Internal problem ID [2538]
Internal file name [OUTPUT/2030_Sunday_June_05_2022_02_45_29_AM_29896145/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 16.11.
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]]
\[ \boxed {z y^{\prime \prime }+\left (2 z -3\right ) y^{\prime }+\frac {4 y}{z}=0} \] 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. \[ z y^{\prime \prime }+\left (2 z -3\right ) y^{\prime }+\frac {4 y}{z} = 0 \] 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 -3}{z}\\ q(z) &= \frac {4}{z^{2}}\\ \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 \[ z^{2} y^{\prime \prime }+\left (2 z^{2}-3 z \right ) y^{\prime }+4 y = 0 \] 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} z^{n +r -2}\right ) z^{2}+\left (2 z^{2}-3 z \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} z^{n +r -1}\right )+4 \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} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 z^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 z^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} z^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(z\) be \(n +r\) in each summation term. Going over each summation term above with power of \(z\) in it which is not already \(z^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 z^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) z^{n +r} \\ \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 \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}z^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) z^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 z^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} z^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ z^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 z^{n +r} a_{n} \left (n +r \right )+4 a_{n} z^{n +r} = 0 \] When \(n = 0\) the above becomes \[ z^{r} a_{0} r \left (-1+r \right )-3 z^{r} a_{0} r +4 a_{0} z^{r} = 0 \] Or \[ \left (z^{r} r \left (-1+r \right )-3 z^{r} r +4 z^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r -2\right )^{2} z^{r} = 0 \] Since the above is true for all \(z\) then the indicial equation becomes \[ \left (r -2\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= 2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r -2\right )^{2} z^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([2, 2]\).
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 (z \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (z \right ) &= y_{1}\left (z \right ) \ln \left (z \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} z^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (z \right )+c_{2} y_{2}\left (z \right ) \] 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, \(r = 2\), Eqs (1A,1B) become \begin {align*} y_{1}\left (z \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n +2}\\ y_{2}\left (z \right ) &= y_{1}\left (z \right ) \ln \left (z \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} z^{n +2}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (z \right )\). 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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -1} \left (n +r -1\right )-3 a_{n} \left (n +r \right )+4 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 a_{n -1} \left (n +r -1\right )}{n^{2}+2 n r +r^{2}-4 n -4 r +4}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = -\frac {2 a_{n -1} \left (1+n \right )}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=-\frac {2 r}{\left (-1+r \right )^{2}} \] Which for the root \(r = 2\) becomes \[ a_{1}=-4 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{\left (-1+r \right )^{2}}\) | \(-4\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4+4 r}{r \left (-1+r \right )^{2}} \] Which for the root \(r = 2\) becomes \[ a_{2}=6 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{\left (-1+r \right )^{2}}\) | \(-4\) |
| \(a_{2}\) | \(\frac {4+4 r}{r \left (-1+r \right )^{2}}\) | \(6\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-16-8 r}{\left (1+r \right ) r \left (-1+r \right )^{2}} \] Which for the root \(r = 2\) becomes \[ a_{3}=-{\frac {16}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{\left (-1+r \right )^{2}}\) | \(-4\) |
| \(a_{2}\) | \(\frac {4+4 r}{r \left (-1+r \right )^{2}}\) | \(6\) |
| \(a_{3}\) | \(\frac {-16-8 r}{\left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(-{\frac {16}{3}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {48+16 r}{\left (2+r \right ) \left (1+r \right ) r \left (-1+r \right )^{2}} \] Which for the root \(r = 2\) becomes \[ a_{4}={\frac {10}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{\left (-1+r \right )^{2}}\) | \(-4\) |
| \(a_{2}\) | \(\frac {4+4 r}{r \left (-1+r \right )^{2}}\) | \(6\) |
| \(a_{3}\) | \(\frac {-16-8 r}{\left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(-{\frac {16}{3}}\) |
| \(a_{4}\) | \(\frac {48+16 r}{\left (2+r \right ) \left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(\frac {10}{3}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-128-32 r}{\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (-1+r \right )^{2}} \] Which for the root \(r = 2\) becomes \[ a_{5}=-{\frac {8}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{\left (-1+r \right )^{2}}\) | \(-4\) |
| \(a_{2}\) | \(\frac {4+4 r}{r \left (-1+r \right )^{2}}\) | \(6\) |
| \(a_{3}\) | \(\frac {-16-8 r}{\left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(-{\frac {16}{3}}\) |
| \(a_{4}\) | \(\frac {48+16 r}{\left (2+r \right ) \left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(\frac {10}{3}\) |
| \(a_{5}\) | \(\frac {-128-32 r}{\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(-{\frac {8}{5}}\) |
