarly, for for \(r=-\dfrac{1}{2}\), one obtains the second solution, \(y_{2}(x)=\) \(\dfrac{d_{0}}{\sqrt{x}} \cos x .\) Setting \(c+0=1\) and \(d_{0}=1\), give the two linearly independent solutions. This differential equation is the Bessel equation of order one half and the solutions are Bessel functions of order one half:

\[\begin{aligned} J_{\dfrac{1}{2}}(x) &=\sqrt{\dfrac{2}{\pi x}} \sin x, x>0 \\ J_{-\dfrac{1}{2}}(x) &=\sqrt{\dfrac{2}{\pi x}} \cos x, x>0 \end{aligned} \nonumber \]

Example \(\PageIndex{2}\)

\(x^{2} y^{\prime \prime}+3 x y^{\prime}+(1-2 x) y=0, x>0\) For this problem \(x a(x)=3\) and \(x^{2} b(x)=1-2 x .\) Thus, the indicial equation is

\[0=r(r-1)+3 r+1=(r+1)^{2} \nonumber \]

This is a case with two equal roots, \(r=-1\). A solution of the form

\[y(x)=\sum_{n=0}^{\infty} c_{n} x^{n+r}\nonumber \]

will only result in one solution. We will use this solution form to arrive at a second linearly independent solution.

We will not insert \(r=-1\) into the solution form yet. Writing the differential equation in operator form, \(L[y]=0\), we have

\[\begin{aligned} L[y]=& x^{2} y^{\prime \prime}+3 x y^{\prime}+(1-2 x) y \\ =& x^{2} \sum_{n=0}^{\infty} c_{n}(n+r)(n+r-1) x^{n+r-2}+3 x \sum_{n=0}^{\infty} c_{n}(n+r) x^{n+r-1} \\ &+(1-2 x) \sum_{n=0}^{\infty} c_{n} x^{n+r} \\ =& \sum_{n=0}^{\infty}[(n+r)(n+r-1)+2(n+r)+1] c_{n} x^{n+r}-\sum_{n=0}^{\infty} 2 c_{n} x^{n+r+1} \\ =& \sum_{n=0}^{\infty}(n+r+1)^{2} c_{n} x^{n+r}-\sum_{n=1}^{\infty} 2 c_{n-1} x^{n+r} \end{aligned} \nonumber \]

Setting the coefficients of like terms equal to zero for \(n \geq 1\), we have

\[c_{n}=\dfrac{2}{(n+r+1)^{2}} a_{n-1}, \quad n \geq 1.\nonumber \]

Iterating, we find

\[c_{n}=\dfrac{2^{n}}{[(r+2)(r+3) \cdots(r+n+1)]^{2}} c_{0}, \quad n \geq 1 \nonumber \]

Setting \(c_{0}=1\), we have the expression

\[y(x, r)=x^{r}+\sum_{n=1}^{\infty} \dfrac{2^{n}}{[(r+2)(r+3) \cdots(r+n+1)]^{2}} x^{n+r}.\nonumber \]

This is not a solution of the differential equation because we did not use the root \(r=-1\). Instead, we have

\[L[y(x, c)]=(r+1)^{2} x^{r} \nonumber \]

from the \(n=0\) term. If \(r=-1\), then \(y(x,-1)\) is one solution of the differential equation. Namely,

\[y_{1}(x)=x^{-1}+\sum_{n=1}^{\infty} c_{n}(-1) x^{n-1}\nonumber \]

Now consider what happens if we differentiate Equation \(\PageIndex{4}\) with respect to \(r\) :

\[\begin{aligned} \dfrac{\partial}{\partial r} L[y(x, r)] &=\left[\dfrac{\partial y(x, r)}{\partial r}\right] \\ &=2(r+1) x^{r}+(r+1)^{2} x^{r} \ln x, \quad x>0 \end{aligned} \nonumber \]

Therefore, \(\dfrac{\partial y(x, r)}{\partial r}\) is also a solution to the differential equation when \(r=-1\).

