Internal problem ID [5565]
Internal file name [OUTPUT/4813_Sunday_June_05_2022_03_06_32_PM_30756957/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.2 page
239
Problem number: 10.
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]]
\[ \boxed {\left (x^{3}-2 x^{2}+3 x \right )^{2} y^{\prime \prime }+x \left (x -3\right )^{2} y^{\prime }-y \left (1+x \right )=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{6}-4 x^{5}+10 x^{4}-12 x^{3}+9 x^{2}\right ) y^{\prime \prime }+\left (x^{3}-6 x^{2}+9 x \right ) y^{\prime }+\left (-1-x \right ) y = 0 \] 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 {\left (x -3\right )^{2}}{x \left (x^{2}-2 x +3\right )^{2}}\\ q(x) &= -\frac {1+x}{x^{2} \left (x^{2}-2 x +3\right )^{2}}\\ \end {align*}
| \(p(x)=\frac {\left (x -3\right )^{2}}{x \left (x^{2}-2 x +3\right )^{2}}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -i \sqrt {2}+1\) | \(\text {``irregular''}\) |
| \(x = i \sqrt {2}+1\) | \(\text {``irregular''}\) |
| \(q(x)=-\frac {1+x}{x^{2} \left (x^{2}-2 x +3\right )^{2}}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -i \sqrt {2}+1\) | \(\text {``regular''}\) |
| \(x = i \sqrt {2}+1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \(\left [-i \sqrt {2}+1, i \sqrt {2}+1\right ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\right ) y^{\prime \prime }+\left (x^{3}-6 x^{2}+9 x \right ) y^{\prime }+\left (-1-x \right ) y = 0 \] 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} x^{2} \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\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 (x^{3}-6 x^{2}+9 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-1-x \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 }}x^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-12 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 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 +2} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 }}x^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}a_{n -4} \left (n -4+r \right ) \left (n -5+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-4 a_{n -3} \left (-3+n +r \right ) \left (n -4+r \right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}10 a_{n -2} \left (n +r -2\right ) \left (-3+n +r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-12 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-12 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-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 =4}{\sum }}a_{n -4} \left (n -4+r \right ) \left (n -5+r \right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-4 a_{n -3} \left (-3+n +r \right ) \left (n -4+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}10 a_{n -2} \left (n +r -2\right ) \left (-3+n +r \right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-12 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+9 x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 9 x^{r} a_{0} r \left (-1+r \right )+9 x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (9 x^{r} r \left (-1+r \right )+9 x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (9 r^{2}-1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 9 r^{2}-1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{3}}\\ r_2 &= -{\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (9 r^{2}-1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{3}}, -{\frac {1}{3}}\right ]\).
