Internal problem ID [1394]
Internal file name [OUTPUT/1395_Sunday_June_05_2022_02_14_51_AM_85027384/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: 42.
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]]
\[ \boxed {x^{2} \left (x^{2}+2\right ) y^{\prime \prime }+x \left (-x^{2}+14\right ) y^{\prime }+2 \left (x^{2}+9\right ) y=0} \] 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. \[ \left (x^{4}+2 x^{2}\right ) y^{\prime \prime }+\left (-x^{3}+14 x \right ) y^{\prime }+\left (2 x^{2}+18\right ) y = 0 \] 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 {x^{2}-14}{x \left (x^{2}+2\right )}\\ q(x) &= \frac {2 x^{2}+18}{x^{2} \left (x^{2}+2\right )}\\ \end {align*}
| \(p(x)=-\frac {x^{2}-14}{x \left (x^{2}+2\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -i \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = i \sqrt {2}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {2 x^{2}+18}{x^{2} \left (x^{2}+2\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -i \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = i \sqrt {2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -i \sqrt {2}, i \sqrt {2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}+2\right ) y^{\prime \prime }+\left (-x^{3}+14 x \right ) y^{\prime }+\left (2 x^{2}+18\right ) y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (x^{2}+2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-x^{3}+14 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (2 x^{2}+18\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}18 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 }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} 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 ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}18 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+14 x^{n +r} a_{n} \left (n +r \right )+18 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{r} a_{0} r \left (-1+r \right )+14 x^{r} a_{0} r +18 a_{0} x^{r} = 0 \] Or \[ \left (2 x^{r} r \left (-1+r \right )+14 x^{r} r +18 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 2 x^{r} \left (3+r \right )^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 \left (3+r \right )^{2} = 0 \] 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 \[ 2 x^{r} \left (3+r \right )^{2} = 0 \] 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 )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -2} \left (n +r -2\right )+14 a_{n} \left (n +r \right )+2 a_{n -2}+18 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -2} \left (n^{2}+2 n r +r^{2}-6 n -6 r +10\right )}{2 \left (n^{2}+2 n r +r^{2}+6 n +6 r +9\right )}\tag {4} \] Which for the root \(r = -3\) becomes \[ a_{n} = -\frac {a_{n -2} \left (n^{2}-12 n +37\right )}{2 n^{2}}\tag {5} \] 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 \[ a_{2}=\frac {-r^{2}+2 r -2}{2 \left (5+r \right )^{2}} \] Which for the root \(r = -3\) becomes \[ a_{2}=-{\frac {17}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-r^{2}+2 r -2}{2 \left (5+r \right )^{2}}\) | \(-{\frac {17}{8}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-r^{2}+2 r -2}{2 \left (5+r \right )^{2}}\) | \(-{\frac {17}{8}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{4}+4}{4 \left (5+r \right )^{2} \left (r +7\right )^{2}} \] Which for the root \(r = -3\) becomes \[ a_{4}={\frac {85}{256}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-r^{2}+2 r -2}{2 \left (5+r \right )^{2}}\) | \(-{\frac {17}{8}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {r^{4}+4}{4 \left (5+r \right )^{2} \left (r +7\right )^{2}}\) | \(\frac {85}{256}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-r^{2}+2 r -2}{2 \left (5+r \right )^{2}}\) | \(-{\frac {17}{8}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {r^{4}+4}{4 \left (5+r \right )^{2} \left (r +7\right )^{2}}\) | \(\frac {85}{256}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{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 ) \\ &= \frac {1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )}{x^{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 \(r = -3\). 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 |
| \(b_{1}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-r^{2}+2 r -2}{2 \left (5+r \right )^{2}}\) | \(-{\frac {17}{8}}\) | \(\frac {-6 r +7}{\left (5+r \right )^{3}}\) | \(\frac {25}{8}\) |
| \(b_{3}\) | \(0\) | \(0\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {r^{4}+4}{4 \left (5+r \right )^{2} \left (r +7\right )^{2}}\) | \(\frac {85}{256}\) | \(\frac {6 r^{4}+35 r^{3}-4 r -24}{\left (5+r \right )^{3} \left (r +7\right )^{3}}\) | \(-{\frac {471}{512}}\) |