Since

\[y(x, r)=x^{r}+\sum_{n=1}^{\infty} c_{n}(r) x^{n+r} \nonumber \]

we have

\[ \begin{aligned} \dfrac{\partial y(x, r)}{\partial r} &=x^{r} \ln x+\sum_{n=1}^{\infty} c_{n}(r) x^{n+r} \ln x+\sum_{n=1}^{\infty} c_{n}^{\prime}(r) x^{n+r} \\ &=y(x, r) \ln x+\sum_{n=1}^{\infty} c_{n}^{\prime}(r) x^{n+r} \end{aligned} \label{4.44} \]

Therefore, the second solution is given by

\[y_{2}(x)=\left.\dfrac{\partial y(x, r)}{\partial r}\right|_{r=-1}=y(x,-1) \ln x+\sum_{n=1}^{\infty} c_{n}^{\prime}(-1) x^{n-1} \nonumber \]

In order to determine the solutions, we need to evaluate \(c_{n}(-1)\) and \(c_{n}^{\prime}(-1) .\) Recall that (setting \(c_{0}=1\) )

\[c_{n}(r)=\dfrac{2^{n}}{[(r+2)(r+3) \cdots(r+n+1)]^{2}}, \quad n \geq 1 \nonumber \]

Therefore,

\[ \begin{aligned} c_{n}(-1) &=\dfrac{2^{n}}{[(1)(2) \cdots(n)]^{2}}, \quad n \geq 1 \\ &=\dfrac{2^{n}}{(n !)^{2}} \end{aligned} \label{4.45} \]

Next we compute \(c_{n}^{\prime}(-1)\). This can be done using logarithmic differentiation. We consider

\[\begin{aligned} \ln c_{n}(r) &=\ln \left(\dfrac{2^{n}}{[(r+2)(r+3) \cdots(r+n+1)]^{2}}\right) \\ &=\ln 2^{n}-2 \ln (r+2)-2 \ln (r+3) \cdots-2 \ln (r+n+1) \end{aligned} \nonumber \]

Differentiating with respect to \(r\) and evaluating at \(r=-1\), we have

\[ \begin{aligned} \dfrac{c_{n}^{\prime}(r)}{c_{n}(r)} &=-2\left(\dfrac{1}{r+2}+\dfrac{1}{r+2}+\cdots+\dfrac{1}{r+3}\right) \\ c_{n}^{\prime}(r) &=-2 c_{n}(r)\left(\dfrac{1}{r+2}+\dfrac{1}{r+3}+\cdots+\dfrac{1}{r+n+1}\right) . \\ c_{n}^{\prime}(-1) &=-2 c_{n}(-1)\left(\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right) \\ &=-\dfrac{2^{n+1}}{(n !)^{2}} H_{n} \end{aligned} \label{4.46} \]

where we have defined

\[H_{n}=\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{n}=\sum_{k=1}^{n} \dfrac{1}{k} \nonumber \]

This gives the second solution as

\[y_{2}(x)=y_{1}(x) \ln x-\sum_{n=1}^{\infty} \dfrac{2^{n+1}}{(n !)^{2}} H_{n} x^{n-1} \nonumber \]

Example \(\PageIndex{3}\)

\(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-1\right) y=0 .\)

This equation is similar to the last example, but it is the Bessel equation of order one. The indicial equation is given by

\[0=r(r-1)+r-1=r^{2}-1 \nonumber \]

The roots are \(r_{1}=1, r_{2}=-1 .\) In this case the roots differ by an integer, \(r_{1}-r_{2}=2\).