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 {1}{3}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{3}} \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 \[ a_{1} = \frac {12 r^{2}-6 r +1}{9 r^{2}+18 r +8} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {54 r^{4}+45 r^{3}+70 r^{2}+48 r +7}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {-108 r^{6}-540 r^{5}+528 r^{4}+3732 r^{3}+3489 r^{2}+739 r +224}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )} \] For \(4\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -4} \left (n -4+r \right ) \left (n -5+r \right )-4 a_{n -3} \left (-3+n +r \right ) \left (n -4+r \right )+10 a_{n -2} \left (n +r -2\right ) \left (-3+n +r \right )-12 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+9 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -2} \left (n +r -2\right )-6 a_{n -1} \left (n +r -1\right )+9 a_{n} \left (n +r \right )-a_{n}-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -4}-4 n^{2} a_{n -3}+10 n^{2} a_{n -2}-12 n^{2} a_{n -1}+2 n r a_{n -4}-8 n r a_{n -3}+20 n r a_{n -2}-24 n r a_{n -1}+r^{2} a_{n -4}-4 r^{2} a_{n -3}+10 r^{2} a_{n -2}-12 r^{2} a_{n -1}-9 n a_{n -4}+28 n a_{n -3}-49 n a_{n -2}+30 n a_{n -1}-9 r a_{n -4}+28 r a_{n -3}-49 r a_{n -2}+30 r a_{n -1}+20 a_{n -4}-48 a_{n -3}+58 a_{n -2}-19 a_{n -1}}{9 n^{2}+18 n r +9 r^{2}-1}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{n} = \frac {\left (-9 a_{n -4}+36 a_{n -3}-90 a_{n -2}+108 a_{n -1}\right ) n^{2}+\left (75 a_{n -4}-228 a_{n -3}+381 a_{n -2}-198 a_{n -1}\right ) n -154 a_{n -4}+352 a_{n -3}-385 a_{n -2}+93 a_{n -1}}{81 n^{2}+54 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {12 r^{2}-6 r +1}{9 r^{2}+18 r +8}\) | \(\frac {1}{45}\) |
| \(a_{2}\) | \(\frac {54 r^{4}+45 r^{3}+70 r^{2}+48 r +7}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(\frac {149}{3240}\) |
| \(a_{3}\) | \(\frac {-108 r^{6}-540 r^{5}+528 r^{4}+3732 r^{3}+3489 r^{2}+739 r +224}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {2701}{192456}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {-2997 r^{8}-29079 r^{7}-99054 r^{6}-125883 r^{5}+9826 r^{4}+138853 r^{3}+80233 r^{2}+5477 r +8064}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{4}={\frac {236933}{121247280}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {12 r^{2}-6 r +1}{9 r^{2}+18 r +8}\) | \(\frac {1}{45}\) |
| \(a_{2}\) | \(\frac {54 r^{4}+45 r^{3}+70 r^{2}+48 r +7}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(\frac {149}{3240}\) |
| \(a_{3}\) | \(\frac {-108 r^{6}-540 r^{5}+528 r^{4}+3732 r^{3}+3489 r^{2}+739 r +224}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {2701}{192456}\) |
| \(a_{4}\) | \(\frac {-2997 r^{8}-29079 r^{7}-99054 r^{6}-125883 r^{5}+9826 r^{4}+138853 r^{3}+80233 r^{2}+5477 r +8064}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )}\) | \(\frac {236933}{121247280}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-17496 r^{10}-292572 r^{9}-2032830 r^{8}-7628094 r^{7}-16915518 r^{6}-23128560 r^{5}-20473273 r^{4}-12932200 r^{3}-6220667 r^{2}-1947462 r -14560}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right ) \left (9 r^{2}+90 r +224\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{5}=-{\frac {67092967}{92754169200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {12 r^{2}-6 r +1}{9 r^{2}+18 r +8}\) | \(\frac {1}{45}\) |
| \(a_{2}\) | \(\frac {54 r^{4}+45 r^{3}+70 r^{2}+48 r +7}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(\frac {149}{3240}\) |
| \(a_{3}\) | \(\frac {-108 r^{6}-540 r^{5}+528 r^{4}+3732 r^{3}+3489 r^{2}+739 r +224}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {2701}{192456}\) |
| \(a_{4}\) | \(\frac {-2997 r^{8}-29079 r^{7}-99054 r^{6}-125883 r^{5}+9826 r^{4}+138853 r^{3}+80233 r^{2}+5477 r +8064}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )}\) | \(\frac {236933}{121247280}\) |