| \(b_{5}\) | \(0\) | \(0\) | \(0\) | \(0\) |
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 \\ &= \frac {\left (1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {\frac {25 x^{2}}{8}-\frac {471 x^{4}}{512}+O\left (x^{6}\right )}{x^{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-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right )}{x^{3}} + c_{2} \left (\frac {\left (1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {\frac {25 x^{2}}{8}-\frac {471 x^{4}}{512}+O\left (x^{6}\right )}{x^{3}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right )}{x^{3}}+c_{2} \left (\frac {\left (1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {\frac {25 x^{2}}{8}-\frac {471 x^{4}}{512}+O\left (x^{6}\right )}{x^{3}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right )}{x^{3}}+c_{2} \left (\frac {\left (1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {\frac {25 x^{2}}{8}-\frac {471 x^{4}}{512}+O\left (x^{6}\right )}{x^{3}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right )}{x^{3}}+c_{2} \left (\frac {\left (1-\frac {17 x^{2}}{8}+\frac {85 x^{4}}{256}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {\frac {25 x^{2}}{8}-\frac {471 x^{4}}{512}+O\left (x^{6}\right )}{x^{3}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+2\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-x^{3}+14 x \right ) y^{\prime }+\left (2 x^{2}+18\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 {2 \left (x^{2}+9\right ) y}{x^{2} \left (x^{2}+2\right )}+\frac {\left (x^{2}-14\right ) y^{\prime }}{x \left (x^{2}+2\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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {\left (x^{2}-14\right ) y^{\prime }}{x \left (x^{2}+2\right )}+\frac {2 \left (x^{2}+9\right ) y}{x^{2} \left (x^{2}+2\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {x^{2}-14}{x \left (x^{2}+2\right )}, P_{3}\left (x \right )=\frac {2 \left (x^{2}+9\right )}{x^{2} \left (x^{2}+2\right )}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=7 \\ {} & \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}}}=9 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+2\right ) \left (\frac {d}{d x}y^{\prime }\right )-x \left (x^{2}-14\right ) y^{\prime }+\left (2 x^{2}+18\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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{0} \left (3+r \right )^{2} x^{r}+2 a_{1} \left (4+r \right )^{2} x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (2 a_{k} \left (k +r +3\right )^{2}+a_{k -2} \left (\left (k -2\right )^{2}+2 \left (k -2\right ) r +r^{2}-2 k +6-2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 \left (3+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =-3 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 2 a_{1} \left (4+r \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 a_{k} \left (k +r +3\right )^{2}+\left (k^{2}+\left (2 r -6\right ) k +r^{2}-6 r +10\right ) a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 2 a_{k +2} \left (k +5+r \right )^{2}+\left (\left (k +2\right )^{2}+\left (2 r -6\right ) \left (k +2\right )+r^{2}-6 r +10\right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {\left (k^{2}+2 k r +r^{2}-2 k -2 r +2\right ) a_{k}}{2 \left (k +5+r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-3 \\ {} & {} & a_{k +2}=-\frac {\left (k^{2}-8 k +17\right ) a_{k}}{2 \left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -3}, a_{k +2}=-\frac {\left (k^{2}-8 k +17\right ) a_{k}}{2 \left (k +2\right )^{2}}, a_{1}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful -> solution has integrals; searching for one without integrals... -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric solution without integrals succesful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 51
Order:=6; dsolve(x^2*(2+x^2)*diff(y(x),x$2)+x*(14-x^2)*diff(y(x),x)+2*(9+x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-\frac {17}{8} x^{2}+\frac {85}{256} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+\left (\frac {25}{8} x^{2}-\frac {471}{512} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) c_{2}}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 71
AsymptoticDSolveValue[x^2*(2+x^2)*y''[x]+x*(14-x^2)*y'[x]+2*(9+x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \frac {c_1 \left (\frac {85 x^4}{256}-\frac {17 x^2}{8}+1\right )}{x^3}+c_2 \left (\frac {\frac {25 x^2}{8}-\frac {471 x^4}{512}}{x^3}+\frac {\left (\frac {85 x^4}{256}-\frac {17 x^2}{8}+1\right ) \log (x)}{x^3}\right ) \]