Internal problem ID [1358]
Internal file name [OUTPUT/1359_Sunday_June_05_2022_02_12_59_AM_1175149/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 II. Exercises 7.6. Page 374
Problem number: 6.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {9 x^{2} y^{\prime \prime }+3 x \left (-2 x^{2}+3 x +5\right ) y^{\prime }+\left (-14 x^{2}+12 x +1\right ) y=0} \] 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. \[ 9 x^{2} y^{\prime \prime }+\left (-6 x^{3}+9 x^{2}+15 x \right ) y^{\prime }+\left (-14 x^{2}+12 x +1\right ) y = 0 \] 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 {2 x^{2}-3 x -5}{3 x}\\ q(x) &= -\frac {14 x^{2}-12 x -1}{9 x^{2}}\\ \end {align*}
| \(p(x)=-\frac {2 x^{2}-3 x -5}{3 x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \infty \) | \(\text {``regular''}\) |
| \(x = -\infty \) | \(\text {``regular''}\) |
| \(q(x)=-\frac {14 x^{2}-12 x -1}{9 x^{2}}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 9 x^{2} y^{\prime \prime }+\left (-6 x^{3}+9 x^{2}+15 x \right ) y^{\prime }+\left (-14 x^{2}+12 x +1\right ) y = 0 \] 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} 9 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-6 x^{3}+9 x^{2}+15 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-14 x^{2}+12 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 }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}15 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-14 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}\left (-6 x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-6 a_{n -2} \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}9 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-14 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-14 a_{n -2} x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}12 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}12 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 =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-6 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}15 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-14 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}12 a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+15 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 9 x^{r} a_{0} r \left (-1+r \right )+15 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (9 x^{r} r \left (-1+r \right )+15 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} \left (3 r +1\right )^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (3 r +1\right )^{2} = 0 \] 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 \[ x^{r} \left (3 r +1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [-{\frac {1}{3}}, -{\frac {1}{3}}\right ]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. Using the value of the indicial root found earlier, \(r = -{\frac {1}{3}}\), Eqs (1A,1B) become \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 ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n -\frac {1}{3}}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). 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 \[ a_{1} = -\frac {3}{3 r +4} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 9 a_{n} \left (n +r \right ) \left (n +r -1\right )-6 a_{n -2} \left (n +r -2\right )+9 a_{n -1} \left (n +r -1\right )+15 a_{n} \left (n +r \right )-14 a_{n -2}+12 a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 a_{n -2}-3 a_{n -1}}{1+3 n +3 r}\tag {4} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{n} = \frac {2 a_{n -2}-3 a_{n -1}}{3 n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{3 r +4}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {6 r +17}{9 r^{2}+33 r +28} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{2}={\frac {5}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{3 r +4}\) | \(-1\) |