| \(a_{5}\) | \(\frac {-17496 r^{10}-292572 r^{9}-2032830 r^{8}-7628094 r^{7}-16915518 r^{6}-23128560 r^{5}-20473273 r^{4}-12932200 r^{3}-6220667 r^{2}-1947462 r -14560}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right ) \left (9 r^{2}+90 r +224\right )}\) | \(-{\frac {67092967}{92754169200}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{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 {1}{3}} \left (1+\frac {x}{45}+\frac {149 x^{2}}{3240}+\frac {2701 x^{3}}{192456}+\frac {236933 x^{4}}{121247280}-\frac {67092967 x^{5}}{92754169200}+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 \[ b_{1} = \frac {12 r^{2}-6 r +1}{9 r^{2}+18 r +8} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {54 r^{4}+45 r^{3}+70 r^{2}+48 r +7}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ b_{3} = \frac {-108 r^{6}-540 r^{5}+528 r^{4}+3732 r^{3}+3489 r^{2}+739 r +224}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )} \] For \(4\le n\) the recursive equation is \begin{equation} \tag{3} b_{n -4} \left (n -4+r \right ) \left (n -5+r \right )-4 b_{n -3} \left (-3+n +r \right ) \left (n -4+r \right )+10 b_{n -2} \left (n +r -2\right ) \left (-3+n +r \right )-12 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+9 b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -2} \left (n +r -2\right )-6 b_{n -1} \left (n +r -1\right )+9 b_{n} \left (n +r \right )-b_{n}-b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {n^{2} b_{n -4}-4 n^{2} b_{n -3}+10 n^{2} b_{n -2}-12 n^{2} b_{n -1}+2 n r b_{n -4}-8 n r b_{n -3}+20 n r b_{n -2}-24 n r b_{n -1}+r^{2} b_{n -4}-4 r^{2} b_{n -3}+10 r^{2} b_{n -2}-12 r^{2} b_{n -1}-9 n b_{n -4}+28 n b_{n -3}-49 n b_{n -2}+30 n b_{n -1}-9 r b_{n -4}+28 r b_{n -3}-49 r b_{n -2}+30 r b_{n -1}+20 b_{n -4}-48 b_{n -3}+58 b_{n -2}-19 b_{n -1}}{9 n^{2}+18 n r +9 r^{2}-1}\tag {4} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{n} = \frac {\left (-9 b_{n -4}+36 b_{n -3}-90 b_{n -2}+108 b_{n -1}\right ) n^{2}+\left (87 b_{n -4}-276 b_{n -3}+501 b_{n -2}-342 b_{n -1}\right ) n -208 b_{n -4}+520 b_{n -3}-679 b_{n -2}+273 b_{n -1}}{81 n^{2}-54 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {12 r^{2}-6 r +1}{9 r^{2}+18 r +8}\) | \(\frac {13}{9}\) |
| \(b_{2}\) | \(\frac {54 r^{4}+45 r^{3}+70 r^{2}+48 r +7}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(-{\frac {5}{162}}\) |
| \(b_{3}\) | \(\frac {-108 r^{6}-540 r^{5}+528 r^{4}+3732 r^{3}+3489 r^{2}+739 r +224}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {1591}{30618}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {-2997 r^{8}-29079 r^{7}-99054 r^{6}-125883 r^{5}+9826 r^{4}+138853 r^{3}+80233 r^{2}+5477 r +8064}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{4}={\frac {106583}{5511240}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {12 r^{2}-6 r +1}{9 r^{2}+18 r +8}\) | \(\frac {13}{9}\) |
| \(b_{2}\) | \(\frac {54 r^{4}+45 r^{3}+70 r^{2}+48 r +7}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(-{\frac {5}{162}}\) |
| \(b_{3}\) | \(\frac {-108 r^{6}-540 r^{5}+528 r^{4}+3732 r^{3}+3489 r^{2}+739 r +224}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {1591}{30618}\) |
| \(b_{4}\) | \(\frac {-2997 r^{8}-29079 r^{7}-99054 r^{6}-125883 r^{5}+9826 r^{4}+138853 r^{3}+80233 r^{2}+5477 r +8064}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )}\) | \(\frac {106583}{5511240}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-17496 r^{10}-292572 r^{9}-2032830 r^{8}-7628094 r^{7}-16915518 r^{6}-23128560 r^{5}-20473273 r^{4}-12932200 r^{3}-6220667 r^{2}-1947462 r -14560}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right ) \left (9 r^{2}+90 r +224\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{5}={\frac {7435523}{3224075400}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {12 r^{2}-6 r +1}{9 r^{2}+18 r +8}\) | \(\frac {13}{9}\) |
