Internal problem ID [2407]
Internal file name [OUTPUT/2407_Tuesday_February_27_2024_08_37_00_AM_51977979/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 42, page 206
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]]
\[ \boxed {x^{2} \left (-x^{2}+1\right ) y^{\prime \prime }-5 y^{\prime } x +9 y=0} \] 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. \[ \left (-x^{4}+x^{2}\right ) y^{\prime \prime }-5 y^{\prime } x +9 y = 0 \] 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 {5}{x \left (x^{2}-1\right )}\\ q(x) &= -\frac {9}{x^{2} \left (x^{2}-1\right )}\\ \end {align*}
| \(p(x)=\frac {5}{x \left (x^{2}-1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {9}{x^{2} \left (x^{2}-1\right )}\) | |
| 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 \[ -y^{\prime \prime } x^{2} \left (x^{2}-1\right )-5 y^{\prime } x +9 y = 0 \] 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} -\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 )-5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +9 \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 (-x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 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 (-x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) 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 (-a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-5 x^{n +r} a_{n} \left (n +r \right )+9 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-5 x^{r} a_{0} r +9 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-5 x^{r} r +9 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r -3\right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (r -3\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= 3 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r -3\right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([3, 3]\).
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, \(r = 3\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +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 +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 \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} -a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+a_{n} \left (n +r \right ) \left (n +r -1\right )-5 a_{n} \left (n +r \right )+9 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {\left (n +r -2\right ) a_{n -2}}{n -3+r}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = \frac {\left (n +1\right ) a_{n -2}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 3\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r}{-1+r} \] Which for the root \(r = 3\) becomes \[ a_{2}={\frac {3}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {r}{-1+r}\) | \(\frac {3}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {r}{-1+r}\) | \(\frac {3}{2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (2+r \right )}{r^{2}-1} \] Which for the root \(r = 3\) becomes \[ a_{4}={\frac {15}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {r}{-1+r}\) | \(\frac {3}{2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {r \left (2+r \right )}{r^{2}-1}\) | \(\frac {15}{8}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {r}{-1+r}\) | \(\frac {3}{2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {r \left (2+r \right )}{r^{2}-1}\) | \(\frac {15}{8}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{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^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right ) \\ \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 \(r = 3\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =3\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {r}{-1+r}\) | \(\frac {3}{2}\) | \(-\frac {1}{\left (-1+r \right )^{2}}\) | \(-{\frac {1}{4}}\) |
| \(b_{3}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {r \left (2+r \right )}{r^{2}-1}\) | \(\frac {15}{8}\) | \(\frac {-2 r^{2}-2 r -2}{\left (r^{2}-1\right )^{2}}\) | \(-{\frac {13}{32}}\) |
| \(b_{5}\) | \(0\) | \(0\) | \(0\) | \(0\) |
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}\dots \\ &= x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{3} \left (-\frac {x^{2}}{4}-\frac {13 x^{4}}{32}+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} x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right ) + c_{2} \left (x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{3} \left (-\frac {x^{2}}{4}-\frac {13 x^{4}}{32}+O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right )+c_{2} \left (x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{3} \left (-\frac {x^{2}}{4}-\frac {13 x^{4}}{32}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right )+c_{2} \left (x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{3} \left (-\frac {x^{2}}{4}-\frac {13 x^{4}}{32}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right )+c_{2} \left (x^{3} \left (1+\frac {3 x^{2}}{2}+\frac {15 x^{4}}{8}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x^{3} \left (-\frac {x^{2}}{4}-\frac {13 x^{4}}{32}+O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{\prime \prime } x^{2} \left (x^{2}-1\right )-5 y^{\prime } x +9 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 {9 y}{x^{2} \left (x^{2}-1\right )}-\frac {5 y^{\prime }}{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 {5 y^{\prime }}{x \left (x^{2}-1\right )}-\frac {9 y}{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 {5}{x \left (x^{2}-1\right )}, P_{3}\left (x \right )=-\frac {9}{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 {5}{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} \\ {} & {} & y^{\prime \prime } x^{2} \left (x^{2}-1\right )+5 y^{\prime } x -9 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 (u^{4}-4 u^{3}+5 u^{2}-2 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (5 u -5\right ) \left (\frac {d}{d u}y \left (u \right )\right )-9 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 \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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} \\ {} & {} & -a_{0} r \left (3+2 r \right ) u^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (5+2 r \right )+a_{0} \left (5 r^{2}-9\right )\right ) u^{r}+\left (-a_{2} \left (2+r \right ) \left (7+2 r \right )+a_{1} \left (5 r^{2}+10 r -4\right )-4 a_{0} r \left (-1+r \right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k +5+2 r \right )+a_{k} \left (5 k^{2}+10 k r +5 r^{2}-9\right )-4 a_{k -1} \left (k +r -1\right ) \left (k -2+r \right )+a_{k -2} \left (k -2+r \right ) \left (k -3+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {3}{2}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-a_{1} \left (1+r \right ) \left (5+2 r \right )+a_{0} \left (5 r^{2}-9\right )=0, -a_{2} \left (2+r \right ) \left (7+2 r \right )+a_{1} \left (5 r^{2}+10 r -4\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 (5 r^{2}-9\right )}{2 r^{2}+7 r +5}, a_{2}=\frac {a_{0} \left (17 r^{4}+30 r^{3}-57 r^{2}-70 r +36\right )}{4 r^{4}+36 r^{3}+115 r^{2}+153 r +70}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) k^{2}+\left (2 \left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) r -5 a_{k -2}+12 a_{k -1}-7 a_{k +1}\right ) k +\left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) r^{2}+\left (-5 a_{k -2}+12 a_{k -1}-7 a_{k +1}\right ) r -9 a_{k}+6 a_{k -2}-8 a_{k -1}-5 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) \left (k +2\right )^{2}+\left (2 \left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) r -5 a_{k}+12 a_{k +1}-7 a_{k +3}\right ) \left (k +2\right )+\left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) r^{2}+\left (-5 a_{k}+12 a_{k +1}-7 a_{k +3}\right ) r -9 a_{k +2}+6 a_{k}-8 a_{k +1}-5 a_{k +3}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}+2 k r a_{k}-8 k r a_{k +1}+10 k r a_{k +2}+r^{2} a_{k}-4 r^{2} a_{k +1}+5 r^{2} a_{k +2}-k a_{k}-4 k a_{k +1}+20 k a_{k +2}-r a_{k}-4 r a_{k +1}+20 r a_{k +2}+11 a_{k +2}}{2 k^{2}+4 k r +2 r^{2}+15 k +15 r +27} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-k a_{k}-4 k a_{k +1}+20 k a_{k +2}+11 a_{k +2}}{2 k^{2}+15 k +27} \\ \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 {k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-k a_{k}-4 k a_{k +1}+20 k a_{k +2}+11 a_{k +2}}{2 k^{2}+15 k +27}, a_{1}=-\frac {9 a_{0}}{5}, a_{2}=\frac {18 a_{0}}{35}\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 {k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-k a_{k}-4 k a_{k +1}+20 k a_{k +2}+11 a_{k +2}}{2 k^{2}+15 k +27}, a_{1}=-\frac {9 a_{0}}{5}, a_{2}=\frac {18 a_{0}}{35}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3}{2} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-4 k a_{k}+8 k a_{k +1}+5 k a_{k +2}+\frac {15}{4} a_{k}-3 a_{k +1}-\frac {31}{4} a_{k +2}}{2 k^{2}+9 k +9} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {3}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {3}{2}}, a_{k +3}=\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-4 k a_{k}+8 k a_{k +1}+5 k a_{k +2}+\frac {15}{4} a_{k}-3 a_{k +1}-\frac {31}{4} a_{k +2}}{2 k^{2}+9 k +9}, a_{1}=-\frac {9 a_{0}}{4}, a_{2}=\frac {39 a_{0}}{32}\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}{2}}, a_{k +3}=\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-4 k a_{k}+8 k a_{k +1}+5 k a_{k +2}+\frac {15}{4} a_{k}-3 a_{k +1}-\frac {31}{4} a_{k +2}}{2 k^{2}+9 k +9}, a_{1}=-\frac {9 a_{0}}{4}, a_{2}=\frac {39 a_{0}}{32}\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}{2}}\right ), a_{k +3}=\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-k a_{k}-4 k a_{k +1}+20 k a_{k +2}+11 a_{k +2}}{2 k^{2}+15 k +27}, a_{1}=-\frac {9 a_{0}}{5}, a_{2}=\frac {18 a_{0}}{35}, b_{k +3}=\frac {k^{2} b_{k}-4 k^{2} b_{k +1}+5 k^{2} b_{k +2}-4 k b_{k}+8 k b_{k +1}+5 k b_{k +2}+\frac {15}{4} b_{k}-3 b_{k +1}-\frac {31}{4} b_{k +2}}{2 k^{2}+9 k +9}, b_{1}=-\frac {9 b_{0}}{4}, b_{2}=\frac {39 b_{0}}{32}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 51
Order:=6; dsolve(x^2*(1-x^2)*diff(y(x),x$2)-5*x*diff(y(x),x)+9*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+\frac {3}{2} x^{2}+\frac {15}{8} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+\left (-\frac {1}{4} x^{2}-\frac {13}{32} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \right ) x^{3} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 71
AsymptoticDSolveValue[x^2*(1-x^2)*y''[x]-5*x*y'[x]+9*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {15 x^4}{8}+\frac {3 x^2}{2}+1\right ) x^3+c_2 \left (\left (-\frac {13 x^4}{32}-\frac {x^2}{4}\right ) x^3+\left (\frac {15 x^4}{8}+\frac {3 x^2}{2}+1\right ) x^3 \log (x)\right ) \]