Internal problem ID [2535]
Internal file name [OUTPUT/2027_Sunday_June_05_2022_02_45_18_AM_81816058/index.tex]
Book: Mathematical methods for physics and engineering, Riley, Hobson, Bence, second edition,
2002
Section: Chapter 16, Series solutions of ODEs. Section 16.6 Exercises, page 550
Problem number: Problem 16.8.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[_Lienard]
\[ \boxed {z y^{\prime \prime }-2 y^{\prime }+z y=0} \] With the expansion point for the power series method at \(z = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ z y^{\prime \prime }-2 y^{\prime }+z y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(z) y^{\prime } + q(z) y &=0 \end {align*}
Where \begin {align*} p(z) &= -\frac {2}{z}\\ q(z) &= 1\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(z = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ z y^{\prime \prime }-2 y^{\prime }+z y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} z^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} z^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} z \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} z^{n +r -2}\right )-2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} z^{n +r -1}\right )+z \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}z^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \left (n +r \right ) a_{n} z^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}z^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(z\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(z\) in it which is not already \(z^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}z^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} z^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(z\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}z^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \left (n +r \right ) a_{n} z^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} z^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ z^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 \left (n +r \right ) a_{n} z^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ z^{-1+r} a_{0} r \left (-1+r \right )-2 r a_{0} z^{-1+r} = 0 \] Or \[ \left (z^{-1+r} r \left (-1+r \right )-2 r \,z^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,z^{-1+r} \left (-3+r \right ) = 0 \] Since the above is true for all \(z\) then the indicial equation becomes \[ r \left (-3+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,z^{-1+r} \left (-3+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([3, 0]\).
Since \(r_1 - r_2 = 3\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (z \right ) &= z^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n}\right )\\ y_{2}\left (z \right ) &= C y_{1}\left (z \right ) \ln \left (z \right )+z^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} z^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (z \right ) &= z^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n}\right )\\ y_{2}\left (z \right ) &= C y_{1}\left (z \right ) \ln \left (z \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} z^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (z \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n +3}\\ y_{2}\left (z \right ) &= C y_{1}\left (z \right ) \ln \left (z \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} z^{n}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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} \left (n +r \right ) \left (n +r -1\right )-2 a_{n} \left (n +r \right )+a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -2}}{n^{2}+2 n r +r^{2}-3 n -3 r}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = -\frac {a_{n -2}}{n \left (n +3\right )}\tag {5} \] 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 \[ a_{2}=-\frac {1}{r^{2}+r -2} \] Which for the root \(r = 3\) becomes \[ a_{2}=-{\frac {1}{10}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}+r -2}\) | \(-{\frac {1}{10}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}+r -2}\) | \(-{\frac {1}{10}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{r^{4}+6 r^{3}+7 r^{2}-6 r -8} \] Which for the root \(r = 3\) becomes \[ a_{4}={\frac {1}{280}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}+r -2}\) | \(-{\frac {1}{10}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {1}{r^{4}+6 r^{3}+7 r^{2}-6 r -8}\) | \(\frac {1}{280}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}+r -2}\) | \(-{\frac {1}{10}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {1}{r^{4}+6 r^{3}+7 r^{2}-6 r -8}\) | \(\frac {1}{280}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (z \right )\) is \begin {align*} y_{1}\left (z \right )&= z^{3} \left (a_{0}+a_{1} z +a_{2} z^{2}+a_{3} z^{3}+a_{4} z^{4}+a_{5} z^{5}+a_{6} z^{6}\dots \right ) \\ &= z^{3} \left (1-\frac {z^{2}}{10}+\frac {z^{4}}{280}+O\left (z^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (z \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=3\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{3}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{3} \\ &= 0 \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow 0}0\\ &= 0 \end {align*}
The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (z \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} z^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} z^{n} \end {align*}
