Internal problem ID [1421]
Internal file name [OUTPUT/1422_Sunday_June_05_2022_02_16_21_AM_75049073/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS III. Exercises 7.7. Page 389
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]]
\[ \boxed {2 x^{2} \left (2+3 x \right ) y^{\prime \prime }+x \left (4+21 x \right ) y^{\prime }-\left (1-9 x \right ) y=0} \] 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. \[ \left (6 x^{3}+4 x^{2}\right ) y^{\prime \prime }+\left (21 x^{2}+4 x \right ) y^{\prime }+\left (9 x -1\right ) y = 0 \] 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 {4+21 x}{2 x \left (2+3 x \right )}\\ q(x) &= \frac {9 x -1}{2 x^{2} \left (2+3 x \right )}\\ \end {align*}
| \(p(x)=\frac {4+21 x}{2 x \left (2+3 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {2}{3}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {9 x -1}{2 x^{2} \left (2+3 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {2}{3}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -{\frac {2}{3}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 2 x^{2} \left (2+3 x \right ) y^{\prime \prime }+\left (21 x^{2}+4 x \right ) y^{\prime }+\left (9 x -1\right ) y = 0 \] 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} 2 x^{2} \left (2+3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (21 x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (9 x -1\right ) \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 }}6 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} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}21 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\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 }}6 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}21 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}21 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}9 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}21 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+4 x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+4 x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, -{\frac {1}{2}}\right ]\).
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 ) &= \sqrt {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 )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{\sqrt {x}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{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 -\frac {1}{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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+21 a_{n -1} \left (n +r -1\right )+4 a_{n} \left (n +r \right )+9 a_{n -1}-a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {3 \left (n +r \right ) a_{n -1}}{2 n +2 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = -\frac {3 \left (2 n +1\right ) a_{n -1}}{4 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=\frac {-3-3 r}{1+2 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=-{\frac {9}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{1+2 r}\) | \(-{\frac {9}{4}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9 r^{2}+27 r +18}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {135}{32}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{1+2 r}\) | \(-{\frac {9}{4}}\) |
| \(a_{2}\) | \(\frac {9 r^{2}+27 r +18}{4 r^{2}+8 r +3}\) | \(\frac {135}{32}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-27 r^{3}-162 r^{2}-297 r -162}{8 r^{3}+36 r^{2}+46 r +15} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {945}{128}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{1+2 r}\) | \(-{\frac {9}{4}}\) |
| \(a_{2}\) | \(\frac {9 r^{2}+27 r +18}{4 r^{2}+8 r +3}\) | \(\frac {135}{32}\) |
| \(a_{3}\) | \(\frac {-27 r^{3}-162 r^{2}-297 r -162}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {945}{128}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81 r^{4}+810 r^{3}+2835 r^{2}+4050 r +1944}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {25515}{2048}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{1+2 r}\) | \(-{\frac {9}{4}}\) |
| \(a_{2}\) | \(\frac {9 r^{2}+27 r +18}{4 r^{2}+8 r +3}\) | \(\frac {135}{32}\) |
| \(a_{3}\) | \(\frac {-27 r^{3}-162 r^{2}-297 r -162}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {945}{128}}\) |
| \(a_{4}\) | \(\frac {81 r^{4}+810 r^{3}+2835 r^{2}+4050 r +1944}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {25515}{2048}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {243 \left (5+r \right ) \left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (1+r \right )}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {168399}{8192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{1+2 r}\) | \(-{\frac {9}{4}}\) |
