Internal problem ID [2534]
Internal file name [OUTPUT/2026_Sunday_June_05_2022_02_45_13_AM_50657971/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 16.6.
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]]
\[ \boxed {z^{2} y^{\prime \prime }-\frac {3 z y^{\prime }}{2}+\left (z +1\right ) y=0} \] 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. \[ z^{2} y^{\prime \prime }-\frac {3 z y^{\prime }}{2}+\left (z +1\right ) y = 0 \] 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 {3}{2 z}\\ q(z) &= \frac {z +1}{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 \[ z^{2} y^{\prime \prime }-\frac {3 z y^{\prime }}{2}+\left (z +1\right ) y = 0 \] 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}-\frac {3 z \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} z^{n +r -1}\right )}{2}+\left (z +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 }}z^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {3 z^{n +r} a_{n} \left (n +r \right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}z^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}z^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} 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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {3 z^{n +r} a_{n} \left (n +r \right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} z^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ z^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {3 z^{n +r} a_{n} \left (n +r \right )}{2}+a_{n} z^{n +r} = 0 \] When \(n = 0\) the above becomes \[ z^{r} a_{0} r \left (-1+r \right )-\frac {3 z^{r} a_{0} r}{2}+a_{0} z^{r} = 0 \] Or \[ \left (z^{r} r \left (-1+r \right )-\frac {3 z^{r} r}{2}+z^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \frac {\left (2 r^{2}-5 r +2\right ) z^{r}}{2} = 0 \] Since the above is true for all \(z\) then the indicial equation becomes \[ r^{2}-\frac {5}{2} r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= {\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \frac {\left (2 r^{2}-5 r +2\right ) z^{r}}{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [2, {\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = {\frac {3}{2}}\) 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 +2}\\ y_{2}\left (z \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} z^{n +\frac {1}{2}} \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} a_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {3 a_{n} \left (n +r \right )}{2}+a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 a_{n -1}}{2 n^{2}+4 n r +2 r^{2}-5 n -5 r +2}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = -\frac {2 a_{n -1}}{n \left (2 n +3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) 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 \[ a_{1}=-\frac {2}{2 r^{2}-r -1} \] Which for the root \(r = 2\) becomes \[ a_{1}=-{\frac {2}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(-{\frac {2}{5}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r} \] Which for the root \(r = 2\) becomes \[ a_{2}={\frac {2}{35}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r}\) | \(\frac {2}{35}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {8}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right )} \] Which for the root \(r = 2\) becomes \[ a_{3}=-{\frac {4}{945}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r}\) | \(\frac {2}{35}\) |
| \(a_{3}\) | \(-\frac {8}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right )}\) | \(-{\frac {4}{945}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right )} \] Which for the root \(r = 2\) becomes \[ a_{4}={\frac {2}{10395}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r}\) | \(\frac {2}{35}\) |
| \(a_{3}\) | \(-\frac {8}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right )}\) | \(-{\frac {4}{945}}\) |
| \(a_{4}\) | \(\frac {16}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right )}\) | \(\frac {2}{10395}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {32}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right ) \left (2 r^{2}+15 r +27\right )} \] Which for the root \(r = 2\) becomes \[ a_{5}=-{\frac {4}{675675}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r}\) | \(\frac {2}{35}\) |
| \(a_{3}\) | \(-\frac {8}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right )}\) | \(-{\frac {4}{945}}\) |
| \(a_{4}\) | \(\frac {16}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right )}\) | \(\frac {2}{10395}\) |
| \(a_{5}\) | \(-\frac {32}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right ) \left (2 r^{2}+15 r +27\right )}\) | \(-{\frac {4}{675675}}\) |
Using the above table, then the 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 (1-\frac {2 z}{5}+\frac {2 z^{2}}{35}-\frac {4 z^{3}}{945}+\frac {2 z^{4}}{10395}-\frac {4 z^{5}}{675675}+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} b_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {3 b_{n} \left (n +r \right )}{2}+b_{n -1}+b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {2 b_{n -1}}{2 n^{2}+4 n r +2 r^{2}-5 n -5 r +2}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{n} = -\frac {2 b_{n -1}}{n \left (2 n -3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ b_{1}=-\frac {2}{2 r^{2}-r -1} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{1}=2 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{2}=-2 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(2\) |
| \(b_{2}\) | \(\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r}\) | \(-2\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {8}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{3}={\frac {4}{9}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(2\) |
| \(b_{2}\) | \(\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r}\) | \(-2\) |
| \(b_{3}\) | \(-\frac {8}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right )}\) | \(\frac {4}{9}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {16}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{4}=-{\frac {2}{45}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(2\) |
