Internal problem ID [1323]
Internal file name [OUTPUT/1324_Sunday_June_05_2022_02_10_21_AM_82739945/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: 34.
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]]
\[ \boxed {8 x^{2} \left (-x^{2}+1\right ) y^{\prime \prime }+2 x \left (-13 x^{2}+1\right ) y^{\prime }+\left (-9 x^{2}+1\right ) y=0} \] 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. \[ \left (-8 x^{4}+8 x^{2}\right ) y^{\prime \prime }+\left (-26 x^{3}+2 x \right ) y^{\prime }+\left (-9 x^{2}+1\right ) y = 0 \] 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 {13 x^{2}-1}{4 x \left (x^{2}-1\right )}\\ q(x) &= \frac {9 x^{2}-1}{8 x^{2} \left (x^{2}-1\right )}\\ \end {align*}
| \(p(x)=\frac {13 x^{2}-1}{4 x \left (x^{2}-1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=\frac {9 x^{2}-1}{8 x^{2} \left (x^{2}-1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0, 1, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -8 y^{\prime \prime } x^{2} \left (x^{2}-1\right )+\left (-26 x^{3}+2 x \right ) y^{\prime }+\left (-9 x^{2}+1\right ) y = 0 \] 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} -8 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2} \left (x^{2}-1\right )+\left (-26 x^{3}+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-9 x^{2}+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} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-26 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-9 x^{n +r +2} 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 (-8 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-8 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-26 x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-26 a_{n -2} \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-9 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-9 a_{n -2} 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} \moverset {\infty }{\munderset {n =2}{\sum }}\left (-8 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-26 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-9 a_{n -2} 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 \[ 8 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 8 x^{r} a_{0} r \left (-1+r \right )+2 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (8 x^{r} r \left (-1+r \right )+2 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (8 r^{2}-6 r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 8 r^{2}-6 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= {\frac {1}{4}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (8 r^{2}-6 r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, {\frac {1}{4}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{4}}\) 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}{2}}\\ 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 \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} -8 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+8 a_{n} \left (n +r \right ) \left (n +r -1\right )-26 a_{n -2} \left (n +r -2\right )+2 a_{n} \left (n +r \right )-9 a_{n -2}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {\left (4 n +4 r -5\right ) a_{n -2}}{4 n +4 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {\left (4 n -3\right ) a_{n -2}}{4 n +1}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {3+4 r}{7+4 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {5}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {3+4 r}{7+4 r}\) | \(\frac {5}{9}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {3+4 r}{7+4 r}\) | \(\frac {5}{9}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 r^{2}+56 r +33}{16 r^{2}+88 r +105} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {65}{153}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {3+4 r}{7+4 r}\) | \(\frac {5}{9}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {16 r^{2}+56 r +33}{16 r^{2}+88 r +105}\) | \(\frac {65}{153}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {3+4 r}{7+4 r}\) | \(\frac {5}{9}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {16 r^{2}+56 r +33}{16 r^{2}+88 r +105}\) | \(\frac {65}{153}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \sqrt {x} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= \sqrt {x}\, \left (1+\frac {5 x^{2}}{9}+\frac {65 x^{4}}{153}+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 \[ b_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} -8 b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+8 b_{n} \left (n +r \right ) \left (n +r -1\right )-26 b_{n -2} \left (n +r -2\right )+2 b_{n} \left (n +r \right )-9 b_{n -2}+b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {\left (4 n +4 r -5\right ) b_{n -2}}{4 n +4 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{n} = \frac {\left (n -1\right ) b_{n -2}}{n}\tag {5} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {3+4 r}{7+4 r} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {3+4 r}{7+4 r}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {3+4 r}{7+4 r}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {16 r^{2}+56 r +33}{16 r^{2}+88 r +105} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{4}={\frac {3}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {3+4 r}{7+4 r}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {16 r^{2}+56 r +33}{16 r^{2}+88 r +105}\) | \(\frac {3}{8}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {3+4 r}{7+4 r}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {16 r^{2}+56 r +33}{16 r^{2}+88 r +105}\) | \(\frac {3}{8}\) |
| \(b_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= \sqrt {x} \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 ) \\ &= x^{\frac {1}{4}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{8}+O\left (x^{6}\right )\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \sqrt {x}\, \left (1+\frac {5 x^{2}}{9}+\frac {65 x^{4}}{153}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {1}{4}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{8}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1+\frac {5 x^{2}}{9}+\frac {65 x^{4}}{153}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{8}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1+\frac {5 x^{2}}{9}+\frac {65 x^{4}}{153}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{8}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1+\frac {5 x^{2}}{9}+\frac {65 x^{4}}{153}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{8}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -8 y^{\prime \prime } x^{2} \left (x^{2}-1\right )+\left (-26 x^{3}+2 x \right ) y^{\prime }+\left (-9 x^{2}+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 (9 x^{2}-1\right ) y}{8 x^{2} \left (x^{2}-1\right )}-\frac {\left (13 x^{2}-1\right ) y^{\prime }}{4 x \left (x^{2}-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 (13 x^{2}-1\right ) y^{\prime }}{4 x \left (x^{2}-1\right )}+\frac {\left (9 x^{2}-1\right ) y}{8 x^{2} \left (x^{2}-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 {13 x^{2}-1}{4 x \left (x^{2}-1\right )}, P_{3}\left (x \right )=\frac {9 x^{2}-1}{8 x^{2} \left (x^{2}-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 {3}{2} \\ {} & \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} \\ {} & {} & 8 y^{\prime \prime } x^{2} \left (x^{2}-1\right )+2 x \left (13 x^{2}-1\right ) y^{\prime }+y \left (9 x^{2}-1\right )=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 (8 u^{4}-32 u^{3}+40 u^{2}-16 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (26 u^{3}-78 u^{2}+76 u -24\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (9 u^{2}-18 u +8\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..4 \\ {} & {} & 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} \\ {} & {} & -8 a_{0} r \left (1+2 r \right ) u^{-1+r}+\left (-8 a_{1} \left (1+r \right ) \left (3+2 r \right )+4 a_{0} \left (1+2 r \right ) \left (2+5 r \right )\right ) u^{r}+\left (-8 a_{2} \left (2+r \right ) \left (5+2 r \right )+4 a_{1} \left (3+2 r \right ) \left (7+5 r \right )-2 a_{0} \left (16 r^{2}+23 r +9\right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-8 a_{k +1} \left (k +1+r \right ) \left (2 k +2 r +3\right )+4 a_{k} \left (2 k +2 r +1\right ) \left (5 k +5 r +2\right )-2 a_{k -1} \left (16 \left (k -1\right )^{2}+32 \left (k -1\right ) r +16 r^{2}+23 k -14+23 r \right )+a_{k -2} \left (2 k -1+2 r \right ) \left (4 k -5+4 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -8 r \left (1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {1}{2}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-8 a_{1} \left (1+r \right ) \left (3+2 r \right )+4 a_{0} \left (1+2 r \right ) \left (2+5 r \right )=0, -8 a_{2} \left (2+r \right ) \left (5+2 r \right )+4 a_{1} \left (3+2 r \right ) \left (7+5 r \right )-2 a_{0} \left (16 r^{2}+23 r +9\right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (10 r^{2}+9 r +2\right )}{2 \left (2 r^{2}+5 r +3\right )}, a_{2}=\frac {a_{0} \left (34 r^{3}+76 r^{2}+41 r +5\right )}{4 \left (2 r^{3}+11 r^{2}+19 r +10\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 8 \left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) k^{2}+2 \left (8 \left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) r +18 a_{k}-7 a_{k -2}+9 a_{k -1}-20 a_{k +1}\right ) k +8 \left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) r^{2}+2 \left (18 a_{k}-7 a_{k -2}+9 a_{k -1}-20 a_{k +1}\right ) r +8 a_{k}+5 a_{k -2}-4 a_{k -1}-24 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 8 \left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) \left (k +2\right )^{2}+2 \left (8 \left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) r +18 a_{k +2}-7 a_{k}+9 a_{k +1}-20 a_{k +3}\right ) \left (k +2\right )+8 \left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) r^{2}+2 \left (18 a_{k +2}-7 a_{k}+9 a_{k +1}-20 a_{k +3}\right ) r +8 a_{k +2}+5 a_{k}-4 a_{k +1}-24 a_{k +3}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {8 k^{2} a_{k}-32 k^{2} a_{k +1}+40 k^{2} a_{k +2}+16 k r a_{k}-64 k r a_{k +1}+80 k r a_{k +2}+8 r^{2} a_{k}-32 r^{2} a_{k +1}+40 r^{2} a_{k +2}+18 k a_{k}-110 k a_{k +1}+196 k a_{k +2}+18 r a_{k}-110 r a_{k +1}+196 r a_{k +2}+9 a_{k}-96 a_{k +1}+240 a_{k +2}}{8 \left (2 k^{2}+4 k r +2 r^{2}+13 k +13 r +21\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {8 k^{2} a_{k}-32 k^{2} a_{k +1}+40 k^{2} a_{k +2}+18 k a_{k}-110 k a_{k +1}+196 k a_{k +2}+9 a_{k}-96 a_{k +1}+240 a_{k +2}}{8 \left (2 k^{2}+13 k +21\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 {8 k^{2} a_{k}-32 k^{2} a_{k +1}+40 k^{2} a_{k +2}+18 k a_{k}-110 k a_{k +1}+196 k a_{k +2}+9 a_{k}-96 a_{k +1}+240 a_{k +2}}{8 \left (2 k^{2}+13 k +21\right )}, a_{1}=\frac {a_{0}}{3}, 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}, a_{k +3}=\frac {8 k^{2} a_{k}-32 k^{2} a_{k +1}+40 k^{2} a_{k +2}+18 k a_{k}-110 k a_{k +1}+196 k a_{k +2}+9 a_{k}-96 a_{k +1}+240 a_{k +2}}{8 \left (2 k^{2}+13 k +21\right )}, a_{1}=\frac {a_{0}}{3}, a_{2}=\frac {a_{0}}{8}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +3}=\frac {8 k^{2} a_{k}-32 k^{2} a_{k +1}+40 k^{2} a_{k +2}+10 k a_{k}-78 k a_{k +1}+156 k a_{k +2}+2 a_{k}-49 a_{k +1}+152 a_{k +2}}{8 \left (2 k^{2}+11 k +15\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {1}{2}}, a_{k +3}=\frac {8 k^{2} a_{k}-32 k^{2} a_{k +1}+40 k^{2} a_{k +2}+10 k a_{k}-78 k a_{k +1}+156 k a_{k +2}+2 a_{k}-49 a_{k +1}+152 a_{k +2}}{8 \left (2 k^{2}+11 k +15\right )}, a_{1}=0, a_{2}=-\frac {a_{0}}{16}\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 {1}{2}}, a_{k +3}=\frac {8 k^{2} a_{k}-32 k^{2} a_{k +1}+40 k^{2} a_{k +2}+10 k a_{k}-78 k a_{k +1}+156 k a_{k +2}+2 a_{k}-49 a_{k +1}+152 a_{k +2}}{8 \left (2 k^{2}+11 k +15\right )}, a_{1}=0, a_{2}=-\frac {a_{0}}{16}\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 {1}{2}}\right ), a_{k +3}=\frac {8 k^{2} a_{k}-32 k^{2} a_{1+k}+40 k^{2} a_{k +2}+18 k a_{k}-110 k a_{1+k}+196 k a_{k +2}+9 a_{k}-96 a_{1+k}+240 a_{k +2}}{8 \left (2 k^{2}+13 k +21\right )}, a_{1}=\frac {a_{0}}{3}, a_{2}=\frac {a_{0}}{8}, b_{k +3}=\frac {8 k^{2} b_{k}-32 k^{2} b_{1+k}+40 k^{2} b_{k +2}+10 k b_{k}-78 k b_{1+k}+156 k b_{k +2}+2 b_{k}-49 b_{1+k}+152 b_{k +2}}{8 \left (2 k^{2}+11 k +15\right )}, b_{1}=0, b_{2}=-\frac {b_{0}}{16}\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 <- heuristic approach successful <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 35
Order:=6; dsolve(8*x^2*(1-x^2)*diff(y(x),x$2)+2*x*(1-13*x^2)*diff(y(x),x)+(1-9*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {1}{4}} \left (1+\frac {1}{2} x^{2}+\frac {3}{8} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \sqrt {x}\, \left (1+\frac {5}{9} x^{2}+\frac {65}{153} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 52
AsymptoticDSolveValue[8*x^2*(1-x^2)*y''[x]+2*x*(1-13*x^2)*y'[x]+(1-9*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt {x} \left (\frac {65 x^4}{153}+\frac {5 x^2}{9}+1\right )+c_2 \sqrt [4]{x} \left (\frac {3 x^4}{8}+\frac {x^2}{2}+1\right ) \]