| \(b_{2}\) | \(\frac {54 r^{4}+45 r^{3}+70 r^{2}+48 r +7}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(-{\frac {5}{162}}\) |
| \(b_{3}\) | \(\frac {-108 r^{6}-540 r^{5}+528 r^{4}+3732 r^{3}+3489 r^{2}+739 r +224}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {1591}{30618}\) |
| \(b_{4}\) | \(\frac {-2997 r^{8}-29079 r^{7}-99054 r^{6}-125883 r^{5}+9826 r^{4}+138853 r^{3}+80233 r^{2}+5477 r +8064}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )}\) | \(\frac {106583}{5511240}\) |
| \(b_{5}\) | \(\frac {-17496 r^{10}-292572 r^{9}-2032830 r^{8}-7628094 r^{7}-16915518 r^{6}-23128560 r^{5}-20473273 r^{4}-12932200 r^{3}-6220667 r^{2}-1947462 r -14560}{\left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right ) \left (9 r^{2}+90 r +224\right )}\) | \(\frac {7435523}{3224075400}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{3}} \left (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 \right ) \\ &= \frac {1+\frac {13 x}{9}-\frac {5 x^{2}}{162}+\frac {1591 x^{3}}{30618}+\frac {106583 x^{4}}{5511240}+\frac {7435523 x^{5}}{3224075400}+O\left (x^{6}\right )}{x^{\frac {1}{3}}} \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 {1}{3}} \left (1+\frac {x}{45}+\frac {149 x^{2}}{3240}+\frac {2701 x^{3}}{192456}+\frac {236933 x^{4}}{121247280}-\frac {67092967 x^{5}}{92754169200}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+\frac {13 x}{9}-\frac {5 x^{2}}{162}+\frac {1591 x^{3}}{30618}+\frac {106583 x^{4}}{5511240}+\frac {7435523 x^{5}}{3224075400}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{3}} \left (1+\frac {x}{45}+\frac {149 x^{2}}{3240}+\frac {2701 x^{3}}{192456}+\frac {236933 x^{4}}{121247280}-\frac {67092967 x^{5}}{92754169200}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {13 x}{9}-\frac {5 x^{2}}{162}+\frac {1591 x^{3}}{30618}+\frac {106583 x^{4}}{5511240}+\frac {7435523 x^{5}}{3224075400}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{3}} \left (1+\frac {x}{45}+\frac {149 x^{2}}{3240}+\frac {2701 x^{3}}{192456}+\frac {236933 x^{4}}{121247280}-\frac {67092967 x^{5}}{92754169200}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {13 x}{9}-\frac {5 x^{2}}{162}+\frac {1591 x^{3}}{30618}+\frac {106583 x^{4}}{5511240}+\frac {7435523 x^{5}}{3224075400}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{3}} \left (1+\frac {x}{45}+\frac {149 x^{2}}{3240}+\frac {2701 x^{3}}{192456}+\frac {236933 x^{4}}{121247280}-\frac {67092967 x^{5}}{92754169200}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {13 x}{9}-\frac {5 x^{2}}{162}+\frac {1591 x^{3}}{30618}+\frac {106583 x^{4}}{5511240}+\frac {7435523 x^{5}}{3224075400}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\right ) y^{\prime \prime }+\left (x^{3}-6 x^{2}+9 x \right ) y^{\prime }+\left (-1-x \right ) 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 {\left (1+x \right ) y}{x^{2} \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\right )}-\frac {\left (x^{2}-6 x +9\right ) y^{\prime }}{x \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\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 (x^{2}-6 x +9\right ) y^{\prime }}{x \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\right )}-\frac {\left (1+x \right ) y}{x^{2} \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\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 {x^{2}-6 x +9}{x \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\right )}, P_{3}\left (x \right )=-\frac {1+x}{x^{2} \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\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}{9} \\ {} & \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} \\ {} & {} & x^{2} \left (x^{4}-4 x^{3}+10 x^{2}-12 x +9\right ) y^{\prime \prime }+x \left (x^{2}-6 x +9\right ) y^{\prime }+\left (-1-x \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..3 \\ {} & {} & 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 =2..6 \\ {} & {} & 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} \left (1+3 r \right ) \left (-1+3 r \right ) x^{r}+\left (a_{1} \left (4+3 r \right ) \left (2+3 r \right )-a_{0} \left (12 r^{2}-6 r +1\right )\right ) x^{1+r}+\left (a_{2} \left (7+3 r \right ) \left (5+3 r \right )-a_{1} \left (12 r^{2}+18 r +7\right )+a_{0} r \left (-9+10 r \right )\right ) x^{2+r}+\left (a_{3} \left (10+3 r \right ) \left (8+3 r \right )-a_{2} \left (12 r^{2}+42 r +37\right )+a_{1} \left (1+r \right ) \left (1+10 r \right )-4 a_{0} r \left (-1+r \right )\right ) x^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (a_{k} \left (3 k +3 r +1\right ) \left (3 k +3 r -1\right )-a_{k -1} \left (12 \left (k -1\right )^{2}+24 \left (k -1\right ) r +12 r^{2}-6 k +7-6 r \right )+a_{k -2} \left (k -2+r \right ) \left (10 k -29+10 r \right )-4 a_{k -3} \left (k -3+r \right ) \left (k -4+r \right )+a_{k -4} \left (k -4+r \right ) \left (k -5+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+3 r \right ) \left (-1+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{3}, \frac {1}{3}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (4+3 r \right ) \left (2+3 r \right )-a_{0} \left (12 r^{2}-6 r +1\right )=0, a_{2} \left (7+3 r \right ) \left (5+3 r \right )-a_{1} \left (12 r^{2}+18 r +7\right )+a_{0} r \left (-9+10 r \right )=0, a_{3} \left (10+3 r \right ) \left (8+3 r \right )-a_{2} \left (12 r^{2}+42 r +37\right )+a_{1} \left (1+r \right ) \left (1+10 r \right )-4 a_{0} r \left (-1+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (12 r^{2}-6 r +1\right )}{9 r^{2}+18 r +8}, a_{2}=\frac {a_{0} \left (54 r^{4}+45 r^{3}+70 r^{2}+48 r +7\right )}{81 r^{4}+486 r^{3}+1035 r^{2}+918 r +280}, a_{3}=-\frac {a_{0} \left (108 r^{6}+540 r^{5}-528 r^{4}-3732 r^{3}-3489 r^{2}-739 r -224\right )}{729 r^{6}+8748 r^{5}+42039 r^{4}+103032 r^{3}+134892 r^{2}+88560 r +22400}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (9 a_{k}+a_{k -4}-4 a_{k -3}+10 a_{k -2}-12 a_{k -1}\right ) k^{2}+\left (2 \left (9 a_{k}+a_{k -4}-4 a_{k -3}+10 a_{k -2}-12 a_{k -1}\right ) r -9 a_{k -4}+28 a_{k -3}-49 a_{k -2}+30 a_{k -1}\right ) k +\left (9 a_{k}+a_{k -4}-4 a_{k -3}+10 a_{k -2}-12 a_{k -1}\right ) r^{2}+\left (-9 a_{k -4}+28 a_{k -3}-49 a_{k -2}+30 a_{k -1}\right ) r -a_{k}+20 a_{k -4}-48 a_{k -3}+58 a_{k -2}-19 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & \left (9 a_{k +4}+a_{k}-4 a_{k +1}+10 a_{k +2}-12 a_{k +3}\right ) \left (k +4\right )^{2}+\left (2 \left (9 a_{k +4}+a_{k}-4 a_{k +1}+10 a_{k +2}-12 a_{k +3}\right ) r -9 a_{k}+28 a_{k +1}-49 a_{k +2}+30 a_{k +3}\right ) \left (k +4\right )+\left (9 a_{k +4}+a_{k}-4 a_{k +1}+10 a_{k +2}-12 a_{k +3}\right ) r^{2}+\left (-9 a_{k}+28 a_{k +1}-49 a_{k +2}+30 a_{k +3}\right ) r -a_{k +4}+20 a_{k}-48 a_{k +1}+58 a_{k +2}-19 a_{k +3}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+10 k^{2} a_{k +2}-12 k^{2} a_{k +3}+2 k r a_{k}-8 k r a_{k +1}+20 k r a_{k +2}-24 k r a_{k +3}+r^{2} a_{k}-4 r^{2} a_{k +1}+10 r^{2} a_{k +2}-12 r^{2} a_{k +3}-k a_{k}-4 k a_{k +1}+31 k a_{k +2}-66 k a_{k +3}-r