| \(a_{2}\) | \(\frac {9 r^{2}+27 r +18}{4 r^{2}+8 r +3}\) | \(\frac {135}{32}\) |
| \(a_{3}\) | \(\frac {-27 r^{3}-162 r^{2}-297 r -162}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {945}{128}}\) |
| \(a_{4}\) | \(\frac {81 r^{4}+810 r^{3}+2835 r^{2}+4050 r +1944}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {25515}{2048}\) |
| \(a_{5}\) | \(-\frac {243 \left (5+r \right ) \left (4+r \right ) \left (3+r \right ) \left (2+r \right ) \left (1+r \right )}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(-{\frac {168399}{8192}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \sqrt {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 ) \\ &= \sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+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 {-3-3 r}{1+2 r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-3-3 r}{1+2 r}&= \lim _{r\rightarrow -{\frac {1}{2}}}\frac {-3-3 r}{1+2 r}\\ &= \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 \(2 x^{2} \left (2+3 x \right ) y^{\prime \prime }+\left (21 x^{2}+4 x \right ) y^{\prime }+\left (9 x -1\right ) y = 0\) gives \[ 2 x^{2} \left (2+3 x \right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (21 x^{2}+4 x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+\left (9 x -1\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (2 x^{2} \left (2+3 x \right ) y_{1}^{\prime \prime }\left (x \right )+\left (21 x^{2}+4 x \right ) y_{1}^{\prime }\left (x \right )+\left (9 x -1\right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+2 x^{2} \left (2+3 x \right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (21 x^{2}+4 x \right ) y_{1}\left (x \right )}{x}\right ) C +2 x^{2} \left (2+3 x \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 (21 x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (9 x -1\right ) \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 \[ 2 x^{2} \left (2+3 x \right ) y_{1}^{\prime \prime }\left (x \right )+\left (21 x^{2}+4 x \right ) y_{1}^{\prime }\left (x \right )+\left (9 x -1\right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (2 x^{2} \left (2+3 x \right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (21 x^{2}+4 x \right ) y_{1}\left (x \right )}{x}\right ) C +2 x^{2} \left (2+3 x \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 (21 x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (9 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \left (12 \left (\frac {2}{3}+x \right ) x \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+15 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right ) x \right ) C +\left (6 x^{3}+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 )+\left (21 x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+\left (9 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = {\frac {1}{2}}\) and \(r_{2} = -{\frac {1}{2}}\) then the above becomes \begin{equation} \tag{10} \left (12 \left (\frac {2}{3}+x \right ) x \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {1}{2}} a_{n} \left (n +\frac {1}{2}\right )\right )+15 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{2}}\right ) x \right ) C +\left (6 x^{3}+4 x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {5}{2}+n} b_{n} \left (n -\frac {1}{2}\right ) \left (-\frac {3}{2}+n \right )\right )+\left (21 x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {3}{2}+n} b_{n} \left (n -\frac {1}{2}\right )\right )+\left (9 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}}\right ) = 0 \end{equation} Expanding \(4 C \sqrt {x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 4 C \sqrt {x} &= 4 C \sqrt {x} + \dots \\ &= 4 C \sqrt {x} \end {align*}
Expanding \(\frac {3 \sqrt {x}}{2}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {3 \sqrt {x}}{2} &= \frac {3 \sqrt {x}}{2} + \dots \\ &= \frac {3 \sqrt {x}}{2} \end {align*}
Expanding \(\frac {1}{\sqrt {x}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {1}{\sqrt {x}} &= \frac {1}{\sqrt {x}} + \dots \\ &= \frac {1}{\sqrt {x}} \end {align*}
Expanding \(\frac {21 \sqrt {x}}{2}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {21 \sqrt {x}}{2} &= \frac {21 \sqrt {x}}{2} + \dots \\ &= \frac {21 \sqrt {x}}{2} \end {align*}
