Internal problem ID [2934]
Internal file name [OUTPUT/2426_Sunday_June_05_2022_03_09_22_AM_60454255/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.4. page
758
Problem number: 19.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {3 x^{2} y^{\prime \prime }+x \left (3 x^{2}+1\right ) y^{\prime }-2 y x=0} \] 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. \[ 3 x^{2} y^{\prime \prime }+\left (3 x^{3}+x \right ) y^{\prime }-2 y x = 0 \] 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 {3 x^{2}+1}{3 x}\\ q(x) &= -\frac {2}{3 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 \[ 3 x^{2} y^{\prime \prime }+\left (3 x^{3}+x \right ) y^{\prime }-2 y x = 0 \] 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} 3 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (3 x^{3}+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 ) x = 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 }}3 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right ) = 0 \] When \(n = 0\) the above becomes \[ 3 x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r = 0 \] Or \[ \left (3 x^{r} r \left (-1+r \right )+x^{r} r \right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} r \left (-2+3 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r^{2}-2 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {2}{3}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} r \left (-2+3 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {2}{3}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {2}{3}}\) is not 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 ) &= 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 +\frac {2}{3}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 \[ a_{1} = \frac {2}{3 r^{2}+4 r +1} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 3 a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n -2} \left (n +r -2\right )+a_{n} \left (n +r \right )-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {3 n a_{n -2}+3 r a_{n -2}-6 a_{n -2}-2 a_{n -1}}{3 n^{2}+6 n r +3 r^{2}-2 n -2 r}\tag {4} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{n} = \frac {-3 n a_{n -2}+4 a_{n -2}+2 a_{n -1}}{n \left (3 n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {2}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(\frac {2}{5}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{2}=-{\frac {3}{40}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8}\) | \(-{\frac {3}{40}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-36 r^{3}-102 r^{2}-114 r -40}{27 r^{6}+270 r^{5}+1062 r^{4}+2080 r^{3}+2103 r^{2}+1010 r +168} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{3}=-{\frac {43}{660}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8}\) | \(-{\frac {3}{40}}\) |
| \(a_{3}\) | \(\frac {-36 r^{3}-102 r^{2}-114 r -40}{27 r^{6}+270 r^{5}+1062 r^{4}+2080 r^{3}+2103 r^{2}+1010 r +168}\) | \(-{\frac {43}{660}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81 r^{6}+702 r^{5}+2250 r^{4}+3132 r^{3}+1521 r^{2}-486 r -584}{81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{4}={\frac {31}{3696}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8}\) | \(-{\frac {3}{40}}\) |
| \(a_{3}\) | \(\frac {-36 r^{3}-102 r^{2}-114 r -40}{27 r^{6}+270 r^{5}+1062 r^{4}+2080 r^{3}+2103 r^{2}+1010 r +168}\) | \(-{\frac {43}{660}}\) |
| \(a_{4}\) | \(\frac {81 r^{6}+702 r^{5}+2250 r^{4}+3132 r^{3}+1521 r^{2}-486 r -584}{81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720}\) | \(\frac {31}{3696}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {486 r^{6}+5670 r^{5}+26460 r^{4}+62622 r^{3}+79734 r^{2}+52788 r +13232}{\left (81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720\right ) \left (3 r^{2}+28 r +65\right )} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{5}={\frac {2259}{261800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8}\) | \(-{\frac {3}{40}}\) |