The first solution can be obtained using

\[ \begin{aligned} y(x) &=\sum_{n=0}^{\infty} c_{n} x^{n+1} \\ y^{\prime}(x) &=\sum_{n=0}^{\infty} c_{n}(n+1) x^{n} \\ y^{\prime \prime}(x) &=\sum_{n=0}^{\infty} c_{n}(n+1)(n) x^{n-1} \end{aligned} \label{4.47} \]

Inserting these series into the differential equation, we have

\[ \begin{aligned} 0 &=x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-1\right) y \\ &=x^{2} \sum_{n=0}^{\infty} c_{n} n(n+1) x^{n-1}+x \sum_{n=0}^{\infty} c_{n}(n+1) x^{n} \\ &+\left(x^{2}-1\right) \sum_{n=0}^{\infty} c_{n} x^{n+1} \\ &=\sum_{n=0}^{\infty} c_{n} n(n+1) x^{n+1}+\sum_{n=0}^{\infty} c_{n}(n+1) x^{n+1} \\ & \sum_{n=0}^{\infty} c_{n} x^{n+3}-\sum_{n=0}^{\infty} c_{n} x^{n+1} \\ &=\sum_{n=0}^{\infty}[n(n+1)+(n+1)-1] c_{n} x^{n+1}+\sum_{n=0}^{\infty} c_{n} x^{n+3} \\ &=\sum_{n=0}^{\infty}\left[(n+1)^{2}-1\right] c_{n} x^{n+1}+\sum_{n=0}^{\infty} c_{n} x^{n+3} \end{aligned} \label{4.48} \]

We re-index the last sum with \(k=n+2\), or \(n=k-2\), to obtain

\[\sum_{n=2}^{\infty}\left(\left[(k+1)^{2}-1\right] c_{k}+c_{k-2}\right) x^{k+1}+3 c_{1} x^{2}=0 \nonumber \]

Thus, \(c_{1}=0\) and

\[c_{k}=-\dfrac{1}{(k+1)^{2}-1} c_{k-2}=-\dfrac{1}{k(k+2)} c_{k-2}, \quad k \geq 2 \nonumber \]

Since \(c_{1}=0\), all \(c_{k}{ }^{\prime} \mathrm{s}\) vanish for odd \(k .\) For \(k=2 n\), we have

\[c_{2 n}=-\dfrac{1}{2 n(2 n+2)} c_{2 n-2}=-\dfrac{1}{4 n(n+1)} c_{2(n-1)}, \quad n \geq 1 .\nonumber \]

\[\begin{array}{ll} n=1: & c_{2}=-\dfrac{1}{4(1)(2)} c_{0} \\ n= & 2: \quad c_{4}=-\dfrac{1}{4(2)(3)} c_{2}=\dfrac{1}{4^{2} 2 ! 3 !} c_{0} \\ n= 3: & c_{6}=-\dfrac{1}{4(3)(4)} c_{4}=\dfrac{1}{4^{3} 3 ! 4 !} c_{0} . \end{array} \nonumber \]

Continuing, this gives

\[c_{2 n}=\dfrac{(-1)^{n}}{4^{n} n !(n+1) !} c_{0}\nonumber \]

and the first solution is

\[y_{1}(x)=c_{0} \sum_{n=0}^{\infty} \dfrac{(-1)^{n}}{4^{n} n !(n+1) !} x^{2 n+1}\nonumber \]

Now we look for a second linearly independent solution of the form

\[y_{2}(x)=\sum_{n=0}^{\infty} d_{n} x^{n-1}+\alpha y_{1}(x) \ln x, \quad x>0 \nonumber \]

The function and its derivatives are given by

\[\begin{aligned} y_{2}(x)=& \sum_{n=0}^{\infty} d_{n} x^{n-1}+\alpha y_{1}(x) \ln x \\ y_{2}^{\prime}(x)=& \sum_{n=0}^{\infty}(n-1) d_{n} x^{n-2}+\alpha\left[y_{1}^{\prime}(x) \ln x+y_{1}(x) x^{-1}\right] \\ y_{2}^{\prime \prime}(x)=& \sum_{n=0}^{\infty}(n-1) d_{n}(n-2) x^{n-3} \\ &+\alpha\left[y_{1}^{\prime \prime}(x) \ln x+2 y_{1}^{\prime}(x) x^{-1}-y_{1}(x) x^{-2}\right] \end{aligned} \nonumber \]

