Internal problem ID [2537]
Internal file name [OUTPUT/2029_Sunday_June_05_2022_02_45_24_AM_69342567/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 16.10.
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
[_Jacobi]
\[ \boxed {z \left (1-z \right ) y^{\prime \prime }+\left (1-z \right ) y^{\prime }+\lambda y=0} \] 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. \[ \left (-z^{2}+z \right ) y^{\prime \prime }+\left (1-z \right ) y^{\prime }+\lambda y = 0 \] 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 {1}{z}\\ q(z) &= -\frac {\lambda }{z \left (z -1\right )}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1, \infty ]\)
Irregular singular points : \([]\)
Since \(z = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -y^{\prime \prime } z \left (z -1\right )+\left (1-z \right ) y^{\prime }+\lambda y = 0 \] 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 \left (z -1\right )+\left (1-z \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} z^{n +r -1}\right )+\lambda \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} z^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-z^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\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 (-z^{n +r} a_{n} \left (n +r \right )\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 }}\lambda 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 (-z^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) z^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-z^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) z^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\lambda a_{n} z^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}\lambda a_{n -1} 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} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) z^{n +r -1}\right )+\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 =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) z^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} z^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\lambda a_{n -1} z^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ z^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} z^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ z^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} z^{-1+r} = 0 \] Or \[ \left (z^{-1+r} r \left (-1+r \right )+r \,z^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ z^{-1+r} r^{2} = 0 \] Since the above is true for all \(z\) then the indicial equation becomes \[ r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ z^{-1+r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
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. 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 -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+\lambda a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (-n^{2}-2 n r -r^{2}+\lambda +2 n +2 r -1\right )}{n^{2}+2 n r +r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {a_{n -1} \left (n^{2}-\lambda -2 n +1\right )}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) 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 \[ a_{1}=\frac {r^{2}-\lambda }{\left (r +1\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{1}=-\lambda \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}-\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}=\frac {\left (\lambda -1\right ) \lambda }{4} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}-\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
| \(a_{2}\) | \(\frac {\left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {\left (r^{2}-\lambda +4 r +4\right ) \left (r^{2}-\lambda +2 r +1\right ) \left (r^{2}-\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{3}=-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{36} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}-\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
| \(a_{2}\) | \(\frac {\left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}-\lambda +4 r +4\right ) \left (r^{2}-\lambda +2 r +1\right ) \left (r^{2}-\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{36}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (-r^{2}+\lambda -6 r -9\right ) \left (-r^{2}+\lambda -4 r -4\right ) \left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{4}=\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{576} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}-\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
| \(a_{2}\) | \(\frac {\left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}-\lambda +4 r +4\right ) \left (r^{2}-\lambda +2 r +1\right ) \left (r^{2}-\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{36}\) |
