Internal problem ID [6444]
Internal file name [OUTPUT/5692_Sunday_June_05_2022_03_47_14_PM_30898659/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Section 4.4. REGULAR
SINGULAR POINTS. Page 175
Problem number: 1(d).
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]]
\[ \boxed {\left (3 x +1\right ) x y^{\prime \prime }-\left (1+x \right ) y^{\prime }+2 y=0} \] 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. \[ \left (3 x^{2}+x \right ) y^{\prime \prime }+\left (-1-x \right ) y^{\prime }+2 y = 0 \] 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 {1+x}{x \left (3 x +1\right )}\\ q(x) &= \frac {2}{\left (3 x +1\right ) x}\\ \end {align*}
| \(p(x)=-\frac {1+x}{x \left (3 x +1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{3}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {2}{\left (3 x +1\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{3}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -{\frac {1}{3}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ \left (3 x +1\right ) x y^{\prime \prime }+\left (-1-x \right ) y^{\prime }+2 y = 0 \] 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 (3 x +1\right ) 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 (-1-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \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 }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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 }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 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 =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )-r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )-r \,x^{-1+r}\right ) a_{0} = 0 \] 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 &= 2\\ r_2 &= 0 \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 \([2, 0]\).
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 ) &= x^{2} \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 +2}\\ 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} 3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (3 n^{2}+6 n r +3 r^{2}-10 n -10 r +9\right )}{n^{2}+2 n r +r^{2}-2 n -2 r}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = -\frac {a_{n -1} \left (3 n^{2}+2 n +1\right )}{n \left (n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) 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 \[ a_{1}=\frac {-3 r^{2}+4 r -2}{r^{2}-1} \] Which for the root \(r = 2\) becomes \[ a_{1}=-2 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r -2}{r^{2}-1}\) | \(-2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )} \] Which for the root \(r = 2\) becomes \[ a_{2}={\frac {17}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r -2}{r^{2}-1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )}\) | \(\frac {17}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-27 r^{6}-54 r^{5}-9 r^{4}+28 r^{3}-12 r^{2}-16 r -12}{r \left (-1+r \right ) \left (r +2\right ) \left (r +1\right )^{2} \left (r +3\right )} \] Which for the root \(r = 2\) becomes \[ a_{3}=-{\frac {289}{30}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r -2}{r^{2}-1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )}\) | \(\frac {17}{4}\) |
| \(a_{3}\) | \(\frac {-27 r^{6}-54 r^{5}-9 r^{4}+28 r^{3}-12 r^{2}-16 r -12}{r \left (-1+r \right ) \left (r +2\right ) \left (r +1\right )^{2} \left (r +3\right )}\) | \(-{\frac {289}{30}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81 r^{8}+540 r^{7}+1242 r^{6}+960 r^{5}-203 r^{4}-260 r^{3}+464 r^{2}+440 r +204}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right ) \left (r +4\right )} \] Which for the root \(r = 2\) becomes \[ a_{4}={\frac {5491}{240}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r -2}{r^{2}-1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )}\) | \(\frac {17}{4}\) |
| \(a_{3}\) | \(\frac {-27 r^{6}-54 r^{5}-9 r^{4}+28 r^{3}-12 r^{2}-16 r -12}{r \left (-1+r \right ) \left (r +2\right ) \left (r +1\right )^{2} \left (r +3\right )}\) | \(-{\frac {289}{30}}\) |
| \(a_{4}\) | \(\frac {81 r^{8}+540 r^{7}+1242 r^{6}+960 r^{5}-203 r^{4}-260 r^{3}+464 r^{2}+440 r +204}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right ) \left (r +4\right )}\) | \(\frac {5491}{240}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-243 r^{10}-3240 r^{9}-17280 r^{8}-46080 r^{7}-60819 r^{6}-27800 r^{5}+10710 r^{4}-1760 r^{3}-25188 r^{2}-19040 r -6936}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right ) \left (r +5\right )} \] Which for the root \(r = 2\) becomes \[ a_{5}=-{\frac {236113}{4200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r -2}{r^{2}-1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )}\) | \(\frac {17}{4}\) |
