Internal problem ID [2531]
Internal file name [OUTPUT/2023_Sunday_June_05_2022_02_45_01_AM_49575234/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 16.2.
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]]
\[ \boxed {4 z y^{\prime \prime }+2 \left (1-z \right ) y^{\prime }-y=0} \] 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. \[ 4 z y^{\prime \prime }+\left (-2 z +2\right ) y^{\prime }-y = 0 \] 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 {z -1}{2 z}\\ q(z) &= -\frac {1}{4 z}\\ \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 \[ 4 z y^{\prime \prime }+\left (-2 z +2\right ) y^{\prime }-y = 0 \] 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} 4 z \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} z^{n +r -2}\right )+\left (-2 z +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} z^{n +r -1}\right )-\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 }}4 z^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 z^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +r \right ) a_{n} z^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} z^{n +r}\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 }}\left (-2 z^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) z^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} z^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} z^{n +r -1}\right ) \\ \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 }}4 z^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) z^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +r \right ) a_{n} z^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} z^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 z^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 \left (n +r \right ) a_{n} z^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 4 z^{-1+r} a_{0} r \left (-1+r \right )+2 r a_{0} z^{-1+r} = 0 \] Or \[ \left (4 z^{-1+r} r \left (-1+r \right )+2 r \,z^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-2 r \right ) z^{-1+r} = 0 \] Since the above is true for all \(z\) then the indicial equation becomes \[ 4 r^{2}-2 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-2 r \right ) z^{-1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) is not 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 ) &= 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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n +\frac {1}{2}}\\ y_{2}\left (z \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} z^{n} \end {align*}
We start by finding \(y_{1}\left (z \right )\). 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} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )+2 a_{n} \left (n +r \right )-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1}}{2 n +2 r}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {a_{n -1}}{2 n +1}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {1}{2+2 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{4 \left (1+r \right ) \left (2+r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {1}{15}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {1}{4 \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{15}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {1}{8 \left (1+r \right ) \left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}={\frac {1}{105}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {1}{4 \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{15}\) |
| \(a_{3}\) | \(\frac {1}{8 \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{105}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{16 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {1}{945}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {1}{4 \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{15}\) |
| \(a_{3}\) | \(\frac {1}{8 \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{105}\) |
| \(a_{4}\) | \(\frac {1}{16 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}\) | \(\frac {1}{945}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1}{32 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}={\frac {1}{10395}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {1}{4 \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{15}\) |
| \(a_{3}\) | \(\frac {1}{8 \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{105}\) |
| \(a_{4}\) | \(\frac {1}{16 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}\) | \(\frac {1}{945}\) |
| \(a_{5}\) | \(\frac {1}{32 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(\frac {1}{10395}\) |
Using the above table, then the solution \(y_{1}\left (z \right )\) is \begin {align*} y_{1}\left (z \right )&= \sqrt {z} \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 ) \\ &= \sqrt {z}\, \left (1+\frac {z}{3}+\frac {z^{2}}{15}+\frac {z^{3}}{105}+\frac {z^{4}}{945}+\frac {z^{5}}{10395}+O\left (z^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (z \right )\) 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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 b_{n} \left (n +r \right ) \left (n +r -1\right )-2 b_{n -1} \left (n +r -1\right )+2 \left (n +r \right ) b_{n}-b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -1}}{2 n +2 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {b_{n -1}}{2 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=\frac {1}{2+2 r} \] Which for the root \(r = 0\) becomes \[ b_{1}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{2}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1}{4 \left (1+r \right ) \left (2+r \right )} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {1}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{8}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {1}{8 \left (1+r \right ) \left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = 0\) becomes \[ b_{3}={\frac {1}{48}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(\frac {1}{8 \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{48}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{16 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {1}{384}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(\frac {1}{8 \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{48}\) |
