Internal problem ID [1299]
Internal file name [OUTPUT/1300_Sunday_June_05_2022_02_09_01_AM_25601244/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. 7.5 THE METHOD OF
FROBENIUS I. Exercises 7.5. Page 358
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 not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {12 x^{2} \left (x +1\right ) y^{\prime \prime }+x \left (3 x^{2}+35 x +11\right ) y^{\prime }-\left (-5 x^{2}-10 x +1\right ) y=0} \] 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. \[ \left (12 x^{3}+12 x^{2}\right ) y^{\prime \prime }+\left (3 x^{3}+35 x^{2}+11 x \right ) y^{\prime }+\left (5 x^{2}+10 x -1\right ) y = 0 \] 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 {3 x^{2}+35 x +11}{12 x \left (x +1\right )}\\ q(x) &= \frac {5 x^{2}+10 x -1}{12 x^{2} \left (x +1\right )}\\ \end {align*}
| \(p(x)=\frac {3 x^{2}+35 x +11}{12 x \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {5 x^{2}+10 x -1}{12 x^{2} \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 12 x^{2} \left (x +1\right ) y^{\prime \prime }+\left (3 x^{3}+35 x^{2}+11 x \right ) y^{\prime }+\left (5 x^{2}+10 x -1\right ) y = 0 \] 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} 12 x^{2} \left (x +1\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 (3 x^{3}+35 x^{2}+11 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (5 x^{2}+10 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 }}12 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 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 }}35 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 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 }}12 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}12 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \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 }}35 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}35 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}5 a_{n -2} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}10 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 }}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 }}12 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 =1}{\sum }}35 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}5 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}10 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 \[ 12 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+11 x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 12 x^{r} a_{0} r \left (-1+r \right )+11 x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (12 x^{r} r \left (-1+r \right )+11 x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (12 r^{2}-r -1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 12 r^{2}-r -1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{3}}\\ r_2 &= -{\frac {1}{4}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (12 r^{2}-r -1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{3}}, -{\frac {1}{4}}\right ]\).
Since \(r_1 - r_2 = {\frac {7}{12}}\) 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}{4}} \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 \[ a_{1} = -1 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 12 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+12 a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n -2} \left (n +r -2\right )+35 a_{n -1} \left (n +r -1\right )+11 a_{n} \left (n +r \right )+5 a_{n -2}+10 a_{n -1}-a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {4 n a_{n -1}+4 r a_{n -1}+a_{n -2}+a_{n -1}}{4 n +4 r +1}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{n} = \frac {\left (-12 n -7\right ) a_{n -1}-3 a_{n -2}}{12 n +7}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r +8}{9+4 r} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{2}={\frac {28}{31}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r +8}{9+4 r}\) | \(\frac {28}{31}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-16 r^{2}-80 r -95}{16 r^{2}+88 r +117} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{3}=-{\frac {1111}{1333}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r +8}{9+4 r}\) | \(\frac {28}{31}\) |
