Internal problem ID [6981]
Internal file name [OUTPUT/6224_Friday_August_12_2022_11_06_04_PM_12870851/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.9 Indicial Equation with Difference of Roots a
Positive Integer: Logarithmic Case. Exercises page 384
Problem number: 1.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {x y^{\prime \prime }+y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x y^{\prime \prime }+y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= 0\\ q(x) &= \frac {1}{x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x y^{\prime \prime }+y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{-1+r} r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 0]\).
Since \(r_1 - r_2 = 1\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +1}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1}}{\left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = -\frac {a_{n -1}}{\left (n +1\right ) n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=-\frac {1}{\left (1+r \right ) r} \] Which for the root \(r = 1\) becomes \[ a_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (1+r \right ) r}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {1}{12}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (1+r \right ) r}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {1}{144}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (1+r \right ) r}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {1}{144}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {1}{2880}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (1+r \right ) r}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {1}{144}}\) |
| \(a_{4}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(\frac {1}{2880}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {1}{86400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (1+r \right ) r}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {1}{144}}\) |
| \(a_{4}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(\frac {1}{2880}\) |
| \(a_{5}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(-{\frac {1}{86400}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{6}={\frac {1}{3628800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (1+r \right ) r}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {1}{144}}\) |
| \(a_{4}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(\frac {1}{2880}\) |
| \(a_{5}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(-{\frac {1}{86400}}\) |
| \(a_{6}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )}\) | \(\frac {1}{3628800}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{7}=-{\frac {1}{203212800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (1+r \right ) r}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {1}{144}}\) |
| \(a_{4}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(\frac {1}{2880}\) |
| \(a_{5}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(-{\frac {1}{86400}}\) |
| \(a_{6}\) | \(\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )}\) | \(\frac {1}{3628800}\) |
| \(a_{7}\) | \(-\frac {1}{\left (1+r \right )^{2} r \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )}\) | \(-{\frac {1}{203212800}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{1} \\ &= -\frac {1}{\left (1+r \right ) r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {1}{\left (1+r \right ) r}&= \lim _{r\rightarrow 0}-\frac {1}{\left (1+r \right ) r}\\ &= \textit {undefined} \end {align*}
Since the limit does not exist then the log term is needed. Therefore the second solution has the form \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Therefore \begin{align*} \frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\ &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*} Substituting these back into the given ode \(x y^{\prime \prime }+y = 0\) gives \[ x \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x +y_{1}\left (x \right )\right ) \ln \left (x \right )+x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )\right ) C +x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ y_{1}^{\prime \prime }\left (x \right ) x +y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) C +x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x}{x} = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (n +1\right )\right ) x -\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +1}\right )\right ) C}{x}+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) x}{x} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n} a_{n} \left (n +1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n} a_{n} \left (n +1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -1}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the difference between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives \[ C +1 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-1 \] For \(n=2\), Eq (2B) gives \[ 3 C a_{1}+b_{1}+2 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 2 b_{2}+\frac {3}{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {3}{4}} \] For \(n=3\), Eq (2B) gives \[ 5 C a_{2}+b_{2}+6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 6 b_{3}-\frac {7}{6} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {7}{36}} \] For \(n=4\), Eq (2B) gives \[ 7 C a_{3}+b_{3}+12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 12 b_{4}+\frac {35}{144} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {35}{1728}} \] For \(n=5\), Eq (2B) gives \[ 9 C a_{4}+b_{4}+20 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 20 b_{5}-\frac {101}{4320} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {101}{86400}} \] For \(n=6\), Eq (2B) gives \[ 11 C a_{5}+b_{5}+30 b_{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 30 b_{6}+\frac {7}{5400} = 0 \] Solving the above for \(b_{6}\) gives \[ b_{6}=-{\frac {7}{162000}} \] For \(n=7\), Eq (2B) gives \[ 13 C a_{6}+b_{6}+42 b_{7} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 42 b_{7}-\frac {283}{6048000} = 0 \] Solving the above for \(b_{7}\) gives \[ b_{7}={\frac {283}{254016000}} \] Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Using the above value found for \(C=-1\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \left (-1\right )\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {7 x^{3}}{36}-\frac {35 x^{4}}{1728}+\frac {101 x^{5}}{86400}-\frac {7 x^{6}}{162000}+\frac {283 x^{7}}{254016000}+O\left (x^{8}\right ) \] Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right ) + c_{2} \left (\left (-1\right )\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {7 x^{3}}{36}-\frac {35 x^{4}}{1728}+\frac {101 x^{5}}{86400}-\frac {7 x^{6}}{162000}+\frac {283 x^{7}}{254016000}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {7 x^{3}}{36}-\frac {35 x^{4}}{1728}+\frac {101 x^{5}}{86400}-\frac {7 x^{6}}{162000}+\frac {283 x^{7}}{254016000}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {7 x^{3}}{36}-\frac {35 x^{4}}{1728}+\frac {101 x^{5}}{86400}-\frac {7 x^{6}}{162000}+\frac {283 x^{7}}{254016000}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}-\frac {x^{3}}{144}+\frac {x^{4}}{2880}-\frac {x^{5}}{86400}+\frac {x^{6}}{3628800}-\frac {x^{7}}{203212800}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {7 x^{3}}{36}-\frac {35 x^{4}}{1728}+\frac {101 x^{5}}{86400}-\frac {7 x^{6}}{162000}+\frac {283 x^{7}}{254016000}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {y}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {y}{x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=0, P_{3}\left (x \right )=\frac {1}{x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r \right )+a_{k}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +r \right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{\left (k +1+r \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k}}{\left (k +1\right ) k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {a_{k}}{\left (k +1\right ) k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=-\frac {a_{k}}{\left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=-\frac {a_{k}}{\left (k +2\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right ), a_{k +1}=-\frac {a_{k}}{\left (k +1\right ) k}, b_{k +1}=-\frac {b_{k}}{\left (k +2\right ) \left (k +1\right )}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 70
Order:=8; dsolve(x*diff(y(x),x$2)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1-\frac {1}{2} x +\frac {1}{12} x^{2}-\frac {1}{144} x^{3}+\frac {1}{2880} x^{4}-\frac {1}{86400} x^{5}+\frac {1}{3628800} x^{6}-\frac {1}{203212800} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-x +\frac {1}{2} x^{2}-\frac {1}{12} x^{3}+\frac {1}{144} x^{4}-\frac {1}{2880} x^{5}+\frac {1}{86400} x^{6}-\frac {1}{3628800} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (1-\frac {3}{4} x^{2}+\frac {7}{36} x^{3}-\frac {35}{1728} x^{4}+\frac {101}{86400} x^{5}-\frac {7}{162000} x^{6}+\frac {283}{254016000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.036 (sec). Leaf size: 119
AsymptoticDSolveValue[x*y''[x]+y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {x \left (x^5-30 x^4+600 x^3-7200 x^2+43200 x-86400\right ) \log (x)}{86400}+\frac {-71 x^6+1965 x^5-35250 x^4+360000 x^3-1620000 x^2+1296000 x+1296000}{1296000}\right )+c_2 \left (\frac {x^7}{3628800}-\frac {x^6}{86400}+\frac {x^5}{2880}-\frac {x^4}{144}+\frac {x^3}{12}-\frac {x^2}{2}+x\right ) \]