Using the above table, then the first solution \(y_{1}\left (z \right )\) is \begin{align*} y_{1}\left (z \right )&= z^{2} \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^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (z \right ) = y_{1}\left (z \right ) \ln \left (z \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} z^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 2\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =2\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(-\frac {2 r}{\left (-1+r \right )^{2}}\) | \(-4\) | \(\frac {2 r +2}{\left (-1+r \right )^{3}}\) | \(6\) |
| \(b_{2}\) | \(\frac {4+4 r}{r \left (-1+r \right )^{2}}\) | \(6\) | \(\frac {-8 r^{2}-12 r +4}{r^{2} \left (-1+r \right )^{3}}\) | \(-13\) |
| \(b_{3}\) | \(\frac {-16-8 r}{\left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(-{\frac {16}{3}}\) | \(\frac {24 r^{3}+72 r^{2}+16 r -16}{\left (1+r \right )^{2} r^{2} \left (-1+r \right )^{3}}\) | \(\frac {124}{9}\) |
| \(b_{4}\) | \(\frac {48+16 r}{\left (2+r \right ) \left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(\frac {10}{3}\) | \(\frac {-64 r^{4}-352 r^{3}-448 r^{2}+96}{\left (2+r \right )^{2} \left (1+r \right )^{2} r^{2} \left (-1+r \right )^{3}}\) | \(-{\frac {173}{18}}\) |
| \(b_{5}\) | \(\frac {-128-32 r}{\left (3+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (-1+r \right )^{2}}\) | \(-{\frac {8}{5}}\) | \(\frac {160 r^{5}+1440 r^{4}+4000 r^{3}+3360 r^{2}-512 r -768}{\left (3+r \right )^{2} \left (2+r \right )^{2} \left (1+r \right )^{2} r^{2} \left (-1+r \right )^{3}}\) | \(\frac {374}{75}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (z \right )&=y_{1}\left (z \right ) \ln \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 \\ &= z^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right ) \ln \left (z \right )+z^{2} \left (-13 z^{2}+6 z +\frac {124 z^{3}}{9}-\frac {173 z^{4}}{18}+\frac {374 z^{5}}{75}+O\left (z^{6}\right )\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^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right ) + c_{2} \left (z^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right ) \ln \left (z \right )+z^{2} \left (-13 z^{2}+6 z +\frac {124 z^{3}}{9}-\frac {173 z^{4}}{18}+\frac {374 z^{5}}{75}+O\left (z^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} z^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right )+c_{2} \left (z^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right ) \ln \left (z \right )+z^{2} \left (-13 z^{2}+6 z +\frac {124 z^{3}}{9}-\frac {173 z^{4}}{18}+\frac {374 z^{5}}{75}+O\left (z^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} z^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right )+c_{2} \left (z^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right ) \ln \left (z \right )+z^{2} \left (-13 z^{2}+6 z +\frac {124 z^{3}}{9}-\frac {173 z^{4}}{18}+\frac {374 z^{5}}{75}+O\left (z^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} z^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right )+c_{2} \left (z^{2} \left (6 z^{2}-4 z +1-\frac {16 z^{3}}{3}+\frac {10 z^{4}}{3}-\frac {8 z^{5}}{5}+O\left (z^{6}\right )\right ) \ln \left (z \right )+z^{2} \left (-13 z^{2}+6 z +\frac {124 z^{3}}{9}-\frac {173 z^{4}}{18}+\frac {374 z^{5}}{75}+O\left (z^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } z^{2}+\left (2 z^{2}-3 z \right ) y^{\prime }+4 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 {4 y}{z^{2}}-\frac {\left (2 z -3\right ) y^{\prime }}{z} \\ \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 (2 z -3\right ) y^{\prime }}{z}+\frac {4 y}{z^{2}}=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 -3}{z}, P_{3}\left (z \right )=\frac {4}{z^{2}}\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}}}=-3 \\ {} & \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}}}=4 \\ {} & \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} \\ {} & {} & y^{\prime \prime } z^{2}+\left (2 z -3\right ) y^{\prime } z +4 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^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & z^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) z^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & z^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) z^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} z^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & z^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) z^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-2+r \right )^{2} z^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r -2\right )^{2}+2 a_{k -1} \left (k +r -1\right )\right ) z^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-2+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =2 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r -2\right )^{2}+2 a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (k +r -1\right )^{2}+2 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +r \right )}{\left (k +r -1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +2\right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} z^{k +2}, a_{k +1}=-\frac {2 a_{k} \left (k +2\right )}{\left (k +1\right )^{2}}\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: 0.0 (sec). Leaf size: 69
Order:=6; dsolve(z*diff(y(z),z$2)+(2*z-3)*diff(y(z),z)+4/z*y(z)=0,y(z),type='series',z=0);
\[ y \left (z \right ) = \left (\left (c_{2} \ln \left (z \right )+c_{1} \right ) \left (1-4 z +6 z^{2}-\frac {16}{3} z^{3}+\frac {10}{3} z^{4}-\frac {8}{5} z^{5}+\operatorname {O}\left (z^{6}\right )\right )+\left (6 z -13 z^{2}+\frac {124}{9} z^{3}-\frac {173}{18} z^{4}+\frac {374}{75} z^{5}+\operatorname {O}\left (z^{6}\right )\right ) c_{2} \right ) z^{2} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 116
AsymptoticDSolveValue[z*y''[z]+(2*z-3)*y'[z]+4/z*y[z]==0,y[z],{z,0,5}]
\[ y(z)\to c_1 \left (-\frac {8 z^5}{5}+\frac {10 z^4}{3}-\frac {16 z^3}{3}+6 z^2-4 z+1\right ) z^2+c_2 \left (\left (\frac {374 z^5}{75}-\frac {173 z^4}{18}+\frac {124 z^3}{9}-13 z^2+6 z\right ) z^2+\left (-\frac {8 z^5}{5}+\frac {10 z^4}{3}-\frac {16 z^3}{3}+6 z^2-4 z+1\right ) z^2 \log (z)\right ) \]