| \(b_{2}\) | \(\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r}\) | \(-2\) |
| \(b_{3}\) | \(-\frac {8}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right )}\) | \(\frac {4}{9}\) |
| \(b_{4}\) | \(\frac {16}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right )}\) | \(-{\frac {2}{45}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {32}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right ) \left (2 r^{2}+15 r +27\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{5}={\frac {4}{1575}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{2 r^{2}-r -1}\) | \(2\) |
| \(b_{2}\) | \(\frac {4}{4 r^{4}+4 r^{3}-5 r^{2}-3 r}\) | \(-2\) |
| \(b_{3}\) | \(-\frac {8}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right )}\) | \(\frac {4}{9}\) |
| \(b_{4}\) | \(\frac {16}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right )}\) | \(-{\frac {2}{45}}\) |
| \(b_{5}\) | \(-\frac {32}{r \left (4 r^{3}+4 r^{2}-5 r -3\right ) \left (2 r^{2}+7 r +5\right ) \left (2 r^{2}+11 r +14\right ) \left (2 r^{2}+15 r +27\right )}\) | \(\frac {4}{1575}\) |
Using the above table, then the solution \(y_{2}\left (z \right )\) is \begin {align*} y_{2}\left (z \right )&= z^{2} \left (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 \right ) \\ &= \sqrt {z}\, \left (1+2 z -2 z^{2}+\frac {4 z^{3}}{9}-\frac {2 z^{4}}{45}+\frac {4 z^{5}}{1575}+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 (1-\frac {2 z}{5}+\frac {2 z^{2}}{35}-\frac {4 z^{3}}{945}+\frac {2 z^{4}}{10395}-\frac {4 z^{5}}{675675}+O\left (z^{6}\right )\right ) + c_{2} \sqrt {z}\, \left (1+2 z -2 z^{2}+\frac {4 z^{3}}{9}-\frac {2 z^{4}}{45}+\frac {4 z^{5}}{1575}+O\left (z^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} z^{2} \left (1-\frac {2 z}{5}+\frac {2 z^{2}}{35}-\frac {4 z^{3}}{945}+\frac {2 z^{4}}{10395}-\frac {4 z^{5}}{675675}+O\left (z^{6}\right )\right )+c_{2} \sqrt {z}\, \left (1+2 z -2 z^{2}+\frac {4 z^{3}}{9}-\frac {2 z^{4}}{45}+\frac {4 z^{5}}{1575}+O\left (z^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} z^{2} \left (1-\frac {2 z}{5}+\frac {2 z^{2}}{35}-\frac {4 z^{3}}{945}+\frac {2 z^{4}}{10395}-\frac {4 z^{5}}{675675}+O\left (z^{6}\right )\right )+c_{2} \sqrt {z}\, \left (1+2 z -2 z^{2}+\frac {4 z^{3}}{9}-\frac {2 z^{4}}{45}+\frac {4 z^{5}}{1575}+O\left (z^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} z^{2} \left (1-\frac {2 z}{5}+\frac {2 z^{2}}{35}-\frac {4 z^{3}}{945}+\frac {2 z^{4}}{10395}-\frac {4 z^{5}}{675675}+O\left (z^{6}\right )\right )+c_{2} \sqrt {z}\, \left (1+2 z -2 z^{2}+\frac {4 z^{3}}{9}-\frac {2 z^{4}}{45}+\frac {4 z^{5}}{1575}+O\left (z^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } z^{2}-\frac {3 z y^{\prime }}{2}+\left (z +1\right ) 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 {3 y^{\prime }}{2 z}-\frac {\left (z +1\right ) y}{z^{2}} \\ \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 {3 y^{\prime }}{2 z}+\frac {\left (z +1\right ) 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 {3}{2 z}, P_{3}\left (z \right )=\frac {z +1}{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}}}=-\frac {3}{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}}}=1 \\ {} & \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} \\ {} & {} & 2 y^{\prime \prime } z^{2}-3 z y^{\prime }+\left (2 z +2\right ) 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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & z^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} z^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & z^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} z^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} z \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & z \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +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 (-1+2 r \right ) \left (-2+r \right ) z^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r -1\right ) \left (k +r -2\right )+2 a_{k -1}\right ) z^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+2 r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{2, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +r -\frac {1}{2}\right ) \left (k +r -2\right ) a_{k}+2 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (k +\frac {1}{2}+r \right ) \left (k +r -1\right ) a_{k +1}+2 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 a_{k}}{\left (2 k +1+2 r \right ) \left (k +r -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=-\frac {2 a_{k}}{\left (2 k +5\right ) \left (k +1\right )} \\ \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 (2 k +5\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {2 a_{k}}{\left (2 k +2\right ) \left (k -\frac {1}{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 {2 a_{k}}{\left (2 k +2\right ) \left (k -\frac {1}{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 +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} z^{k +\frac {1}{2}}\right ), a_{k +1}=-\frac {2 a_{k}}{\left (2 k +5\right ) \left (k +1\right )}, b_{k +1}=-\frac {2 b_{k}}{\left (2 k +2\right ) \left (k -\frac {1}{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 Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 47
Order:=6; dsolve(z^2*diff(y(z),z$2)-3/2*z*diff(y(z),z)+(1+z)*y(z)=0,y(z),type='series',z=0);
\[ y \left (z \right ) = c_{1} \sqrt {z}\, \left (1+2 z -2 z^{2}+\frac {4}{9} z^{3}-\frac {2}{45} z^{4}+\frac {4}{1575} z^{5}+\operatorname {O}\left (z^{6}\right )\right )+c_{2} z^{2} \left (1-\frac {2}{5} z +\frac {2}{35} z^{2}-\frac {4}{945} z^{3}+\frac {2}{10395} z^{4}-\frac {4}{675675} z^{5}+\operatorname {O}\left (z^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 84
AsymptoticDSolveValue[z^2*y''[z]-3/2*z*y'[z]+(1+z)*y[z]==0,y[z],{z,0,5}]
\[ y(z)\to c_1 \left (-\frac {4 z^5}{675675}+\frac {2 z^4}{10395}-\frac {4 z^3}{945}+\frac {2 z^2}{35}-\frac {2 z}{5}+1\right ) z^2+c_2 \left (\frac {4 z^5}{1575}-\frac {2 z^4}{45}+\frac {4 z^3}{9}-2 z^2+2 z+1\right ) \sqrt {z} \]