Internal problem ID [5560]
Internal file name [OUTPUT/4808_Sunday_June_05_2022_03_06_21_PM_37861751/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.2 page
239
Problem number: 5.
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]]
\[ \boxed {\left (x^{3}+4 x \right ) y^{\prime \prime }-2 y^{\prime } x +6 y=0} \] 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. \[ \left (x^{3}+4 x \right ) y^{\prime \prime }-2 y^{\prime } x +6 y = 0 \] 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) &= -\frac {2}{x^{2}+4}\\ q(x) &= \frac {6}{\left (x^{2}+4\right ) x}\\ \end {align*}
| \(p(x)=-\frac {2}{x^{2}+4}\) | |
| singularity | type |
| \(x = -2 i\) | \(\text {``regular''}\) |
| \(x = 2 i\) | \(\text {``regular''}\) |
| \(q(x)=\frac {6}{\left (x^{2}+4\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -2 i\) | \(\text {``regular''}\) |
| \(x = 2 i\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-2 i, 2 i, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } \left (x^{2}+4\right ) x -2 y^{\prime } x +6 y = 0 \] 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) \left (x^{2}+4\right ) x -2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +6 \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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 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 }}x^{1+n +r} 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 -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}6 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ 4 x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ 4 x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 4 x^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r \left (-1+r \right ) = 0 \] 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 \[ 4 x^{-1+r} r \left (-1+r \right ) = 0 \] 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^{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 \[ a_{1} = \frac {r -3}{2 r \left (1+r \right )} \] 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 )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )+6 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -2}+2 n r a_{n -2}+r^{2} a_{n -2}-5 n a_{n -2}-2 n a_{n -1}-5 r a_{n -2}-2 r a_{n -1}+6 a_{n -2}+8 a_{n -1}}{4 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {-n^{2} a_{n -2}+\left (3 a_{n -2}+2 a_{n -1}\right ) n -2 a_{n -2}-6 a_{n -1}}{4 n \left (1+n \right )}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -3}{2 r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-r^{4}+2 r^{2}-5 r +6}{4 r \left (1+r \right )^{2} \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -3}{2 r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-r^{4}+2 r^{2}-5 r +6}{4 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{24}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-2 r^{5}+9 r^{3}+6 r^{2}+17 r -6}{8 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}={\frac {1}{48}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -3}{2 r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-r^{4}+2 r^{2}-5 r +6}{4 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {-2 r^{5}+9 r^{3}+6 r^{2}+17 r -6}{8 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{48}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{8}+8 r^{7}+19 r^{6}+17 r^{5}+9 r^{4}+17 r^{3}-5 r^{2}-114 r -72}{16 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{4}=-{\frac {1}{384}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -3}{2 r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-r^{4}+2 r^{2}-5 r +6}{4 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {-2 r^{5}+9 r^{3}+6 r^{2}+17 r -6}{8 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{48}\) |
| \(a_{4}\) | \(\frac {r^{8}+8 r^{7}+19 r^{6}+17 r^{5}+9 r^{4}+17 r^{3}-5 r^{2}-114 r -72}{16 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {1}{384}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {3 r^{9}+33 r^{8}+124 r^{7}+126 r^{6}-396 r^{5}-1408 r^{4}-2077 r^{3}-1967 r^{2}-798 r +360}{32 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {5}{2304}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -3}{2 r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-r^{4}+2 r^{2}-5 r +6}{4 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {-2 r^{5}+9 r^{3}+6 r^{2}+17 r -6}{8 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{48}\) |
| \(a_{4}\) | \(\frac {r^{8}+8 r^{7}+19 r^{6}+17 r^{5}+9 r^{4}+17 r^{3}-5 r^{2}-114 r -72}{16 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {1}{384}}\) |
| \(a_{5}\) | \(\frac {3 r^{9}+33 r^{8}+124 r^{7}+126 r^{6}-396 r^{5}-1408 r^{4}-2077 r^{3}-1967 r^{2}-798 r +360}{32 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(-{\frac {5}{2304}}\) |