| \(a_{3}\) | \(\frac {-27 r^{6}-54 r^{5}-9 r^{4}+28 r^{3}-12 r^{2}-16 r -12}{r \left (-1+r \right ) \left (r +2\right ) \left (r +1\right )^{2} \left (r +3\right )}\) | \(-{\frac {289}{30}}\) |
| \(a_{4}\) | \(\frac {81 r^{8}+540 r^{7}+1242 r^{6}+960 r^{5}-203 r^{4}-260 r^{3}+464 r^{2}+440 r +204}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right ) \left (r +4\right )}\) | \(\frac {5491}{240}\) |
| \(a_{5}\) | \(\frac {-243 r^{10}-3240 r^{9}-17280 r^{8}-46080 r^{7}-60819 r^{6}-27800 r^{5}+10710 r^{4}-1760 r^{3}-25188 r^{2}-19040 r -6936}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right ) \left (r +5\right )}\) | \(-{\frac {236113}{4200}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {\left (3 r^{2}-4 r +2\right ) \left (3 r^{2}+8 r +6\right ) \left (3 r^{2}+20 r +34\right ) \left (3 r^{2}+2 r +1\right ) \left (3 r^{2}+14 r +17\right ) \left (3 r^{2}+26 r +57\right )}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right ) \left (r +6\right )} \] Which for the root \(r = 2\) becomes \[ a_{6}={\frac {28569673}{201600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r -2}{r^{2}-1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )}\) | \(\frac {17}{4}\) |
| \(a_{3}\) | \(\frac {-27 r^{6}-54 r^{5}-9 r^{4}+28 r^{3}-12 r^{2}-16 r -12}{r \left (-1+r \right ) \left (r +2\right ) \left (r +1\right )^{2} \left (r +3\right )}\) | \(-{\frac {289}{30}}\) |
| \(a_{4}\) | \(\frac {81 r^{8}+540 r^{7}+1242 r^{6}+960 r^{5}-203 r^{4}-260 r^{3}+464 r^{2}+440 r +204}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right ) \left (r +4\right )}\) | \(\frac {5491}{240}\) |
| \(a_{5}\) | \(\frac {-243 r^{10}-3240 r^{9}-17280 r^{8}-46080 r^{7}-60819 r^{6}-27800 r^{5}+10710 r^{4}-1760 r^{3}-25188 r^{2}-19040 r -6936}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right ) \left (r +5\right )}\) | \(-{\frac {236113}{4200}}\) |
| \(a_{6}\) | \(\frac {\left (3 r^{2}-4 r +2\right ) \left (3 r^{2}+8 r +6\right ) \left (3 r^{2}+20 r +34\right ) \left (3 r^{2}+2 r +1\right ) \left (3 r^{2}+14 r +17\right ) \left (3 r^{2}+26 r +57\right )}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right ) \left (r +6\right )}\) | \(\frac {28569673}{201600}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {\left (3 r^{2}-4 r +2\right ) \left (3 r^{2}+8 r +6\right ) \left (3 r^{2}+20 r +34\right ) \left (3 r^{2}+2 r +1\right ) \left (3 r^{2}+14 r +17\right ) \left (3 r^{2}+26 r +57\right ) \left (3 r^{2}+32 r +86\right )}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right ) \left (7+r \right )} \] Which for the root \(r = 2\) becomes \[ a_{7}=-{\frac {28569673}{78400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r -2}{r^{2}-1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )}\) | \(\frac {17}{4}\) |
| \(a_{3}\) | \(\frac {-27 r^{6}-54 r^{5}-9 r^{4}+28 r^{3}-12 r^{2}-16 r -12}{r \left (-1+r \right ) \left (r +2\right ) \left (r +1\right )^{2} \left (r +3\right )}\) | \(-{\frac {289}{30}}\) |
| \(a_{4}\) | \(\frac {81 r^{8}+540 r^{7}+1242 r^{6}+960 r^{5}-203 r^{4}-260 r^{3}+464 r^{2}+440 r +204}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right ) \left (r +4\right )}\) | \(\frac {5491}{240}\) |
| \(a_{5}\) | \(\frac {-243 r^{10}-3240 r^{9}-17280 r^{8}-46080 r^{7}-60819 r^{6}-27800 r^{5}+10710 r^{4}-1760 r^{3}-25188 r^{2}-19040 r -6936}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right ) \left (r +5\right )}\) | \(-{\frac {236113}{4200}}\) |
| \(a_{6}\) | \(\frac {\left (3 r^{2}-4 r +2\right ) \left (3 r^{2}+8 r +6\right ) \left (3 r^{2}+20 r +34\right ) \left (3 r^{2}+2 r +1\right ) \left (3 r^{2}+14 r +17\right ) \left (3 r^{2}+26 r +57\right )}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right ) \left (r +6\right )}\) | \(\frac {28569673}{201600}\) |
| \(a_{7}\) | \(-\frac {\left (3 r^{2}-4 r +2\right ) \left (3 r^{2}+8 r +6\right ) \left (3 r^{2}+20 r +34\right ) \left (3 r^{2}+2 r +1\right ) \left (3 r^{2}+14 r +17\right ) \left (3 r^{2}+26 r +57\right ) \left (3 r^{2}+32 r +86\right )}{r \left (-1+r \right ) \left (r +2\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right ) \left (7+r \right )}\) | \(-{\frac {28569673}{78400}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{2} \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^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+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 \(N=2\). 