| \(a_{3}\) | \(\frac {-16 r^{2}-80 r -95}{16 r^{2}+88 r +117}\) | \(-{\frac {1111}{1333}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {64 r^{3}+576 r^{2}+1656 r +1511}{64 r^{3}+624 r^{2}+1964 r +1989} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{4}={\frac {57493}{73315}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r +8}{9+4 r}\) | \(\frac {28}{31}\) |
| \(a_{3}\) | \(\frac {-16 r^{2}-80 r -95}{16 r^{2}+88 r +117}\) | \(-{\frac {1111}{1333}}\) |
| \(a_{4}\) | \(\frac {64 r^{3}+576 r^{2}+1656 r +1511}{64 r^{3}+624 r^{2}+1964 r +1989}\) | \(\frac {57493}{73315}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {4 \left (64 r^{4}+896 r^{3}+4532 r^{2}+9770 r +7529\right )}{\left (16 r^{2}+88 r +117\right ) \left (17+4 r \right ) \left (21+4 r \right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{5}=-{\frac {3668716}{4912105}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {4 r +8}{9+4 r}\) | \(\frac {28}{31}\) |
| \(a_{3}\) | \(\frac {-16 r^{2}-80 r -95}{16 r^{2}+88 r +117}\) | \(-{\frac {1111}{1333}}\) |
| \(a_{4}\) | \(\frac {64 r^{3}+576 r^{2}+1656 r +1511}{64 r^{3}+624 r^{2}+1964 r +1989}\) | \(\frac {57493}{73315}\) |
| \(a_{5}\) | \(-\frac {4 \left (64 r^{4}+896 r^{3}+4532 r^{2}+9770 r +7529\right )}{\left (16 r^{2}+88 r +117\right ) \left (17+4 r \right ) \left (21+4 r \right )}\) | \(-{\frac {3668716}{4912105}}\) |
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-x +\frac {28 x^{2}}{31}-\frac {1111 x^{3}}{1333}+\frac {57493 x^{4}}{73315}-\frac {3668716 x^{5}}{4912105}+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 \[ b_{1} = -1 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 12 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+12 b_{n} \left (n +r \right ) \left (n +r -1\right )+3 b_{n -2} \left (n +r -2\right )+35 b_{n -1} \left (n +r -1\right )+11 b_{n} \left (n +r \right )+5 b_{n -2}+10 b_{n -1}-b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {4 n b_{n -1}+4 r b_{n -1}+b_{n -2}+b_{n -1}}{4 n +4 r +1}\tag {4} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ b_{n} = \frac {-4 n b_{n -1}-b_{n -2}}{4 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{4}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4 r +8}{9+4 r} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ b_{2}={\frac {7}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(\frac {4 r +8}{9+4 r}\) | \(\frac {7}{8}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-16 r^{2}-80 r -95}{16 r^{2}+88 r +117} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ b_{3}=-{\frac {19}{24}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(\frac {4 r +8}{9+4 r}\) | \(\frac {7}{8}\) |
| \(b_{3}\) | \(\frac {-16 r^{2}-80 r -95}{16 r^{2}+88 r +117}\) | \(-{\frac {19}{24}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {64 r^{3}+576 r^{2}+1656 r +1511}{64 r^{3}+624 r^{2}+1964 r +1989} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ b_{4}={\frac {283}{384}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(\frac {4 r +8}{9+4 r}\) | \(\frac {7}{8}\) |
| \(b_{3}\) | \(\frac {-16 r^{2}-80 r -95}{16 r^{2}+88 r +117}\) | \(-{\frac {19}{24}}\) |
| \(b_{4}\) | \(\frac {64 r^{3}+576 r^{2}+1656 r +1511}{64 r^{3}+624 r^{2}+1964 r +1989}\) | \(\frac {283}{384}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {4 \left (64 r^{4}+896 r^{3}+4532 r^{2}+9770 r +7529\right )}{\left (16 r^{2}+88 r +117\right ) \left (17+4 r \right ) \left (21+4 r \right )} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ b_{5}=-{\frac {1339}{1920}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-1\) | \(-1\) |
| \(b_{2}\) | \(\frac {4 r +8}{9+4 r}\) | \(\frac {7}{8}\) |
| \(b_{3}\) | \(\frac {-16 r^{2}-80 r -95}{16 r^{2}+88 r +117}\) | \(-{\frac {19}{24}}\) |
| \(b_{4}\) | \(\frac {64 r^{3}+576 r^{2}+1656 r +1511}{64 r^{3}+624 r^{2}+1964 r +1989}\) | \(\frac {283}{384}\) |
| \(b_{5}\) | \(-\frac {4 \left (64 r^{4}+896 r^{3}+4532 r^{2}+9770 r +7529\right )}{\left (16 r^{2}+88 r +117\right ) \left (17+4 r \right ) \left (21+4 r \right )}\) | \(-{\frac {1339}{1920}}\) |