| \(a_{3}\) | \(\frac {-36 r^{3}-102 r^{2}-114 r -40}{27 r^{6}+270 r^{5}+1062 r^{4}+2080 r^{3}+2103 r^{2}+1010 r +168}\) | \(-{\frac {43}{660}}\) |
| \(a_{4}\) | \(\frac {81 r^{6}+702 r^{5}+2250 r^{4}+3132 r^{3}+1521 r^{2}-486 r -584}{81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720}\) | \(\frac {31}{3696}\) |
| \(a_{5}\) | \(\frac {486 r^{6}+5670 r^{5}+26460 r^{4}+62622 r^{3}+79734 r^{2}+52788 r +13232}{\left (81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720\right ) \left (3 r^{2}+28 r +65\right )}\) | \(\frac {2259}{261800}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {2}{3}} \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^{\frac {2}{3}} \left (1+\frac {2 x}{5}-\frac {3 x^{2}}{40}-\frac {43 x^{3}}{660}+\frac {31 x^{4}}{3696}+\frac {2259 x^{5}}{261800}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = \frac {2}{3 r^{2}+4 r +1} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 3 b_{n} \left (n +r \right ) \left (n +r -1\right )+3 b_{n -2} \left (n +r -2\right )+b_{n} \left (n +r \right )-2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {3 n b_{n -2}+3 r b_{n -2}-6 b_{n -2}-2 b_{n -1}}{3 n^{2}+6 n r +3 r^{2}-2 n -2 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {-3 n b_{n -2}+6 b_{n -2}+2 b_{n -1}}{n \left (3 n -2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-36 r^{3}-102 r^{2}-114 r -40}{27 r^{6}+270 r^{5}+1062 r^{4}+2080 r^{3}+2103 r^{2}+1010 r +168} \] Which for the root \(r = 0\) becomes \[ b_{3}=-{\frac {5}{21}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {-36 r^{3}-102 r^{2}-114 r -40}{27 r^{6}+270 r^{5}+1062 r^{4}+2080 r^{3}+2103 r^{2}+1010 r +168}\) | \(-{\frac {5}{21}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {81 r^{6}+702 r^{5}+2250 r^{4}+3132 r^{3}+1521 r^{2}-486 r -584}{81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {73}{840}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {-36 r^{3}-102 r^{2}-114 r -40}{27 r^{6}+270 r^{5}+1062 r^{4}+2080 r^{3}+2103 r^{2}+1010 r +168}\) | \(-{\frac {5}{21}}\) |
| \(b_{4}\) | \(\frac {81 r^{6}+702 r^{5}+2250 r^{4}+3132 r^{3}+1521 r^{2}-486 r -584}{81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720}\) | \(-{\frac {73}{840}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {486 r^{6}+5670 r^{5}+26460 r^{4}+62622 r^{3}+79734 r^{2}+52788 r +13232}{\left (81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720\right ) \left (3 r^{2}+28 r +65\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}={\frac {827}{27300}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{3 r^{2}+4 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {-9 r^{3}-12 r^{2}-3 r +4}{9 r^{4}+42 r^{3}+67 r^{2}+42 r +8}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {-36 r^{3}-102 r^{2}-114 r -40}{27 r^{6}+270 r^{5}+1062 r^{4}+2080 r^{3}+2103 r^{2}+1010 r +168}\) | \(-{\frac {5}{21}}\) |
| \(b_{4}\) | \(\frac {81 r^{6}+702 r^{5}+2250 r^{4}+3132 r^{3}+1521 r^{2}-486 r -584}{81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720}\) | \(-{\frac {73}{840}}\) |
| \(b_{5}\) | \(\frac {486 r^{6}+5670 r^{5}+26460 r^{4}+62622 r^{3}+79734 r^{2}+52788 r +13232}{\left (81 r^{8}+1404 r^{7}+10206 r^{6}+40404 r^{5}+94549 r^{4}+132496 r^{3}+106844 r^{2}+44096 r +6720\right ) \left (3 r^{2}+28 r +65\right )}\) | \(\frac {827}{27300}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\ &= 1+2 x +\frac {x^{2}}{2}-\frac {5 x^{3}}{21}-\frac {73 x^{4}}{840}+\frac {827 x^{5}}{27300}+O\left (x^{6}\right ) \end {align*}