Inserting these series into the differential equation, we have

\[\begin{aligned} 0=& x^{2} y_{2}^{\prime \prime}+x y_{2}^{\prime}+\left(x^{2}-1\right) y_{2} \\ =& x^{2} \sum_{n=0}^{\infty}(n-1)(n-2) d_{n} x^{n-3} \\ &+\alpha x^{2}\left[y_{1}^{\prime \prime}(x) \ln x+2 y_{1}^{\prime}(x) x^{-1}-y_{1}(x) x^{-2}\right] \\ &+x \sum_{n=0}^{\infty}(n-1) d_{n} x^{n-2}+\alpha x\left[y_{1}^{\prime}(x) \ln x+y_{1}(x) x^{-1}\right] \\ &+\left(x^{2}-1\right)\left[\sum_{n=0}^{\infty} d_{n} x^{n-1}+\alpha y_{1}(x) \ln x\right] \\ =& \sum_{n=0}^{\infty}[(n-1)(n-2)+(n-1)-1] d_{n} x^{n-1}+\sum_{n=0}^{\infty} d_{n} x^{n+1} \\ &+\alpha\left[x^{2} y_{1}^{\prime \prime}(x)+x y_{1}^{\prime}(x)+\left(x^{2}-1\right) y_{1}(x)\right] \ln x \\ &+\alpha\left[2 x y_{1}^{\prime}(x)-y_{1}(x)\right]+\alpha y_{1}(x) \\ =& \sum_{n=0}^{\infty} n(n-2) d_{n} x^{n-1}+\sum_{n=0}^{\infty} d_{n} x^{n+1}+2 \alpha x y_{1}^{\prime}(x) \\ & \sum_{n=0}^{\infty} n(n-2) d_{n} x^{n-1}+\sum_{n=0}^{\infty} d_{n} x^{n+1}+2 \alpha \sum_{n=0}^{\infty} \dfrac{(-1)^{n}(2 n+1)}{4^{n}(n+1) ! n !} x^{2 n+1} \end{aligned} \nonumber \]

We now try to combine like powers of \(x\). First, we combine the terms involving \(d_{n}^{\prime}\) s,

\[-d_{1}+d_{0} x+\sum_{k=3}^{\infty}\left[k(k-2) d_{k}+d_{k-2}\right] x^{k-1}=-2 \alpha \sum_{n=0}^{\infty} \dfrac{(-1)^{n}(2 n+1)}{4^{n} n !(n+1) !} x^{2 n+1} \nonumber \]

Since there are no even powers on the right-hand side of the equation, we find \(d_{1}=0\), and \(k(k-2) d_{k}+d_{k-2}=0, k \geq 3\) and \(k\) odd. Therefore, all odd \(d_{k}\) ’s vanish.

Next, we set \(k-1=2 n+1\), or \(k=2 n+2\), in the remaining terms, obtaining

\[d_{0} x+\sum_{n=1}^{\infty}\left[(2 n+2)(2 n) d_{2 n+2}+d_{2 n}\right] x^{2 n+1}=-2 \alpha x-2 \alpha \sum_{n=1}^{\infty} \dfrac{(-1)^{n}(2 n+1)}{4^{n}(n+1) ! n !} x^{2 n+1} \nonumber \]

Thus, \(d_{0}=-2 \alpha .\) We choose \(\alpha=-\dfrac{1}{2}\), making \(d_{0}=1 .\) The remaining terms satisfy the relation

\[(2 n+2)(2 n) d_{2 n+2}+d_{2 n}=\dfrac{(-1)^{n}(2 n+1)}{4^{n} n !(n+1) !}, \quad n \geq 1\nonumber \]