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 {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\right )}&= \lim _{r\rightarrow 0}\frac {9 r^{4}-6 r^{3}+r^{2}+2}{\left (r^{2}-1\right ) r \left (r +2\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 \(\left (3 x +1\right ) x y^{\prime \prime }+\left (-1-x \right ) y^{\prime }+2 y = 0\) gives \[ \left (3 x +1\right ) x \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (-1-x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+2 C y_{1}\left (x \right ) \ln \left (x \right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x \left (3 x +1\right ) y_{1}^{\prime \prime }\left (x \right )+\left (-1-x \right ) y_{1}^{\prime }\left (x \right )+2 y_{1}\left (x \right )\right ) \ln \left (x \right )+x \left (3 x +1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (-1-x \right ) y_{1}\left (x \right )}{x}\right ) C +x \left (3 x +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 )+\left (-1-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+2 \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 \[ x \left (3 x +1\right ) y_{1}^{\prime \prime }\left (x \right )+\left (-1-x \right ) y_{1}^{\prime }\left (x \right )+2 y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x \left (3 x +1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (-1-x \right ) y_{1}\left (x \right )}{x}\right ) C +x \left (3 x +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 )+\left (-1-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+2 \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 (6 \left (\frac {1}{3}+x \right ) x \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+2 \left (-2 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {\left (3 x^{3}+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 )+\left (-x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x}{x} = 0 \end{equation} Since \(r_{1} = 2\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (6 \left (\frac {1}{3}+x \right ) x \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} a_{n} \left (n +2\right )\right )+2 \left (-2 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +2}\right )\right ) C}{x}+\frac {\left (3 x^{3}+x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right )+\left (-x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right )+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^{1+n} a_{n} \left (n +2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 C \,x^{n +2} a_{n} \left (n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{1+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} b_{n} n \right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n -1} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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^{1+n} a_{n} \left (n +2\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} n \,x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 C \,x^{n +2} a_{n} \left (n +2\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}6 C a_{n -3} \left (n -1\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-4 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_{-2+n} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} n \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} b_{n} n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-\left (n -1\right ) b_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 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 =2}{\sum }}2 C a_{-2+n} n \,x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}6 C a_{n -3} \left (n -1\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-4 C a_{n -3} x^{n -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{-2+n} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-\left (n -1\right ) b_{n -1} x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n -1} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 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 \(n=1\), 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 \[ \left (8 a_{0}+4 a_{1}\right ) C +6 b_{2}+3 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 3 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=0 \] For \(n=4\), Eq (2B) gives \[ \left (14 a_{1}+6 a_{2}\right ) C +17 b_{3}+8 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {5}{2}+8 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {5}{16}} \] For \(n=5\), Eq (2B) gives \[ \left (20 a_{2}+8 a_{3}\right ) C +34 b_{4}+15 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {2227}{120}+15 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {2227}{1800}} \] For \(n=6\), Eq (2B) gives \[ \left (26 a_{3}+10 a_{4}\right ) C +57 b_{5}+24 b_{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {27659}{300}+24 b_{6} = 0 \] Solving the above for \(b_{6}\) gives \[ b_{6}=-{\frac {27659}{7200}} \] For \(n=7\), Eq (2B) gives \[ \left (32 a_{4}+12 a_{5}\right ) C +86 b_{6}+35 b_{7} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {9774983}{25200}+35 b_{7} = 0 \] Solving the above for \(b_{7}\) gives \[ b_{7}={\frac {9774983}{882000}} \] 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 \[ y_{2}\left (x \right )= \left (-1\right )\eslowast \left (x^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1+2 x -\frac {5 x^{4}}{16}+\frac {2227 x^{5}}{1800}-\frac {27659 x^{6}}{7200}+\frac {9774983 x^{7}}{882000}+O\left (x^{8}\right ) \] 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^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right ) + c_{2} \left (\left (-1\right )\eslowast \left (x^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1+2 x -\frac {5 x^{4}}{16}+\frac {2227 x^{5}}{1800}-\frac {27659 x^{6}}{7200}+\frac {9774983 x^{7}}{882000}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right )+c_{2} \left (-x^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1+2 x -\frac {5 x^{4}}{16}+\frac {2227 x^{5}}{1800}-\frac {27659 x^{6}}{7200}+\frac {9774983 x^{7}}{882000}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right )+c_{2} \left (-x^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1+2 x -\frac {5 x^{4}}{16}+\frac {2227 x^{5}}{1800}-\frac {27659 x^{6}}{7200}+\frac {9774983 x^{7}}{882000}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right )+c_{2} \left (-x^{2} \left (1-2 x +\frac {17 x^{2}}{4}-\frac {289 x^{3}}{30}+\frac {5491 x^{4}}{240}-\frac {236113 x^{5}}{4200}+\frac {28569673 x^{6}}{201600}-\frac {28569673 x^{7}}{78400}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1+2 x -\frac {5 x^{4}}{16}+\frac {2227 x^{5}}{1800}-\frac {27659 x^{6}}{7200}+\frac {9774983 x^{7}}{882000}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (3 x +1\right ) x y^{\prime \prime }+\left (-1-x \right ) y^{\prime }+2 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 {2 y}{x \left (3 x +1\right )}+\frac {\left (1+x \right ) y^{\prime }}{x \left (3 x +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} \\ {} & {} & y^{\prime \prime }-\frac {\left (1+x \right ) y^{\prime }}{x \left (3 x +1\right )}+\frac {2 y}{x \left (3 x +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 )=-\frac {1+x}{x \left (3 x +1\right )}, P_{3}\left (x \right )=\frac {2}{\left (3 x +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}}}=-1 \\ {} & \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 (3 x +1\right ) x y^{\prime \prime }+\left (-1-x \right ) y^{\prime }+2 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 y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +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..2 \\ {} & {} & 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} \\ {} & {} & a_{0} r \left (-2+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )+a_{k} \left (3 k^{2}+6 k r +3 r^{2}-4 k -4 r +2\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (3 k^{2}+\left (6 r -4\right ) k +3 r^{2}-4 r +2\right ) a_{k}+a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {\left (3 k^{2}+6 k r +3 r^{2}-4 k -4 r +2\right ) a_{k}}{\left (k +1+r \right ) \left (k +r -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {\left (3 k^{2}-4 k +2\right ) a_{k}}{\left (k +1\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=-\frac {\left (3 k^{2}-4 k +2\right ) a_{k}}{\left (k +1\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=-\frac {\left (3 k^{2}+8 k +6\right ) a_{k}}{\left (k +3\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +1}=-\frac {\left (3 k^{2}+8 k +6\right ) a_{k}}{\left (k +3\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 -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful -> solution has integrals; searching for one without integrals... -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric solution without integrals succesful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 72
Order:=8; dsolve((3*x+1)*x*diff(y(x),x$2)-(x+1)*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{2} \left (1-2 x +\frac {17}{4} x^{2}-\frac {289}{30} x^{3}+\frac {5491}{240} x^{4}-\frac {236113}{4200} x^{5}+\frac {28569673}{201600} x^{6}-\frac {28569673}{78400} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (2 x^{2}-4 x^{3}+\frac {17}{2} x^{4}-\frac {289}{15} x^{5}+\frac {5491}{120} x^{6}-\frac {236113}{2100} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (-2-4 x +6 x^{2}-12 x^{3}+\frac {209}{8} x^{4}-\frac {54247}{900} x^{5}+\frac {521849}{3600} x^{6}-\frac {158526173}{441000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.421 (sec). Leaf size: 118
AsymptoticDSolveValue[(3*x+1)*x*y''[x]-(x+1)*y'[x]+2*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {27353 x^6-12886 x^5+6525 x^4-3600 x^3+1800 x^2+7200 x+3600}{3600}-\frac {1}{240} x^2 \left (5491 x^4-2312 x^3+1020 x^2-480 x+240\right ) \log (x)\right )+c_2 \left (\frac {28569673 x^8}{201600}-\frac {236113 x^7}{4200}+\frac {5491 x^6}{240}-\frac {289 x^5}{30}+\frac {17 x^4}{4}-2 x^3+x^2\right ) \]