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}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+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 {r -3}{2 r \left (1+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {r -3}{2 r \left (1+r \right )}&= \lim _{r\rightarrow 0}\frac {r -3}{2 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 \(y^{\prime \prime } \left (x^{2}+4\right ) x -2 y^{\prime } x +6 y = 0\) gives \[ \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) \left (x^{2}+4\right ) x -2 \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right ) x +6 C y_{1}\left (x \right ) \ln \left (x \right )+6 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) \left (x^{2}+4\right ) x -2 y_{1}^{\prime }\left (x \right ) x +6 y_{1}\left (x \right )\right ) \ln \left (x \right )+\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) \left (x^{2}+4\right ) x -2 y_{1}\left (x \right )\right ) C +\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 (x^{2}+4\right ) x -2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +6 \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 \[ y_{1}^{\prime \prime }\left (x \right ) \left (x^{2}+4\right ) x -2 y_{1}^{\prime }\left (x \right ) x +6 y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) \left (x^{2}+4\right ) x -2 y_{1}\left (x \right )\right ) C +\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 (x^{2}+4\right ) x -2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +6 \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 x \left (x^{2}+4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )-\left (x^{2}+2 x +4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {\left (x^{4}+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 )-2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x^{2}+6 \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 x \left (x^{2}+4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (1+n \right )\right )-\left (x^{2}+2 x +4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\right )\right ) C}{x}+\frac {\left (x^{4}+4 x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} n \left (n -1\right )\right )-2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right ) x^{2}+6 \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 +2} a_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 C \,x^{n} a_{n} \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{1+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{1+n} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 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 +2} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (n -2\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}8 C \,x^{n} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}8 C a_{n -1} n \,x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{n -3} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{n -2} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C a_{n} x^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 C a_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{1+n} 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 -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 \left (n -1\right ) b_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}6 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 =3}{\sum }}2 C a_{n -3} \left (n -2\right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}8 C a_{n -1} n \,x^{n -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{n -3} x^{n -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{n -2} x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 C a_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 \left (n -1\right ) b_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}6 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 \[ 4 C +6 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {3}{2}} \] For \(n=2\), Eq (2B) gives \[ \left (-2 a_{0}+12 a_{1}\right ) C +4 b_{1}+8 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 12+8 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {3}{2}} \] For \(n=3\), Eq (2B) gives \[ \left (a_{0}-2 a_{1}+20 a_{2}\right ) C +2 b_{2}+24 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {29}{4}+24 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {29}{96}} \] For \(n=4\), Eq (2B) gives \[ \left (3 a_{1}-2 a_{2}+28 a_{3}\right ) C +2 b_{2}+48 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {3}{2}+48 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {1}{32}} \] For \(n=5\), Eq (2B) gives \[ \left (5 a_{2}-2 a_{3}+36 a_{4}\right ) C +6 b_{3}-2 b_{4}+80 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {105}{64}+80 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {21}{1024}} \] 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 \(C=-{\frac {3}{2}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {3}{2}\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{2}+\frac {29 