a_{k}-4 r a_{k +1}+31 r a_{k +2}-66 r a_{k +3}+22 a_{k +2}-91 a_{k +3}}{9 k^{2}+18 k r +9 r^{2}+72 k +72 r +143} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & a_{k +4}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+10 k^{2} a_{k +2}-12 k^{2} a_{k +3}-\frac {5}{3} k a_{k}-\frac {4}{3} k a_{k +1}+\frac {73}{3} k a_{k +2}-58 k a_{k +3}+\frac {4}{9} a_{k}+\frac {8}{9} a_{k +1}+\frac {115}{9} a_{k +2}-\frac {211}{3} a_{k +3}}{9 k^{2}+66 k +120} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{3}}, a_{k +4}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+10 k^{2} a_{k +2}-12 k^{2} a_{k +3}-\frac {5}{3} k a_{k}-\frac {4}{3} k a_{k +1}+\frac {73}{3} k a_{k +2}-58 k a_{k +3}+\frac {4}{9} a_{k}+\frac {8}{9} a_{k +1}+\frac {115}{9} a_{k +2}-\frac {211}{3} a_{k +3}}{9 k^{2}+66 k +120}, a_{1}=\frac {13 a_{0}}{9}, a_{2}=-\frac {5 a_{0}}{162}, a_{3}=\frac {1591 a_{0}}{30618}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +4}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+10 k^{2} a_{k +2}-12 k^{2} a_{k +3}-\frac {1}{3} k a_{k}-\frac {20}{3} k a_{k +1}+\frac {113}{3} k a_{k +2}-74 k a_{k +3}-\frac {2}{9} a_{k}-\frac {16}{9} a_{k +1}+\frac {301}{9} a_{k +2}-\frac {343}{3} a_{k +3}}{9 k^{2}+78 k +168} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{3}}, a_{k +4}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+10 k^{2} a_{k +2}-12 k^{2} a_{k +3}-\frac {1}{3} k a_{k}-\frac {20}{3} k a_{k +1}+\frac {113}{3} k a_{k +2}-74 k a_{k +3}-\frac {2}{9} a_{k}-\frac {16}{9} a_{k +1}+\frac {301}{9} a_{k +2}-\frac {343}{3} a_{k +3}}{9 k^{2}+78 k +168}, a_{1}=\frac {a_{0}}{45}, a_{2}=\frac {149 a_{0}}{3240}, a_{3}=\frac {2701 a_{0}}{192456}\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}{3}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{3}}\right ), a_{k +4}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+10 k^{2} a_{k +2}-12 k^{2} a_{k +3}-\frac {5}{3} k a_{k}-\frac {4}{3} k a_{k +1}+\frac {73}{3} k a_{k +2}-58 k a_{k +3}+\frac {4}{9} a_{k}+\frac {8}{9} a_{k +1}+\frac {115}{9} a_{k +2}-\frac {211}{3} a_{k +3}}{9 k^{2}+66 k +120}, a_{1}=\frac {13 a_{0}}{9}, a_{2}=-\frac {5 a_{0}}{162}, a_{3}=\frac {1591 a_{0}}{30618}, b_{k +4}=-\frac {k^{2} b_{k}-4 k^{2} b_{k +1}+10 k^{2} b_{k +2}-12 k^{2} b_{k +3}-\frac {1}{3} k b_{k}-\frac {20}{3} k b_{k +1}+\frac {113}{3} k b_{k +2}-74 k b_{k +3}-\frac {2}{9} b_{k}-\frac {16}{9} b_{k +1}+\frac {301}{9} b_{k +2}-\frac {343}{3} b_{k +3}}{9 k^{2}+78 k +168}, b_{1}=\frac {b_{0}}{45}, b_{2}=\frac {149 b_{0}}{3240}, b_{3}=\frac {2701 b_{0}}{192456}\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 -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 47
Order:=6; dsolve((x^3-2*x^2+3*x)^2*diff(y(x),x$2)+x*(x-3)^2*diff(y(x),x)-(x+1)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {2}{3}} \left (1+\frac {1}{45} x +\frac {149}{3240} x^{2}+\frac {2701}{192456} x^{3}+\frac {236933}{121247280} x^{4}-\frac {67092967}{92754169200} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1+\frac {13}{9} x -\frac {5}{162} x^{2}+\frac {1591}{30618} x^{3}+\frac {106583}{5511240} x^{4}+\frac {7435523}{3224075400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 90
AsymptoticDSolveValue[(x^3-2*x^2+3*x)^2*y''[x]+x*(x-3)^2*y'[x]-(x+1)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [3]{x} \left (-\frac {67092967 x^5}{92754169200}+\frac {236933 x^4}{121247280}+\frac {2701 x^3}{192456}+\frac {149 x^2}{3240}+\frac {x}{45}+1\right )+\frac {c_2 \left (\frac {7435523 x^5}{3224075400}+\frac {106583 x^4}{5511240}+\frac {1591 x^3}{30618}-\frac {5 x^2}{162}+\frac {13 x}{9}+1\right )}{\sqrt [3]{x}} \]