| \(b_{4}\) | \(\frac {1}{16 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}\) | \(\frac {1}{384}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {1}{32 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = 0\) becomes \[ b_{5}={\frac {1}{3840}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(\frac {1}{8 \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{48}\) |
| \(b_{4}\) | \(\frac {1}{16 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}\) | \(\frac {1}{384}\) |
| \(b_{5}\) | \(\frac {1}{32 \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(\frac {1}{3840}\) |
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}+\frac {z^{2}}{8}+\frac {z^{3}}{48}+\frac {z^{4}}{384}+\frac {z^{5}}{3840}+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} \sqrt {z}\, \left (1+\frac {z}{3}+\frac {z^{2}}{15}+\frac {z^{3}}{105}+\frac {z^{4}}{945}+\frac {z^{5}}{10395}+O\left (z^{6}\right )\right ) + c_{2} \left (1+\frac {z}{2}+\frac {z^{2}}{8}+\frac {z^{3}}{48}+\frac {z^{4}}{384}+\frac {z^{5}}{3840}+O\left (z^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {z}\, \left (1+\frac {z}{3}+\frac {z^{2}}{15}+\frac {z^{3}}{105}+\frac {z^{4}}{945}+\frac {z^{5}}{10395}+O\left (z^{6}\right )\right )+c_{2} \left (1+\frac {z}{2}+\frac {z^{2}}{8}+\frac {z^{3}}{48}+\frac {z^{4}}{384}+\frac {z^{5}}{3840}+O\left (z^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {z}\, \left (1+\frac {z}{3}+\frac {z^{2}}{15}+\frac {z^{3}}{105}+\frac {z^{4}}{945}+\frac {z^{5}}{10395}+O\left (z^{6}\right )\right )+c_{2} \left (1+\frac {z}{2}+\frac {z^{2}}{8}+\frac {z^{3}}{48}+\frac {z^{4}}{384}+\frac {z^{5}}{3840}+O\left (z^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {z}\, \left (1+\frac {z}{3}+\frac {z^{2}}{15}+\frac {z^{3}}{105}+\frac {z^{4}}{945}+\frac {z^{5}}{10395}+O\left (z^{6}\right )\right )+c_{2} \left (1+\frac {z}{2}+\frac {z^{2}}{8}+\frac {z^{3}}{48}+\frac {z^{4}}{384}+\frac {z^{5}}{3840}+O\left (z^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 z y^{\prime \prime }+\left (-2 z +2\right ) y^{\prime }-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 {y}{4 z}+\frac {\left (z -1\right ) y^{\prime }}{2 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 (z -1\right ) y^{\prime }}{2 z}-\frac {y}{4 z}=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 {z -1}{2 z}, P_{3}\left (z \right )=-\frac {1}{4 z}\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}}}=\frac {1}{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} \\ {} & {} & 4 z y^{\prime \prime }+\left (-2 z +2\right ) y^{\prime }-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 =0..1 \\ {} & {} & 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 \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 +1+r \right ) \left (k +r \right ) z^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{0} r \left (-1+2 r \right ) z^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (2 a_{k +1} \left (k +1+r \right ) \left (2 k +2 r +1\right )-a_{k} \left (2 k +2 r +1\right )\right ) z^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 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 {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (a_{k +1} \left (k +1+r \right )-\frac {a_{k}}{2}\right ) \left (k +r +\frac {1}{2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{2 \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k}}{2 \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 +1}=\frac {a_{k}}{2 \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {a_{k}}{2 \left (k +\frac {3}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} z^{k +\frac {1}{2}}, a_{k +1}=\frac {a_{k}}{2 \left (k +\frac {3}{2}\right )}\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 +\frac {1}{2}}\right ), a_{k +1}=\frac {a_{k}}{2 \left (k +1\right )}, b_{k +1}=\frac {b_{k}}{2 \left (k +\frac {3}{2}\right )}\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 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: 0.016 (sec). Leaf size: 44
Order:=6; dsolve(4*z*diff(y(z),z$2)+2*(1-z)*diff(y(z),z)-y(z)=0,y(z),type='series',z=0);
\[ y \left (z \right ) = c_{1} \sqrt {z}\, \left (1+\frac {1}{3} z +\frac {1}{15} z^{2}+\frac {1}{105} z^{3}+\frac {1}{945} z^{4}+\frac {1}{10395} z^{5}+\operatorname {O}\left (z^{6}\right )\right )+c_{2} \left (1+\frac {1}{2} z +\frac {1}{8} z^{2}+\frac {1}{48} z^{3}+\frac {1}{384} z^{4}+\frac {1}{3840} z^{5}+\operatorname {O}\left (z^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 85
AsymptoticDSolveValue[4*z*y''[z]+2*(1-z)*y'[z]-y[z]==0,y[z],{z,0,5}]
\[ y(z)\to c_1 \sqrt {z} \left (\frac {z^5}{10395}+\frac {z^4}{945}+\frac {z^3}{105}+\frac {z^2}{15}+\frac {z}{3}+1\right )+c_2 \left (\frac {z^5}{3840}+\frac {z^4}{384}+\frac {z^3}{48}+\frac {z^2}{8}+\frac {z}{2}+1\right ) \]