Expanding \(\frac {2}{\sqrt {x}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {2}{\sqrt {x}} &= \frac {2}{\sqrt {x}} + \dots \\ &= \frac {2}{\sqrt {x}} \end {align*}
Expanding \(9 \sqrt {x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 9 \sqrt {x} &= 9 \sqrt {x} + \dots \\ &= 9 \sqrt {x} \end {align*}
Expanding \(-\frac {1}{\sqrt {x}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\frac {1}{\sqrt {x}} &= -\frac {1}{\sqrt {x}} + \dots \\ &= -\frac {1}{\sqrt {x}} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n +4\right ) C a_{n} x^{n +\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (12 n +6\right ) C a_{n} x^{\frac {3}{2}+n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}15 C \,x^{\frac {3}{2}+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {3 x^{n +\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {21 x^{n +\frac {1}{2}} b_{n} \left (2 n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (4 n -2\right ) b_{n} x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +\frac {1}{2}} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n -\frac {1}{2}}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -\frac {1}{2}\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -\frac {1}{2}}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n +4\right ) C a_{n} x^{n +\frac {1}{2}} &= \moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} \left (8 n -4\right ) x^{n -\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (12 n +6\right ) C a_{n} x^{\frac {3}{2}+n} &= \moverset {\infty }{\munderset {n =2}{\sum }}C a_{n -2} \left (12 n -18\right ) x^{n -\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}15 C \,x^{\frac {3}{2}+n} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}15 C a_{n -2} x^{n -\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {3 x^{n +\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )}{2} &= \moverset {\infty }{\munderset {n =1}{\sum }}\frac {3 b_{n -1} \left (4 \left (n -1\right )^{2}-8 n +11\right ) x^{n -\frac {1}{2}}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {21 x^{n +\frac {1}{2}} b_{n} \left (2 n -1\right )}{2} &= \moverset {\infty }{\munderset {n =1}{\sum }}\frac {21 b_{n -1} \left (-3+2 n \right ) x^{n -\frac {1}{2}}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +\frac {1}{2}} b_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}9 b_{n -1} x^{n -\frac {1}{2}} \\ \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 \(n -\frac {1}{2}\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} \left (8 n -4\right ) x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}C a_{n -2} \left (12 n -18\right ) x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}15 C a_{n -2} x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {3 b_{n -1} \left (4 \left (n -1\right )^{2}-8 n +11\right ) x^{n -\frac {1}{2}}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {21 b_{n -1} \left (-3+2 n \right ) x^{n -\frac {1}{2}}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (4 n -2\right ) b_{n} x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}9 b_{n -1} x^{n -\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n -\frac {1}{2}}\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 \[ 4 C +3 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {3}{4}} \] For \(n=2\), Eq (2B) gives \[ \left (21 a_{0}+12 a_{1}\right ) C +18 b_{1}+8 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {9}{2}+8 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {9}{16}} \] For \(n=3\), Eq (2B) gives \[ \left (33 a_{1}+20 a_{2}\right ) C +45 b_{2}+24 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {1053}{32}+24 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {351}{256}} \] For \(n=4\), Eq (2B) gives \[ \left (45 a_{2}+28 a_{3}\right ) C +84 b_{3}+48 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {8181}{64}+48 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {2727}{1024}} \] For \(n=5\), Eq (2B) gives \[ \left (57 a_{3}+36 a_{4}\right ) C +135 b_{4}+80 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {778815}{2048}+80 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {155763}{32768}} \] 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 \(C=-{\frac {3}{4}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {3}{4}\eslowast \left (\sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1-\frac {9 x^{2}}{16}+\frac {351 x^{3}}{256}-\frac {2727 x^{4}}{1024}+\frac {155763 x^{5}}{32768}+O\left (x^{6}\right )}{\sqrt {x}} \] 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} \sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right ) + c_{2} \left (-\frac {3}{4}\eslowast \left (\sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1-\frac {9 x^{2}}{16}+\frac {351 x^{3}}{256}-\frac {2727 x^{4}}{1024}+\frac {155763 x^{5}}{32768}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {3 \sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{4}+\frac {1-\frac {9 x^{2}}{16}+\frac {351 x^{3}}{256}-\frac {2727 x^{4}}{1024}+\frac {155763 x^{5}}{32768}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {3 \sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{4}+\frac {1-\frac {9 x^{2}}{16}+\frac {351 x^{3}}{256}-\frac {2727 x^{4}}{1024}+\frac {155763 x^{5}}{32768}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {3 \sqrt {x}\, \left (1-\frac {9 x}{4}+\frac {135 x^{2}}{32}-\frac {945 x^{3}}{128}+\frac {25515 x^{4}}{2048}-\frac {168399 x^{5}}{8192}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{4}+\frac {1-\frac {9 x^{2}}{16}+\frac {351 x^{3}}{256}-\frac {2727 x^{4}}{1024}+\frac {155763 x^{5}}{32768}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (2+3 x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (21 x^{2}+4 x \right ) y^{\prime }+\left (9 x -1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (9 x -1\right ) y}{2 x^{2} \left (2+3 x \right )}-\frac {\left (4+21 x \right ) y^{\prime }}{2 x \left (2+3 x \right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (4+21 x \right ) y^{\prime }}{2 x \left (2+3 x \right )}+\frac {\left (9 x -1\right ) y}{2 x^{2} \left (2+3 x \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 {4+21 x}{2 x \left (2+3 x \right )}, P_{3}\left (x \right )=\frac {9 x -1}{2 x^{2} \left (2+3 x \right )}\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}}}=-\frac {1}{4} \\ {} & \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} \\ {} & {} & 2 x^{2} \left (2+3 x \right ) \left (\frac {d}{d x}y^{\prime }\right )+x \left (4+21 x \right ) y^{\prime }+\left (9 x -1\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+2 r \right ) \left (-1+2 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r +1\right ) \left (2 k +2 r -1\right )+3 a_{k -1} \left (2 k +2 r +1\right ) \left (k +r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+2 r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (\left (k +r -\frac {1}{2}\right ) a_{k}+\frac {3 a_{k -1} \left (k +r \right )}{2}\right ) \left (k +r +\frac {1}{2}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 4 \left (\left (k +r +\frac {1}{2}\right ) a_{k +1}+\frac {3 a_{k} \left (k +r +1\right )}{2}\right ) \left (k +\frac {3}{2}+r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +r +1\right )}{2 k +2 r +1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +\frac {1}{2}\right )}{2 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}}, a_{k +1}=-\frac {3 a_{k} \left (k +\frac {1}{2}\right )}{2 k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +\frac {3}{2}\right )}{2 k +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +1}=-\frac {3 a_{k} \left (k +\frac {3}{2}\right )}{2 k +2}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}}\right ), a_{k +1}=-\frac {3 a_{k} \left (k +\frac {1}{2}\right )}{2 k}, b_{k +1}=-\frac {3 b_{k} \left (k +\frac {3}{2}\right )}{2 k +2}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 69
Order:=6; dsolve(2*x^2*(2+3*x)*diff(y(x),x$2)+x*(4+21*x)*diff(y(x),x)-(1-9*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} x \left (1-\frac {9}{4} x +\frac {135}{32} x^{2}-\frac {945}{128} x^{3}+\frac {25515}{2048} x^{4}-\frac {168399}{8192} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-\frac {3}{4} x +\frac {27}{16} x^{2}-\frac {405}{128} x^{3}+\frac {2835}{512} x^{4}-\frac {76545}{8192} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1-\frac {15}{4} x +\frac {63}{8} x^{2}-\frac {3699}{256} x^{3}+\frac {25623}{1024} x^{4}-\frac {1375137}{32768} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.045 (sec). Leaf size: 108
AsymptoticDSolveValue[2*x^2*(2+3*x)*y''[x]+x*(4+21*x)*y'[x]-(1-9*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {25515 x^{9/2}}{2048}-\frac {945 x^{7/2}}{128}+\frac {135 x^{5/2}}{32}-\frac {9 x^{3/2}}{4}+\sqrt {x}\right )+c_1 \left (\frac {3}{512} \sqrt {x} \left (945 x^3-540 x^2+288 x-128\right ) \log (x)+\frac {8613 x^4-5076 x^3+2880 x^2-1536 x+1024}{1024 \sqrt {x}}\right ) \]