Internal problem ID [5520]
Internal file name [OUTPUT/4768_Sunday_June_05_2022_03_05_21_PM_17524057/index.tex]
Book: Notes on Diffy Qs. Differential Equations for Engineers. By by Jiri Lebl, 2013.
Section: Chapter 7. POWER SERIES METHODS. 7.3.2 The method of Frobenius. Exercises.
page 300
Problem number: 7.3.8 (a).
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+x y=0} \] With the expansion point for the power series method at \(x = 0\).
The ODE is \[ \left (x^{4}+x^{2}\right ) y^{\prime \prime }+x y = 0 \] Or \[ x \left (y^{\prime \prime } x^{3}+y^{\prime \prime } x +y\right ) = 0 \] For \(x \neq 0\) the above simplifies to \[ \left (x^{3}+x \right ) y^{\prime \prime }+y = 0 \] The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{4}+x^{2}\right ) y^{\prime \prime }+x y = 0 \] 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}{\left (x^{2}+1\right ) x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, -i, i, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+x y = 0 \] 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^{2} \left (x^{2}+1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+x \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 +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r} \\ \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 \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} 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}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ x^{r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} r \left (-1+r \right ) = 0 \] 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 \[ x^{r} r \left (-1+r \right ) = 0 \] 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^{1+n}\\ 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -\frac {1}{r \left (1+r \right )} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+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 \[ a_{n} = -\frac {n^{2} a_{n -2}+2 n r a_{n -2}+r^{2} a_{n -2}-5 n a_{n -2}-5 r a_{n -2}+6 a_{n -2}+a_{n -1}}{\left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {-n^{2} a_{n -2}+3 n a_{n -2}-2 a_{n -2}-a_{n -1}}{\left (1+n \right ) n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-r^{4}+r^{2}+1}{r \left (1+r \right )^{2} \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {1}{12}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{2}+1}{r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{12}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {2 r^{4}+4 r^{3}+4 r^{2}+2 r -1}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}={\frac {11}{144}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{2}+1}{r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+4 r^{3}+4 r^{2}+2 r -1}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {11}{144}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{8}+8 r^{7}+22 r^{6}+20 r^{5}-14 r^{4}-40 r^{3}-39 r^{2}-30 r -11}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{4}=-{\frac {83}{2880}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{2}+1}{r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+4 r^{3}+4 r^{2}+2 r -1}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {11}{144}\) |
| \(a_{4}\) | \(\frac {r^{8}+8 r^{7}+22 r^{6}+20 r^{5}-14 r^{4}-40 r^{3}-39 r^{2}-30 r -11}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {83}{2880}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-3 r^{8}-36 r^{7}-180 r^{6}-486 r^{5}-773 r^{4}-750 r^{3}-400 r^{2}-12 r +83}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {2557}{86400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{2}+1}{r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+4 r^{3}+4 r^{2}+2 r -1}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {11}{144}\) |
| \(a_{4}\) | \(\frac {r^{8}+8 r^{7}+22 r^{6}+20 r^{5}-14 r^{4}-40 r^{3}-39 r^{2}-30 r -11}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {83}{2880}}\) |
| \(a_{5}\) | \(\frac {-3 r^{8}-36 r^{7}-180 r^{6}-486 r^{5}-773 r^{4}-750 r^{3}-400 r^{2}-12 r +83}{r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(-{\frac {2557}{86400}}\) |
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}\dots \right ) \\ &= x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\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}{r \left (1+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {1}{r \left (1+r \right )}&= \lim _{r\rightarrow 0}-\frac {1}{r \left (1+r \right )}\\ &= \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 \(x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+x y = 0\) gives \[ x^{2} \left (x^{2}+1\right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+x \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x^{2} \left (x^{2}+1\right ) y_{1}^{\prime \prime }\left (x \right )+y_{1}\left (x \right ) x \right ) \ln \left (x \right )+x^{2} \left (x^{2}+1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )\right ) C +x^{2} \left (x^{2}+1\right ) \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 )+x \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 \[ x^{2} \left (x^{2}+1\right ) y_{1}^{\prime \prime }\left (x \right )+y_{1}\left (x \right ) x = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} x^{2} \left (x^{2}+1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) C +x^{2} \left (x^{2}+1\right ) \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 )+x \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} \left (2 x \left (x^{2}+1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (-x^{2}-1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +\left (x^{4}+x^{2}\right ) \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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \left (2 x \left (x^{2}+1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (1+n \right )\right )+\left (-x^{2}-1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\right )\right ) C +\left (x^{4}+x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} n \left (n -1\right )\right )+x \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +3} a_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{1+n} a_{n} \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +3} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{1+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +2} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +3} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (n -2\right ) x^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{1+n} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +3} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{n -3} x^{n}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +2} b_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n} \\ \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 \(n\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (n -2\right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{n -3} x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n}\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 \[ \left (a_{0}+5 a_{2}\right ) C +b_{2}+6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {13}{6}+6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {13}{36}} \] For \(n=4\), Eq (2B) gives \[ \left (3 a_{1}+7 a_{3}\right ) C +2 b_{2}+b_{3}+12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {25}{144}+12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {25}{1728}} \] For \(n=5\), Eq (2B) gives \[ \left (5 a_{2}+9 a_{4}\right ) C +6 b_{3}+b_{4}+20 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {8743}{4320}+20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {8743}{86400}} \] 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 \[ y_{2}\left (x \right )= \left (-1\right )\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {13 x^{3}}{36}+\frac {25 x^{4}}{1728}-\frac {8743 x^{5}}{86400}+O\left (x^{6}\right ) \] 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 {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right ) + c_{2} \left (\left (-1\right )\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {13 x^{3}}{36}+\frac {25 x^{4}}{1728}-\frac {8743 x^{5}}{86400}+O\left (x^{6}\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 {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {13 x^{3}}{36}+\frac {25 x^{4}}{1728}-\frac {8743 x^{5}}{86400}+O\left (x^{6}\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 {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {13 x^{3}}{36}+\frac {25 x^{4}}{1728}-\frac {8743 x^{5}}{86400}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {11 x^{3}}{144}-\frac {83 x^{4}}{2880}-\frac {2557 x^{5}}{86400}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {13 x^{3}}{36}+\frac {25 x^{4}}{1728}-\frac {8743 x^{5}}{86400}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+1\right ) \left (\frac {d}{d x}y^{\prime }\right )+x 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 \left (x^{2}+1\right )} \\ \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 \left (x^{2}+1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{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}{\left (x^{2}+1\right ) 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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (x^{2}+1\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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) r +a_{0}\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r \right )+a_{k}+a_{k -1} \left (k +r -1\right ) \left (k -2+r \right )\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 must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) r +a_{0}=0 \\ \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}+a_{k -1} \left (k +r -1\right ) \left (k -2+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +1+r \right )+a_{k +1}+a_{k} \left (k +r \right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+2 k r a_{k}+r^{2} a_{k}-k a_{k}-r a_{k}+a_{k +1}}{\left (k +2+r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-k a_{k}+a_{k +1}}{\left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {k^{2} a_{k}-k a_{k}+a_{k +1}}{\left (k +2\right ) \left (k +1\right )}, a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+k a_{k}+a_{k +1}}{\left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +2}=-\frac {k^{2} a_{k}+k a_{k}+a_{k +1}}{\left (k +3\right ) \left (k +2\right )}, 2 a_{1}+a_{0}=0\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 +2}=-\frac {k^{2} a_{k}-k a_{k}+a_{k +1}}{\left (k +2\right ) \left (k +1\right )}, a_{0}=0, b_{k +2}=-\frac {k^{2} b_{k}+k b_{k}+b_{k +1}}{\left (k +3\right ) \left (k +2\right )}, 2 b_{1}+b_{0}=0\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 -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 58
Order:=6; dsolve(x^2*(1+x^2)*diff(y(x),x$2)+x*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1-\frac {1}{2} x +\frac {1}{12} x^{2}+\frac {11}{144} x^{3}-\frac {83}{2880} x^{4}-\frac {2557}{86400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-x +\frac {1}{2} x^{2}-\frac {1}{12} x^{3}-\frac {11}{144} x^{4}+\frac {83}{2880} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1-\frac {3}{4} x^{2}+\frac {13}{36} x^{3}+\frac {25}{1728} x^{4}-\frac {8743}{86400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.024 (sec). Leaf size: 87
AsymptoticDSolveValue[x^2*(1+x^2)*y''[x]+x*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {157 x^4+768 x^3-2160 x^2+1728 x+1728}{1728}-\frac {1}{144} x \left (11 x^3+12 x^2-72 x+144\right ) \log (x)\right )+c_2 \left (-\frac {83 x^5}{2880}+\frac {11 x^4}{144}+\frac {x^3}{12}-\frac {x^2}{2}+x\right ) \]