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-x +\frac {7 x^{2}}{8}-\frac {19 x^{3}}{24}+\frac {283 x^{4}}{384}-\frac {1339 x^{5}}{1920}+O\left (x^{6}\right )}{x^{\frac {1}{4}}} \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-x +\frac {28 x^{2}}{31}-\frac {1111 x^{3}}{1333}+\frac {57493 x^{4}}{73315}-\frac {3668716 x^{5}}{4912105}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-x +\frac {7 x^{2}}{8}-\frac {19 x^{3}}{24}+\frac {283 x^{4}}{384}-\frac {1339 x^{5}}{1920}+O\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{3}} \left (1-x +\frac {28 x^{2}}{31}-\frac {1111 x^{3}}{1333}+\frac {57493 x^{4}}{73315}-\frac {3668716 x^{5}}{4912105}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-x +\frac {7 x^{2}}{8}-\frac {19 x^{3}}{24}+\frac {283 x^{4}}{384}-\frac {1339 x^{5}}{1920}+O\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{3}} \left (1-x +\frac {28 x^{2}}{31}-\frac {1111 x^{3}}{1333}+\frac {57493 x^{4}}{73315}-\frac {3668716 x^{5}}{4912105}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-x +\frac {7 x^{2}}{8}-\frac {19 x^{3}}{24}+\frac {283 x^{4}}{384}-\frac {1339 x^{5}}{1920}+O\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{3}} \left (1-x +\frac {28 x^{2}}{31}-\frac {1111 x^{3}}{1333}+\frac {57493 x^{4}}{73315}-\frac {3668716 x^{5}}{4912105}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-x +\frac {7 x^{2}}{8}-\frac {19 x^{3}}{24}+\frac {283 x^{4}}{384}-\frac {1339 x^{5}}{1920}+O\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 12 x^{2} \left (x +1\right ) y^{\prime \prime }+\left (3 x^{3}+35 x^{2}+11 x \right ) y^{\prime }+\left (5 x^{2}+10 x -1\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 (5 x^{2}+10 x -1\right ) y}{12 x^{2} \left (x +1\right )}-\frac {\left (3 x^{2}+35 x +11\right ) y^{\prime }}{12 x \left (x +1\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {\left (3 x^{2}+35 x +11\right ) y^{\prime }}{12 x \left (x +1\right )}+\frac {\left (5 x^{2}+10 x -1\right ) y}{12 x^{2} \left (x +1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {3 x^{2}+35 x +11}{12 x \left (x +1\right )}, P_{3}\left (x \right )=\frac {5 x^{2}+10 x -1}{12 x^{2} \left (x +1\right )}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=\frac {7}{4} \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 12 x^{2} \left (x +1\right ) y^{\prime \prime }+x \left (3 x^{2}+35 x +11\right ) y^{\prime }+\left (5 x^{2}+10 x -1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (12 u^{3}-24 u^{2}+12 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (3 u^{3}+26 u^{2}-50 u +21\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (5 u^{2}-6\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 3 a_{0} r \left (3+4 r \right ) u^{-1+r}+\left (3 a_{1} \left (1+r \right ) \left (7+4 r \right )-2 a_{0} \left (3+4 r \right ) \left (1+3 r \right )\right ) u^{r}+\left (3 a_{2} \left (2+r \right ) \left (11+4 r \right )-2 a_{1} \left (7+4 r \right ) \left (4+3 r \right )+2 a_{0} r \left (7+6 r \right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (3 a_{k +1} \left (k +1+r \right ) \left (4 k +7+4 r \right )-2 a_{k} \left (4 k +4 r +3\right ) \left (3 k +3 r +1\right )+2 a_{k -1} \left (k +r -1\right ) \left (6 k +1+6 r \right )+a_{k -2} \left (3 k -1+3 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 3 r \left (3+4 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {3}{4}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [3 a_{1} \left (1+r \right ) \left (7+4 r \right )-2 a_{0} \left (3+4 r \right ) \left (1+3 r \right )=0, 3 a_{2} \left (2+r \right ) \left (11+4 r \right )-2 a_{1} \left (7+4 r \right ) \left (4+3 r \right )+2 a_{0} r \left (7+6 r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {2 a_{0} \left (12 r^{2}+13 r +3\right )}{3 \left (4 r^{2}+11 r +7\right )}, a_{2}=\frac {2 a_{0} \left (54 r^{3}+135 r^{2}+101 r +24\right )}{9 \left (4 r^{3}+23 r^{2}+41 r +22\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 12 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) k^{2}+\left (24 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r -26 a_{k}+3 a_{k -2}-10 a_{k -1}+33 a_{k +1}\right ) k +12 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r^{2}+\left (-26 a_{k}+3 a_{k -2}-10 a_{k -1}+33 a_{k +1}\right ) r -6 a_{k}-a_{k -2}-2 a_{k -1}+21 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 12 \left (-2 a_{k +2}+a_{k +1}+a_{k +3}\right ) \left (k +2\right )^{2}+\left (24 \left (-2 a_{k +2}+a_{k +1}+a_{k +3}\right ) r -26 a_{k +2}+3 a_{k}-10 a_{k +1}+33 a_{k +3}\right ) \left (k +2\right )+12 \left (-2 a_{k +2}+a_{k +1}+a_{k +3}\right ) r^{2}+\left (-26 a_{k +2}+3 a_{k}-10 a_{k +1}+33 a_{k +3}\right ) r -6 a_{k +2}-a_{k}-2 a_{k +1}+21 a_{k +3}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {12 k^{2} a_{k +1}-24 k^{2} a_{k +2}+24 k r a_{k +1}-48 k r a_{k +2}+12 r^{2} a_{k +1}-24 r^{2} a_{k +2}+3 k a_{k}+38 k a_{k +1}-122 k a_{k +2}+3 r a_{k}+38 r a_{k +1}-122 r a_{k +2}+5 a_{k}+26 a_{k +1}-154 a_{k +2}}{3 \left (4 k^{2}+8 k r +4 r^{2}+27 k +27 r +45\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {12 k^{2} a_{k +1}-24 k^{2} a_{k +2}+3 k a_{k}+38 k a_{k +1}-122 k a_{k +2}+5 a_{k}+26 a_{k +1}-154 a_{k +2}}{3 \left (4 k^{2}+27 k +45\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +3}=-\frac {12 k^{2} a_{k +1}-24 k^{2} a_{k +2}+3 k a_{k}+38 k a_{k +1}-122 k a_{k +2}+5 a_{k}+26 a_{k +1}-154 a_{k +2}}{3 \left (4 k^{2}+27 k +45\right )}, a_{1}=\frac {2 a_{0}}{7}, a_{2}=\frac {8 a_{0}}{33}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +3}=-\frac {12 k^{2} a_{k +1}-24 k^{2} a_{k +2}+3 k a_{k}+38 k a_{k +1}-122 k a_{k +2}+5 a_{k}+26 a_{k +1}-154 a_{k +2}}{3 \left (4 k^{2}+27 k +45\right )}, a_{1}=\frac {2 a_{0}}{7}, a_{2}=\frac {8 a_{0}}{33}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3}{4} \\ {} & {} & a_{k +3}=-\frac {12 k^{2} a_{k +1}-24 k^{2} a_{k +2}+3 k a_{k}+20 k a_{k +1}-86 k a_{k +2}+\frac {11}{4} a_{k}+\frac {17}{4} a_{k +1}-76 a_{k +2}}{3 \left (4 k^{2}+21 k +27\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {3}{4} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {3}{4}}, a_{k +3}=-\frac {12 k^{2} a_{k +1}-24 k^{2} a_{k +2}+3 k a_{k}+20 k a_{k +1}-86 k a_{k +2}+\frac {11}{4} a_{k}+\frac {17}{4} a_{k +1}-76 a_{k +2}}{3 \left (4 k^{2}+21 k +27\right )}, a_{1}=0, a_{2}=\frac {a_{0}}{8}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -\frac {3}{4}}, a_{k +3}=-\frac {12 k^{2} a_{k +1}-24 k^{2} a_{k +2}+3 k a_{k}+20 k a_{k +1}-86 k a_{k +2}+\frac {11}{4} a_{k}+\frac {17}{4} a_{k +1}-76 a_{k +2}}{3 \left (4 k^{2}+21 k +27\right )}, a_{1}=0, a_{2}=\frac {a_{0}}{8}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k -\frac {3}{4}}\right ), a_{k +3}=-\frac {12 k^{2} a_{1+k}-24 k^{2} a_{k +2}+3 k a_{k}+38 k a_{1+k}-122 k a_{k +2}+5 a_{k}+26 a_{1+k}-154 a_{k +2}}{3 \left (4 k^{2}+27 k +45\right )}, a_{1}=\frac {2 a_{0}}{7}, a_{2}=\frac {8 a_{0}}{33}, b_{k +3}=-\frac {12 k^{2} b_{1+k}-24 k^{2} b_{k +2}+3 k b_{k}+20 k b_{1+k}-86 k b_{k +2}+\frac {11}{4} b_{k}+\frac {17}{4} b_{1+k}-76 b_{k +2}}{3 \left (4 k^{2}+21 k +27\right )}, b_{1}=0, b_{2}=\frac {b_{0}}{8}\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 Solution has integrals. Trying a special function solution free of integrals... -> 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 -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 47
Order:=6; dsolve(12*x^2*(1+x)*diff(y(x),x$2)+x*(11+35*x+3*x^2)*diff(y(x),x)-(1-10*x-5*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-x +\frac {7}{8} x^{2}-\frac {19}{24} x^{3}+\frac {283}{384} x^{4}-\frac {1339}{1920} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {1}{4}}}+c_{2} x^{\frac {1}{3}} \left (1-x +\frac {28}{31} x^{2}-\frac {1111}{1333} x^{3}+\frac {57493}{73315} x^{4}-\frac {3668716}{4912105} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 86
AsymptoticDSolveValue[12*x^2*(1+x)*y''[x]+x*(11+35*x+3*x^2)*y'[x]-(1-10*x-5*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [3]{x} \left (-\frac {3668716 x^5}{4912105}+\frac {57493 x^4}{73315}-\frac {1111 x^3}{1333}+\frac {28 x^2}{31}-x+1\right )+\frac {c_2 \left (-\frac {1339 x^5}{1920}+\frac {283 x^4}{384}-\frac {19 x^3}{24}+\frac {7 x^2}{8}-x+1\right )}{\sqrt [4]{x}} \]