| \(a_{2}\) | \(\frac {6 r +17}{9 r^{2}+33 r +28}\) | \(\frac {5}{6}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-36 r -93}{27 r^{3}+189 r^{2}+414 r +280} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{3}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{3 r +4}\) | \(-1\) |
| \(a_{2}\) | \(\frac {6 r +17}{9 r^{2}+33 r +28}\) | \(\frac {5}{6}\) |
| \(a_{3}\) | \(\frac {-36 r -93}{27 r^{3}+189 r^{2}+414 r +280}\) | \(-{\frac {1}{2}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {36 r^{2}+330 r +619}{81 r^{4}+918 r^{3}+3699 r^{2}+6222 r +3640} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{4}={\frac {19}{72}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{3 r +4}\) | \(-1\) |
| \(a_{2}\) | \(\frac {6 r +17}{9 r^{2}+33 r +28}\) | \(\frac {5}{6}\) |
| \(a_{3}\) | \(\frac {-36 r -93}{27 r^{3}+189 r^{2}+414 r +280}\) | \(-{\frac {1}{2}}\) |
| \(a_{4}\) | \(\frac {36 r^{2}+330 r +619}{81 r^{4}+918 r^{3}+3699 r^{2}+6222 r +3640}\) | \(\frac {19}{72}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-324 r^{2}-2484 r -4275}{243 r^{5}+4050 r^{4}+25785 r^{3}+77850 r^{2}+110472 r +58240} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{5}=-{\frac {43}{360}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{3 r +4}\) | \(-1\) |
| \(a_{2}\) | \(\frac {6 r +17}{9 r^{2}+33 r +28}\) | \(\frac {5}{6}\) |
| \(a_{3}\) | \(\frac {-36 r -93}{27 r^{3}+189 r^{2}+414 r +280}\) | \(-{\frac {1}{2}}\) |
| \(a_{4}\) | \(\frac {36 r^{2}+330 r +619}{81 r^{4}+918 r^{3}+3699 r^{2}+6222 r +3640}\) | \(\frac {19}{72}\) |
| \(a_{5}\) | \(\frac {-324 r^{2}-2484 r -4275}{243 r^{5}+4050 r^{4}+25785 r^{3}+77850 r^{2}+110472 r +58240}\) | \(-{\frac {43}{360}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {216 r^{3}+4104 r^{2}+21726 r +32633}{729 r^{6}+16767 r^{5}+154305 r^{4}+723465 r^{3}+1810566 r^{2}+2273688 r +1106560} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{6}={\frac {319}{6480}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{3 r +4}\) | \(-1\) |
| \(a_{2}\) | \(\frac {6 r +17}{9 r^{2}+33 r +28}\) | \(\frac {5}{6}\) |
| \(a_{3}\) | \(\frac {-36 r -93}{27 r^{3}+189 r^{2}+414 r +280}\) | \(-{\frac {1}{2}}\) |
| \(a_{4}\) | \(\frac {36 r^{2}+330 r +619}{81 r^{4}+918 r^{3}+3699 r^{2}+6222 r +3640}\) | \(\frac {19}{72}\) |
| \(a_{5}\) | \(\frac {-324 r^{2}-2484 r -4275}{243 r^{5}+4050 r^{4}+25785 r^{3}+77850 r^{2}+110472 r +58240}\) | \(-{\frac {43}{360}}\) |
| \(a_{6}\) | \(\frac {216 r^{3}+4104 r^{2}+21726 r +32633}{729 r^{6}+16767 r^{5}+154305 r^{4}+723465 r^{3}+1810566 r^{2}+2273688 r +1106560}\) | \(\frac {319}{6480}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-2592 r^{3}-39528 r^{2}-185220 r -260349}{2187 r^{7}+66339 r^{6}+831789 r^{5}+5565105 r^{4}+21347928 r^{3}+46653516 r^{2}+53340816 r +24344320} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{7}=-{\frac {167}{9072}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{3 r +4}\) | \(-1\) |
| \(a_{2}\) | \(\frac {6 r +17}{9 r^{2}+33 r +28}\) | \(\frac {5}{6}\) |
| \(a_{3}\) | \(\frac {-36 r -93}{27 r^{3}+189 r^{2}+414 r +280}\) | \(-{\frac {1}{2}}\) |
| \(a_{4}\) | \(\frac {36 r^{2}+330 r +619}{81 r^{4}+918 r^{3}+3699 r^{2}+6222 r +3640}\) | \(\frac {19}{72}\) |
| \(a_{5}\) | \(\frac {-324 r^{2}-2484 r -4275}{243 r^{5}+4050 r^{4}+25785 r^{3}+77850 r^{2}+110472 r +58240}\) | \(-{\frac {43}{360}}\) |
| \(a_{6}\) | \(\frac {216 r^{3}+4104 r^{2}+21726 r +32633}{729 r^{6}+16767 r^{5}+154305 r^{4}+723465 r^{3}+1810566 r^{2}+2273688 r +1106560}\) | \(\frac {319}{6480}\) |
| \(a_{7}\) | \(\frac {-2592 r^{3}-39528 r^{2}-185220 r -260349}{2187 r^{7}+66339 r^{6}+831789 r^{5}+5565105 r^{4}+21347928 r^{3}+46653516 r^{2}+53340816 r +24344320}\) | \(-{\frac {167}{9072}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= \frac {1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )}{x^{\frac {1}{3}}} \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = -{\frac {1}{3}}\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =-\frac {1}{3}\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(-\frac {3}{3 r +4}\) | \(-1\) | \(\frac {9}{\left (3 r +4\right )^{2}}\) | \(1\) |