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^{\frac {2}{3}} \left (1+\frac {2 x}{5}-\frac {3 x^{2}}{40}-\frac {43 x^{3}}{660}+\frac {31 x^{4}}{3696}+\frac {2259 x^{5}}{261800}+O\left (x^{6}\right )\right ) + c_{2} \left (1+2 x +\frac {x^{2}}{2}-\frac {5 x^{3}}{21}-\frac {73 x^{4}}{840}+\frac {827 x^{5}}{27300}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {2}{3}} \left (1+\frac {2 x}{5}-\frac {3 x^{2}}{40}-\frac {43 x^{3}}{660}+\frac {31 x^{4}}{3696}+\frac {2259 x^{5}}{261800}+O\left (x^{6}\right )\right )+c_{2} \left (1+2 x +\frac {x^{2}}{2}-\frac {5 x^{3}}{21}-\frac {73 x^{4}}{840}+\frac {827 x^{5}}{27300}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {2}{3}} \left (1+\frac {2 x}{5}-\frac {3 x^{2}}{40}-\frac {43 x^{3}}{660}+\frac {31 x^{4}}{3696}+\frac {2259 x^{5}}{261800}+O\left (x^{6}\right )\right )+c_{2} \left (1+2 x +\frac {x^{2}}{2}-\frac {5 x^{3}}{21}-\frac {73 x^{4}}{840}+\frac {827 x^{5}}{27300}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {2}{3}} \left (1+\frac {2 x}{5}-\frac {3 x^{2}}{40}-\frac {43 x^{3}}{660}+\frac {31 x^{4}}{3696}+\frac {2259 x^{5}}{261800}+O\left (x^{6}\right )\right )+c_{2} \left (1+2 x +\frac {x^{2}}{2}-\frac {5 x^{3}}{21}-\frac {73 x^{4}}{840}+\frac {827 x^{5}}{27300}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x^{2} y^{\prime \prime }+\left (3 x^{3}+x \right ) y^{\prime }-2 y x =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}{3 x}-\frac {\left (3 x^{2}+1\right ) y^{\prime }}{3 x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {\left (3 x^{2}+1\right ) y^{\prime }}{3 x}-\frac {2 y}{3 x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {3 x^{2}+1}{3 x}, P_{3}\left (x \right )=-\frac {2}{3 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}}}=\frac {1}{3} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 3 x y^{\prime \prime }+\left (3 x^{2}+1\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..2 \\ {} & {} & 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 \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-2+3 r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (1+3 r \right )-2 a_{0}\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (3 k +1+3 r \right )-2 a_{k}+3 a_{k -1} \left (k +r -1\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+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {2}{3}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (1+3 r \right )-2 a_{0}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (3 k +1+3 r \right )-2 a_{k}+3 a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (3 k +4+3 r \right )-2 a_{k +1}+3 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {3 k a_{k}+3 r a_{k}-2 a_{k +1}}{\left (k +2+r \right ) \left (3 k +4+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {3 k a_{k}-2 a_{k +1}}{\left (k +2\right ) \left (3 k +4\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 {3 k a_{k}-2 a_{k +1}}{\left (k +2\right ) \left (3 k +4\right )}, a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & a_{k +2}=-\frac {3 k a_{k}+2 a_{k}-2 a_{k +1}}{\left (k +\frac {8}{3}\right ) \left (3 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {2}{3}}, a_{k +2}=-\frac {3 k a_{k}+2 a_{k}-2 a_{k +1}}{\left (k +\frac {8}{3}\right ) \left (3 k +6\right )}, 5 a_{1}-2 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 +\frac {2}{3}}\right ), a_{2+k}=-\frac {3 k a_{k}-2 a_{1+k}}{\left (2+k \right ) \left (3 k +4\right )}, a_{1}-2 a_{0}=0, b_{2+k}=-\frac {3 k b_{k}+2 b_{k}-2 b_{1+k}}{\left (k +\frac {8}{3}\right ) \left (3 k +6\right )}, 5 b_{1}-2 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 HeunB ODE, case c = 0 `
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 44
Order:=6; dsolve(3*x^2*diff(y(x),x$2)+x*(1+3*x^2)*diff(y(x),x)-2*x*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {2}{3}} \left (1+\frac {2}{5} x -\frac {3}{40} x^{2}-\frac {43}{660} x^{3}+\frac {31}{3696} x^{4}+\frac {2259}{261800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+2 x +\frac {1}{2} x^{2}-\frac {5}{21} x^{3}-\frac {73}{840} x^{4}+\frac {827}{27300} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 83
AsymptoticDSolveValue[3*x^2*y''[x]+x*(1+3*x^2)*y'[x]-2*x*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {827 x^5}{27300}-\frac {73 x^4}{840}-\frac {5 x^3}{21}+\frac {x^2}{2}+2 x+1\right )+c_1 x^{2/3} \left (\frac {2259 x^5}{261800}+\frac {31 x^4}{3696}-\frac {43 x^3}{660}-\frac {3 x^2}{40}+\frac {2 x}{5}+1\right ) \]