Eq (3) derived above is 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(3) gives \[ b_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )-2 \left (n +r \right ) b_{n}+b_{n -2} = 0 \end{equation} Which for for the root \(r = 0\) becomes \begin{equation} \tag{4A} b_{n} n \left (n -1\right )-2 n b_{n}+b_{n -2} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {b_{n -2}}{n^{2}+2 n r +r^{2}-3 n -3 r}\tag {5} \] Which for the root \(r = 0\) becomes \[ b_{n} = -\frac {b_{n -2}}{n^{2}-3 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) 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 \[ b_{2}=-\frac {1}{r^{2}+r -2} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {1}{r^{2}+r -2}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {1}{r^{2}+r -2}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{\left (r^{2}+r -2\right ) \left (r^{2}+5 r +4\right )} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {1}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {1}{r^{2}+r -2}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {1}{r^{4}+6 r^{3}+7 r^{2}-6 r -8}\) | \(-{\frac {1}{8}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {1}{r^{2}+r -2}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {1}{r^{4}+6 r^{3}+7 r^{2}-6 r -8}\) | \(-{\frac {1}{8}}\) |
| \(b_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (z \right )\) is \begin {align*} y_{2}\left (z \right )&= b_{0}+b_{1} z +b_{2} z^{2}+b_{3} z^{3}+b_{4} z^{4}+b_{5} z^{5}+b_{6} z^{6}\dots \\ &= 1+\frac {z^{2}}{2}-\frac {z^{4}}{8}+O\left (z^{6}\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(z) &= c_{1} y_{1}\left (z \right )+c_{2} y_{2}\left (z \right ) \\ &= c_{1} z^{3} \left (1-\frac {z^{2}}{10}+\frac {z^{4}}{280}+O\left (z^{6}\right )\right ) + c_{2} \left (1+\frac {z^{2}}{2}-\frac {z^{4}}{8}+O\left (z^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} z^{3} \left (1-\frac {z^{2}}{10}+\frac {z^{4}}{280}+O\left (z^{6}\right )\right )+c_{2} \left (1+\frac {z^{2}}{2}-\frac {z^{4}}{8}+O\left (z^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} z^{3} \left (1-\frac {z^{2}}{10}+\frac {z^{4}}{280}+O\left (z^{6}\right )\right )+c_{2} \left (1+\frac {z^{2}}{2}-\frac {z^{4}}{8}+O\left (z^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} z^{3} \left (1-\frac {z^{2}}{10}+\frac {z^{4}}{280}+O\left (z^{6}\right )\right )+c_{2} \left (1+\frac {z^{2}}{2}-\frac {z^{4}}{8}+O\left (z^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & z y^{\prime \prime }-2 y^{\prime }+z 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 {2 y^{\prime }}{z}-y \\ \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 {2 y^{\prime }}{z}+y=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} z_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (z \right )=-\frac {2}{z}, P_{3}\left (z \right )=1\right ] \\ {} & \circ & z \cdot P_{2}\left (z \right )\textrm {is analytic at}\hspace {3pt} z =0 \\ {} & {} & \left (z \cdot P_{2}\left (z \right )\right )\bigg | {\mstack {}{_{z \hiderel {=}0}}}=-2 \\ {} & \circ & z^{2}\cdot P_{3}\left (z \right )\textrm {is analytic at}\hspace {3pt} z =0 \\ {} & {} & \left (z^{2}\cdot P_{3}\left (z \right )\right )\bigg | {\mstack {}{_{z \hiderel {=}0}}}=0 \\ {} & \circ & z =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} z_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & z_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & z y^{\prime \prime }-2 y^{\prime }+z y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} z^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} z \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & z \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} z^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & z \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} z^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) z^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) z^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} z \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & z \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) z^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & z \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) \left (k +r \right ) z^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-3+r \right ) z^{-1+r}+a_{1} \left (1+r \right ) \left (-2+r \right ) z^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (k -2+r \right )+a_{k -1}\right ) z^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 3\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +r +1\right ) \left (k -2+r \right )+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +r -1\right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2+r \right ) \left (k +r -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} z^{k}, a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k -1\right )}, -2 a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +5\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} z^{k +3}, a_{k +2}=-\frac {a_{k}}{\left (k +5\right ) \left (k +2\right )}, 4 a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} z^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} z^{k +3}\right ), a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k -1\right )}, -2 a_{1}=0, b_{k +2}=-\frac {b_{k}}{\left (k +5\right ) \left (k +2\right )}, 4 b_{1}=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 Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 32
Order:=6; dsolve(z*diff(y(z),z$2)-2*diff(y(z),z)+z*y(z)=0,y(z),type='series',z=0);
\[ y \left (z \right ) = c_{1} z^{3} \left (1-\frac {1}{10} z^{2}+\frac {1}{280} z^{4}+\operatorname {O}\left (z^{6}\right )\right )+c_{2} \left (12+6 z^{2}-\frac {3}{2} z^{4}+\operatorname {O}\left (z^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.009 (sec). Leaf size: 44
AsymptoticDSolveValue[z*y''[z]-2*y'[z]+z*y[z]==0,y[z],{z,0,5}]
\[ y(z)\to c_1 \left (-\frac {z^4}{8}+\frac {z^2}{2}+1\right )+c_2 \left (\frac {z^7}{280}-\frac {z^5}{10}+z^3\right ) \]