| \(b_{2}\) | \(\frac {6 r +17}{9 r^{2}+33 r +28}\) | \(\frac {5}{6}\) | \(\frac {-54 r^{2}-306 r -393}{\left (9 r^{2}+33 r +28\right )^{2}}\) | \(-{\frac {11}{12}}\) |
| \(b_{3}\) | \(\frac {-36 r -93}{27 r^{3}+189 r^{2}+414 r +280}\) | \(-{\frac {1}{2}}\) | \(\frac {1944 r^{3}+14337 r^{2}+35154 r +28422}{\left (27 r^{3}+189 r^{2}+414 r +280\right )^{2}}\) | \(\frac {25}{36}\) |
| \(b_{4}\) | \(\frac {36 r^{2}+330 r +619}{81 r^{4}+918 r^{3}+3699 r^{2}+6222 r +3640}\) | \(\frac {19}{72}\) | \(-\frac {6 \left (972 r^{5}+18873 r^{4}+134406 r^{3}+450234 r^{2}+719547 r +441703\right )}{\left (81 r^{4}+918 r^{3}+3699 r^{2}+6222 r +3640\right )^{2}}\) | \(-{\frac {113}{288}}\) |
| \(b_{5}\) | \(\frac {-324 r^{2}-2484 r -4275}{243 r^{5}+4050 r^{4}+25785 r^{3}+77850 r^{2}+110472 r +58240}\) | \(-{\frac {43}{360}}\) | \(\frac {236196 r^{6}+5038848 r^{5}+43729065 r^{4}+197354880 r^{3}+488279097 r^{2}+627877980 r +327599640}{\left (243 r^{5}+4050 r^{4}+25785 r^{3}+77850 r^{2}+110472 r +58240\right )^{2}}\) | \(\frac {4211}{21600}\) |
| \(b_{6}\) | \(\frac {216 r^{3}+4104 r^{2}+21726 r +32633}{729 r^{6}+16767 r^{5}+154305 r^{4}+723465 r^{3}+1810566 r^{2}+2273688 r +1106560}\) | \(\frac {319}{6480}\) | \(-\frac {9 \left (52488 r^{8}+2134512 r^{7}+35439606 r^{6}+318487950 r^{5}+1707899661 r^{4}+5621723136 r^{3}+11123843391 r^{2}+12120639564 r +5572904216\right )}{\left (729 r^{6}+16767 r^{5}+154305 r^{4}+723465 r^{3}+1810566 r^{2}+2273688 r +1106560\right )^{2}}\) | \(-{\frac {32773}{388800}}\) |
| \(b_{7}\) | \(\frac {-2592 r^{3}-39528 r^{2}-185220 r -260349}{2187 r^{7}+66339 r^{6}+831789 r^{5}+5565105 r^{4}+21347928 r^{3}+46653516 r^{2}+53340816 r +24344320}\) | \(-{\frac {167}{9072}}\) | \(\frac {22674816 r^{9}+948090744 r^{8}+17231442984 r^{7}+178483849677 r^{6}+1159838529066 r^{5}+4897998400617 r^{4}+13427085744756 r^{3}+23017142146968 r^{2}+22367827912248 r +9378173154384}{\left (2187 r^{7}+66339 r^{6}+831789 r^{5}+5565105 r^{4}+21347928 r^{3}+46653516 r^{2}+53340816 r +24344320\right )^{2}}\) | \(\frac {126647}{3810240}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= \frac {\left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{3}}}+\frac {x -\frac {11 x^{2}}{12}+\frac {25 x^{3}}{36}-\frac {113 x^{4}}{288}+\frac {4211 x^{5}}{21600}-\frac {32773 x^{6}}{388800}+\frac {126647 x^{7}}{3810240}+O\left (x^{8}\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 ) \\ &= \frac {c_{1} \left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right )}{x^{\frac {1}{3}}} + c_{2} \left (\frac {\left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{3}}}+\frac {x -\frac {11 x^{2}}{12}+\frac {25 x^{3}}{36}-\frac {113 x^{4}}{288}+\frac {4211 x^{5}}{21600}-\frac {32773 x^{6}}{388800}+\frac {126647 x^{7}}{3810240}+O\left (x^{8}\right )}{x^{\frac {1}{3}}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right )}{x^{\frac {1}{3}}}+c_{2} \left (\frac {\left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{3}}}+\frac {x -\frac {11 x^{2}}{12}+\frac {25 x^{3}}{36}-\frac {113 x^{4}}{288}+\frac {4211 x^{5}}{21600}-\frac {32773 x^{6}}{388800}+\frac {126647 x^{7}}{3810240}+O\left (x^{8}\right )}{x^{\frac {1}{3}}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right )}{x^{\frac {1}{3}}}+c_{2} \left (\frac {\left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{3}}}+\frac {x -\frac {11 x^{2}}{12}+\frac {25 x^{3}}{36}-\frac {113 x^{4}}{288}+\frac {4211 x^{5}}{21600}-\frac {32773 x^{6}}{388800}+\frac {126647 x^{7}}{3810240}+O\left (x^{8}\right )}{x^{\frac {1}{3}}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right )}{x^{\frac {1}{3}}}+c_{2} \left (\frac {\left (1-x +\frac {5 x^{2}}{6}-\frac {x^{3}}{2}+\frac {19 x^{4}}{72}-\frac {43 x^{5}}{360}+\frac {319 x^{6}}{6480}-\frac {167 x^{7}}{9072}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{3}}}+\frac {x -\frac {11 x^{2}}{12}+\frac {25 x^{3}}{36}-\frac {113 x^{4}}{288}+\frac {4211 x^{5}}{21600}-\frac {32773 x^{6}}{388800}+\frac {126647 x^{7}}{3810240}+O\left (x^{8}\right )}{x^{\frac {1}{3}}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (-6 x^{3}+9 x^{2}+15 x \right ) y^{\prime }+\left (-14 x^{2}+12 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 (14 x^{2}-12 x -1\right ) y}{9 x^{2}}+\frac {\left (2 x^{2}-3 x -5\right ) y^{\prime }}{3 x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {\left (2 x^{2}-3 x -5\right ) y^{\prime }}{3 x}-\frac {\left (14 x^{2}-12 x -1\right ) y}{9 x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 x^{2}-3 x -5}{3 x}, P_{3}\left (x \right )=-\frac {14 x^{2}-12 x -1}{9 x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {5}{3} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {1}{9} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 9 x^{2} \left (\frac {d}{d x}y^{\prime }\right )-3 x \left (2 x^{2}-3 x -5\right ) y^{\prime }+\left (-14 x^{2}+12 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..2 \\ {} & {} & 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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+3 r \right )^{2} x^{r}+\left (a_{1} \left (4+3 r \right )^{2}+3 a_{0} \left (4+3 r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (3 k +3 r +1\right )^{2}+3 a_{k -1} \left (3 k +3 r +1\right )-2 a_{k -2} \left (3 k +3 r +1\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 )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =-\frac {1}{3} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (4+3 r \right )^{2}+3 a_{0} \left (4+3 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {3 a_{0}}{4+3 r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (3 k +3 r +1\right )^{2}+\left (3 k +3 r +1\right ) \left (-2 a_{k -2}+3 a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (3 k +3 r +7\right )^{2}+\left (3 k +3 r +7\right ) \left (-2 a_{k}+3 a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 a_{k}-3 a_{k +1}}{3 k +3 r +7} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & a_{k +2}=\frac {2 a_{k}-3 a_{k +1}}{3 k +6} \\ \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 +2}=\frac {2 a_{k}-3 a_{k +1}}{3 k +6}, a_{1}=-a_{0}\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 -> 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 -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunB ODE, case c = 0 Special function solution also has integrals. Returning default Liouvillian solution. <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 81
Order:=8; dsolve(9*x^2*diff(y(x),x$2)+3*x*(5+3*x-2*x^2)*diff(y(x),x)+(1+12*x-14*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-x +\frac {5}{6} x^{2}-\frac {1}{2} x^{3}+\frac {19}{72} x^{4}-\frac {43}{360} x^{5}+\frac {319}{6480} x^{6}-\frac {167}{9072} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (x -\frac {11}{12} x^{2}+\frac {25}{36} x^{3}-\frac {113}{288} x^{4}+\frac {4211}{21600} x^{5}-\frac {32773}{388800} x^{6}+\frac {126647}{3810240} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2}}{x^{\frac {1}{3}}} \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 168
AsymptoticDSolveValue[9*x^2*y''[x]+3*x*(5+3*x-2*x^2)*y'[x]+(1+12*x-14*x^2)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \frac {c_1 \left (-\frac {167 x^7}{9072}+\frac {319 x^6}{6480}-\frac {43 x^5}{360}+\frac {19 x^4}{72}-\frac {x^3}{2}+\frac {5 x^2}{6}-x+1\right )}{\sqrt [3]{x}}+c_2 \left (\frac {\frac {126647 x^7}{3810240}-\frac {32773 x^6}{388800}+\frac {4211 x^5}{21600}-\frac {113 x^4}{288}+\frac {25 x^3}{36}-\frac {11 x^2}{12}+x}{\sqrt [3]{x}}+\frac {\left (-\frac {167 x^7}{9072}+\frac {319 x^6}{6480}-\frac {43 x^5}{360}+\frac {19 x^4}{72}-\frac {x^3}{2}+\frac {5 x^2}{6}-x+1\right ) \log (x)}{\sqrt [3]{x}}\right ) \]