Internal problem ID [5567]
Internal file name [OUTPUT/4815_Sunday_June_05_2022_03_06_37_PM_64655398/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: 12.
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]]
\[ \boxed {x y^{\prime \prime }+\left (x +3\right ) y^{\prime }+7 y x^{2}=0} \] 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. \[ x y^{\prime \prime }+\left (x +3\right ) y^{\prime }+7 y x^{2} = 0 \] 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 {x +3}{x}\\ q(x) &= 7 x\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x y^{\prime \prime }+\left (x +3\right ) y^{\prime }+7 y x^{2} = 0 \] 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 (x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) x^{2} = 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 }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{2+n +r} a_{n}\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^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}7 x^{2+n +r} a_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}7 a_{n -3} 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} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}7 a_{n -3} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+3 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+3 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (2+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -2]\).
Since \(r_1 - r_2 = 2\) 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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x^{2}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{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 -2}\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 \[ a_{1} = -\frac {r}{r^{2}+4 r +3} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r}{r^{3}+9 r^{2}+26 r +24} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+3 a_{n} \left (n +r \right )+7 a_{n -3} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -1}+r a_{n -1}+7 a_{n -3}-a_{n -1}}{n^{2}+2 n r +r^{2}+2 n +2 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {-n a_{n -1}-7 a_{n -3}+a_{n -1}}{n \left (n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+4 r +3}\) | \(0\) |
| \(a_{2}\) | \(\frac {r}{r^{3}+9 r^{2}+26 r +24}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-7 r^{2}-50 r -84}{\left (5+r \right ) \left (r +3\right )^{2} \left (r +4\right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {7}{15}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+4 r +3}\) | \(0\) |
| \(a_{2}\) | \(\frac {r}{r^{3}+9 r^{2}+26 r +24}\) | \(0\) |
| \(a_{3}\) | \(\frac {-7 r^{2}-50 r -84}{\left (5+r \right ) \left (r +3\right )^{2} \left (r +4\right )}\) | \(-{\frac {7}{15}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {14 r^{3}+120 r^{2}+274 r +84}{\left (r +6\right ) \left (r +4\right )^{2} \left (r +3\right ) \left (1+r \right ) \left (5+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {7}{120}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+4 r +3}\) | \(0\) |
| \(a_{2}\) | \(\frac {r}{r^{3}+9 r^{2}+26 r +24}\) | \(0\) |
| \(a_{3}\) | \(\frac {-7 r^{2}-50 r -84}{\left (5+r \right ) \left (r +3\right )^{2} \left (r +4\right )}\) | \(-{\frac {7}{15}}\) |
| \(a_{4}\) | \(\frac {14 r^{3}+120 r^{2}+274 r +84}{\left (r +6\right ) \left (r +4\right )^{2} \left (r +3\right ) \left (1+r \right ) \left (5+r \right )}\) | \(\frac {7}{120}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-21 r^{4}-232 r^{3}-801 r^{2}-842 r -168}{\left (r +7\right ) \left (5+r \right )^{2} \left (r +3\right ) \left (r +4\right ) \left (2+r \right ) \left (1+r \right ) \left (r +6\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {1}{150}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+4 r +3}\) | \(0\) |
| \(a_{2}\) | \(\frac {r}{r^{3}+9 r^{2}+26 r +24}\) | \(0\) |
| \(a_{3}\) | \(\frac {-7 r^{2}-50 r -84}{\left (5+r \right ) \left (r +3\right )^{2} \left (r +4\right )}\) | \(-{\frac {7}{15}}\) |
| \(a_{4}\) | \(\frac {14 r^{3}+120 r^{2}+274 r +84}{\left (r +6\right ) \left (r +4\right )^{2} \left (r +3\right ) \left (1+r \right ) \left (5+r \right )}\) | \(\frac {7}{120}\) |
| \(a_{5}\) | \(\frac {-21 r^{4}-232 r^{3}-801 r^{2}-842 r -168}{\left (r +7\right ) \left (5+r \right )^{2} \left (r +3\right ) \left (r +4\right ) \left (2+r \right ) \left (1+r \right ) \left (r +6\right )}\) | \(-{\frac {1}{150}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 \\ &= 1-\frac {7 x^{3}}{15}+\frac {7 x^{4}}{120}-\frac {x^{5}}{150}+O\left (x^{6}\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 \(N=2\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{2}\left (r \right )\). 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_{2} \\ &= \frac {r}{r^{3}+9 r^{2}+26 r +24} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {r}{r^{3}+9 r^{2}+26 r +24}&= \lim _{r\rightarrow -2}\frac {r}{r^{3}+9 r^{2}+26 r +24}\\ &= \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 \(x y^{\prime \prime }+\left (x +3\right ) y^{\prime }+7 y x^{2} = 0\) gives \[ x \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (x +3\right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+7 \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) x^{2} = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x y_{1}^{\prime \prime }\left (x \right )+\left (x +3\right ) y_{1}^{\prime }\left (x \right )+7 y_{1}\left (x \right ) x^{2}\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 )+\frac {\left (x +3\right ) y_{1}\left (x \right )}{x}\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 (x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x^{2} = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ x y_{1}^{\prime \prime }\left (x \right )+\left (x +3\right ) y_{1}^{\prime }\left (x \right )+7 y_{1}\left (x \right ) x^{2} = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (x +3\right ) y_{1}\left (x \right )}{x}\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 (x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x^{2} = 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 (x +2\right ) \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 (x^{2}+3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x^{3}}{x} = 0 \end{equation} Since \(r_{1} = 0\) and \(r_{2} = -2\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} a_{n} n \right ) x +\left (x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\right ) C}{x}+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-4+n} b_{n} \left (n -2\right ) \left (n -3\right )\right ) x^{2}+\left (x^{2}+3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -3} b_{n} \left (n -2\right )\right )+7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2}\right ) x^{3}}{x} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n -1} a_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}C a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n -1} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -3} b_{n} \left (n^{2}-5 n +6\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n -3} b_{n} \left (n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 b_{n} x^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -3\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -3}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n -1} a_{n} n &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C \left (n -2\right ) a_{n -2} x^{n -3} \\ \moverset {\infty }{\munderset {n =0}{\sum }}C a_{n} x^{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}C a_{n -3} x^{n -3} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n -1} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} x^{n -3} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n -2\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (n -3\right ) x^{n -3} \\ \moverset {\infty }{\munderset {n =0}{\sum }}7 b_{n} x^{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}7 b_{n -3} x^{n -3} \\ \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 \(n -3\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C \left (n -2\right ) a_{n -2} x^{n -3}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}C a_{n -3} x^{n -3}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} x^{n -3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -3} b_{n} \left (n^{2}-5 n +6\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (n -3\right ) x^{n -3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n -3} b_{n} \left (n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}7 b_{n -3} x^{n -3}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=1\), Eq (2B) gives \[ -b_{1}-2 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -b_{1}-2 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=-2 \] For \(n=N\), where \(N=2\) which is the difference between the two roots, we are free to choose \(b_{2} = 0\). Hence for \(n=2\), Eq (2B) gives \[ 2 C +2 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-1 \] For \(n=3\), Eq (2B) gives \[ \left (a_{0}+4 a_{1}\right ) C +7 b_{0}+3 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 6+3 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-2 \] For \(n=4\), Eq (2B) gives \[ \left (a_{1}+6 a_{2}\right ) C +7 b_{1}+b_{3}+8 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -16+8 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=2 \] For \(n=5\), Eq (2B) gives \[ \left (a_{2}+8 a_{3}\right ) C +7 b_{2}+2 b_{4}+15 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {116}{15}+15 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {116}{225}} \] 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 \[ y_{2}\left (x \right )= \left (-1\right )\eslowast \left (1-\frac {7 x^{3}}{15}+\frac {7 x^{4}}{120}-\frac {x^{5}}{150}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-2 x -2 x^{3}+2 x^{4}-\frac {116 x^{5}}{225}+O\left (x^{6}\right )}{x^{2}} \] 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} \left (1-\frac {7 x^{3}}{15}+\frac {7 x^{4}}{120}-\frac {x^{5}}{150}+O\left (x^{6}\right )\right ) + c_{2} \left (\left (-1\right )\eslowast \left (1-\frac {7 x^{3}}{15}+\frac {7 x^{4}}{120}-\frac {x^{5}}{150}+O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-2 x -2 x^{3}+2 x^{4}-\frac {116 x^{5}}{225}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-\frac {7 x^{3}}{15}+\frac {7 x^{4}}{120}-\frac {x^{5}}{150}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-1+\frac {7 x^{3}}{15}-\frac {7 x^{4}}{120}+\frac {x^{5}}{150}-O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-2 x -2 x^{3}+2 x^{4}-\frac {116 x^{5}}{225}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {7 x^{3}}{15}+\frac {7 x^{4}}{120}-\frac {x^{5}}{150}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-1+\frac {7 x^{3}}{15}-\frac {7 x^{4}}{120}+\frac {x^{5}}{150}-O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-2 x -2 x^{3}+2 x^{4}-\frac {116 x^{5}}{225}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {7 x^{3}}{15}+\frac {7 x^{4}}{120}-\frac {x^{5}}{150}+O\left (x^{6}\right )\right )+c_{2} \left (\left (-1+\frac {7 x^{3}}{15}-\frac {7 x^{4}}{120}+\frac {x^{5}}{150}-O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-2 x -2 x^{3}+2 x^{4}-\frac {116 x^{5}}{225}+O\left (x^{6}\right )}{x^{2}}\right ) \] Verified OK.
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 -> 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 a symmetry of the form [xi=0, eta=F(x)] 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 2nd order, integrating factor of the form mu(x,y) -> 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 2nd order exact linear 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 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: 56
Order:=6; dsolve(x*diff(y(x),x$2)+(x+3)*diff(y(x),x)+7*x^2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \left (1-\frac {7}{15} x^{3}+\frac {7}{120} x^{4}-\frac {1}{150} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (\ln \left (x \right ) \left (2 x^{2}-\frac {14}{15} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (-2+4 x -3 x^{2}+4 x^{3}-4 x^{4}+\frac {547}{225} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.025 (sec). Leaf size: 55
AsymptoticDSolveValue[x*y''[x]+(x+3)*y'[x]+7*x^2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {7 x^4}{120}-\frac {7 x^3}{15}+1\right )+c_1 \left (\frac {2 x^4-2 x^3+2 x^2-2 x+1}{x^2}-\log (x)\right ) \]