x^{3}}{96}+\frac {x^{4}}{32}-\frac {21 x^{5}}{1024}+O\left (x^{6}\right ) \] 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}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right ) + c_{2} \left (-\frac {3}{2}\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{2}+\frac {29 x^{3}}{96}+\frac {x^{4}}{32}-\frac {21 x^{5}}{1024}+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}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {3 x \left (1-\frac {x}{2}+\frac {x^{2}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1-\frac {3 x^{2}}{2}+\frac {29 x^{3}}{96}+\frac {x^{4}}{32}-\frac {21 x^{5}}{1024}+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}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {3 x \left (1-\frac {x}{2}+\frac {x^{2}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1-\frac {3 x^{2}}{2}+\frac {29 x^{3}}{96}+\frac {x^{4}}{32}-\frac {21 x^{5}}{1024}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {x}{2}+\frac {x^{2}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {3 x \left (1-\frac {x}{2}+\frac {x^{2}}{24}+\frac {x^{3}}{48}-\frac {x^{4}}{384}-\frac {5 x^{5}}{2304}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1-\frac {3 x^{2}}{2}+\frac {29 x^{3}}{96}+\frac {x^{4}}{32}-\frac {21 x^{5}}{1024}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } \left (x^{2}+4\right ) x -2 y^{\prime } x +6 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {6 y}{\left (x^{2}+4\right ) x}+\frac {2 y^{\prime }}{x^{2}+4} \\ \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} \\ {} & {} & y^{\prime \prime }-\frac {2 y^{\prime }}{x^{2}+4}+\frac {6 y}{\left (x^{2}+4\right ) x}=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 )=-\frac {2}{x^{2}+4}, P_{3}\left (x \right )=\frac {6}{\left (x^{2}+4\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} \\ {} & {} & y^{\prime \prime } \left (x^{2}+4\right ) x -2 y^{\prime } x +6 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 y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\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 y^{\prime \prime }=\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} \\ {} & {} & 4 a_{0} r \left (-1+r \right ) x^{-1+r}+\left (4 a_{1} \left (1+r \right ) r -2 a_{0} \left (-3+r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +1+r \right ) \left (k +r \right )-2 a_{k} \left (k +r -3\right )+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} \\ {} & {} & 4 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} \\ {} & {} & 4 a_{1} \left (1+r \right ) r -2 a_{0} \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 a_{k +1} \left (k +1+r \right ) \left (k +r \right )-2 a_{k} \left (k +r -3\right )+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 \\ {} & {} & 4 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right )-2 a_{k +1} \left (k -2+r \right )+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}-a_{k} k -2 k a_{k +1}-a_{k} r -2 r a_{k +1}+4 a_{k +1}}{4 \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}-a_{k} k -2 k a_{k +1}+4 a_{k +1}}{4 \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}-a_{k} k -2 k a_{k +1}+4 a_{k +1}}{4 \left (k +2\right ) \left (k +1\right )}, 6 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+a_{k} k -2 k a_{k +1}+2 a_{k +1}}{4 \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}+a_{k} k -2 k a_{k +1}+2 a_{k +1}}{4 \left (k +3\right ) \left (k +2\right )}, 8 a_{1}+4 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}-2 k a_{k +1}+4 a_{k +1}}{4 \left (k +2\right ) \left (k +1\right )}, 6 a_{0}=0, b_{k +2}=-\frac {k^{2} b_{k}+k b_{k}-2 k b_{k +1}+2 b_{k +1}}{4 \left (k +3\right ) \left (k +2\right )}, 8 b_{1}+4 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 -> Kummer -> 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 <- Heun successful: received ODE is equivalent to the HeunG ODE, case a <> 0, e <> 0, g <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 60
Order:=6; dsolve((x^3+4*x)*diff(y(x),x$2)-2*x*diff(y(x),x)+6*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1-\frac {1}{2} x +\frac {1}{24} x^{2}+\frac {1}{48} x^{3}-\frac {1}{384} x^{4}-\frac {5}{2304} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-\frac {3}{2} x +\frac {3}{4} x^{2}-\frac {1}{16} x^{3}-\frac {1}{32} x^{4}+\frac {1}{256} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1+\frac {1}{2} x -\frac {7}{4} x^{2}+\frac {31}{96} x^{3}+\frac {1}{24} x^{4}-\frac {67}{3072} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.038 (sec). Leaf size: 85
AsymptoticDSolveValue[(x^3+4*x)*y''[x]-2*x*y'[x]+6*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{96} \left (7 x^4+37 x^3-240 x^2+192 x+96\right )-\frac {1}{32} x \left (x^3+2 x^2-24 x+48\right ) \log (x)\right )+c_2 \left (-\frac {x^5}{384}+\frac {x^4}{48}+\frac {x^3}{24}-\frac {x^2}{2}+x\right ) \]