Or

\[d_{2 n+2}=\dfrac{d_{2 n}}{4(n+1)(n)}+\dfrac{(-1)^{n}(2 n+1)}{(n+1)(n) 4^{n+1}(n+1) ! n !}, \quad n \geq 1 \nonumber \]

\[ \begin{aligned} & d_{4}=-\dfrac{1}{4(2)(1)} d_{2}-\dfrac{3}{(2)(1) 4^{2} 2 ! 1 !} \\ & =-\dfrac{1}{4^{2} 2 ! 1 !}\left(4 d_{2}+\dfrac{3}{2}\right) \\ & d_{6}=-\dfrac{1}{4(3)(2)} d_{4}+\dfrac{5}{(3)(2) 4^{3} 3 ! 2 !} \\ & =-\dfrac{1}{4(3)(2)}\left(-\dfrac{1}{4^{2} 2 ! 1 !}\left(4 d_{2}+\dfrac{3}{2}\right)\right)+\dfrac{5}{(3)(2) 4^{3} 3 ! 2 !} \\ & =\dfrac{1}{4^{3} 3 ! 2 !}\left(4 d_{2}+\dfrac{3}{2}+\dfrac{5}{6}\right) \\ & d_{8}=-\dfrac{1}{4(4)(3)} d_{6}-\dfrac{7}{(4)(3) 4^{4} 4 ! 3 !} \\ & =-\dfrac{1}{4^{4} 4 ! 3 !}\left(4 d_{2}+\dfrac{3}{2}+\dfrac{5}{6}\right)-\dfrac{7}{(4)(3) 4^{4} 4 ! 3 !} \\ & =-\dfrac{1}{4^{4} 4 ! 3 !}\left(4 d_{2}+\dfrac{3}{2}+\dfrac{5}{6}+\dfrac{7}{12}\right) . \end{aligned} \label{4.49} \]

Choosing \(4 d_{2}=1\), the coefficients take an interesting form. Namely,

\[ \begin{aligned} 1+\dfrac{3}{2} &=1+\dfrac{1}{2}+1 \\ 1+\dfrac{3}{2}+\dfrac{5}{6} &=1+\dfrac{1}{2} \dfrac{1}{3}+1+\dfrac{1}{2} \\ 1+\dfrac{3}{2}+\dfrac{5}{6}+\dfrac{7}{12} &=1+\dfrac{1}{2} \dfrac{1}{3}+\dfrac{1}{4}+1+\dfrac{1}{2}+\dfrac{1}{3} \end{aligned} \label{4.50} \]

Defining the partial sums of the harmonic series,

\[H_{n}=\sum_{k=1}^{n} \dfrac{1}{k}, \quad H_{0}=0 \nonumber \]

these coefficients become \(H_{n}+H_{n-1}\) and the coefficients in the expansion are

\[d_{2 n}=\dfrac{(-1)^{n-1}\left(H_{n}+H_{n-1}\right)}{4^{n} n !(n-1) !}, n=1,2, \ldots \nonumber \]

We can verify this by computing \(d_{10}\) :

\[\begin{aligned} d_{10} &=-\dfrac{1}{4(4)(3)} d_{8}+\dfrac{9}{(5)(4) 4^{5} 5 ! 4 !} \\ &=\dfrac{1}{4^{5} 5 ! 4 !}\left(4 d_{2}+\dfrac{3}{2}+\dfrac{5}{6}+\dfrac{7}{12}\right)+\dfrac{9}{(5)(4) 4^{5} 5 ! 4 !} \\ &=\dfrac{1}{4^{5} 5 ! 4 !}\left(4 d_{2}+\dfrac{3}{2}+\dfrac{5}{6}+\dfrac{7}{12}+\dfrac{9}{20}\right) \\ &=\dfrac{1}{4^{5} 5 ! 4 !}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right) \\ &=\dfrac{1}{4^{55} 5 ! 4 !}\left(H_{5}+H_{4}\right) \end{aligned} \nonumber \]

This gives the second solution as

\(y_{2}(x)=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}\left(H_{n}+H_{n-1}\right)}{4^{n} n !(n-1) !} x^{2 n-1}-\dfrac{1}{2} y_{1}(x) \ln x+x^{-1}\)