| \(a_{4}\) | \(\frac {\left (-r^{2}+\lambda -6 r -9\right ) \left (-r^{2}+\lambda -4 r -4\right ) \left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{576}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {\left (r^{2}-\lambda +8 r +16\right ) \left (r^{2}-\lambda +6 r +9\right ) \left (r^{2}-\lambda +4 r +4\right ) \left (r^{2}-\lambda +2 r +1\right ) \left (r^{2}-\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{5}=-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{14400} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}-\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
| \(a_{2}\) | \(\frac {\left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}-\lambda +4 r +4\right ) \left (r^{2}-\lambda +2 r +1\right ) \left (r^{2}-\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{36}\) |
| \(a_{4}\) | \(\frac {\left (-r^{2}+\lambda -6 r -9\right ) \left (-r^{2}+\lambda -4 r -4\right ) \left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{576}\) |
| \(a_{5}\) | \(\frac {\left (r^{2}-\lambda +8 r +16\right ) \left (r^{2}-\lambda +6 r +9\right ) \left (r^{2}-\lambda +4 r +4\right ) \left (r^{2}-\lambda +2 r +1\right ) \left (r^{2}-\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{14400}\) |
Using the above table, then the first solution \(y_{1}\left (z \right )\) becomes \begin{align*} y_{1}\left (z \right )&= 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 \\ &= -\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\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 \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {r^{2}-\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) | \(\frac {2 \lambda +2 r}{\left (r +1\right )^{3}}\) | \(2 \lambda \) |
| \(b_{2}\) | \(\frac {\left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) | \(\frac {4 r^{4}+\left (4 \lambda +12\right ) r^{3}+\left (6 \lambda +12\right ) r^{2}+\left (-4 \lambda ^{2}+2 \lambda +4\right ) r -6 \lambda ^{2}+2 \lambda }{\left (r +1\right )^{3} \left (2+r \right )^{3}}\) | \(-\frac {\lambda }{2}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\) |
| \(b_{3}\) | \(\frac {\left (r^{2}-\lambda +4 r +4\right ) \left (r^{2}-\lambda +2 r +1\right ) \left (r^{2}-\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{36}\) | \(\frac {6 r^{7}+\left (6 \lambda +54\right ) r^{6}+\left (36 \lambda +198\right ) r^{5}+\left (-12 \lambda ^{2}+78 \lambda +378\right ) r^{4}+\left (-66 \lambda ^{2}+72 \lambda +396\right ) r^{3}+\left (6 \lambda ^{3}-138 \lambda ^{2}+24 \lambda +216\right ) r^{2}+\left (24 \lambda ^{3}-150 \lambda ^{2}+12 \lambda +48\right ) r +22 \lambda ^{3}-74 \lambda ^{2}+16 \lambda }{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3}}\) | \(-\frac {\left (-\lambda +1\right ) \lambda }{9}-\frac {\left (-\lambda +4\right ) \lambda }{18}+\frac {11 \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{108}\) |
| \(b_{4}\) | \(\frac {\left (-r^{2}+\lambda -6 r -9\right ) \left (-r^{2}+\lambda -4 r -4\right ) \left (-r^{2}+\lambda -2 r -1\right ) \left (-r^{2}+\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{576}\) | \(\frac {8 r^{10}+\left (8 \lambda +144\right ) r^{9}+\left (108 \lambda +1128\right ) r^{8}+\left (-24 \lambda ^{2}+600 \lambda +5040\right ) r^{7}+\left (-300 \lambda ^{2}+1740 \lambda +14136\right ) r^{6}+\left (24 \lambda ^{3}-1596 \lambda ^{2}+2676 \lambda +25776\right ) r^{5}+\left (244 \lambda ^{3}-4772 \lambda ^{2}+1684 \lambda +30520\right ) r^{4}+\left (-8 \lambda ^{4}+992 \lambda ^{3}-8836 \lambda ^{2}-652 \lambda +22608\right ) r^{3}+\left (-60 \lambda ^{4}+2100 \lambda ^{3}-10428 \lambda ^{2}-1260 \lambda +9504\right ) r^{2}+\left (-140 \lambda ^{4}+2368 \lambda ^{3}-7532 \lambda ^{2}+24 \lambda +1728\right ) r -100 \lambda ^{4}+1112 \lambda ^{3}-2596 \lambda ^{2}+432 \lambda }{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{96}-\frac {\left (\lambda -9\right ) \left (\lambda -1\right ) \lambda }{144}-\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \lambda }{288}-\frac {25 \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{3456}\) |
| \(b_{5}\) | \(\frac {\left (r^{2}-\lambda +8 r +16\right ) \left (r^{2}-\lambda +6 r +9\right ) \left (r^{2}-\lambda +4 r +4\right ) \left (r^{2}-\lambda +2 r +1\right ) \left (r^{2}-\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{14400}\) | \(\frac {10 r^{13}+\left (10 \lambda +300\right ) r^{12}+\left (240 \lambda +4050\right ) r^{11}+\left (-40 \lambda ^{2}+2510 \lambda +32500\right ) r^{10}+\left (-900 \lambda ^{2}+14900 \lambda +172470\right ) r^{9}+\left (60 \lambda ^{3}-9000 \lambda ^{2}+54390 \lambda +636900\right ) r^{8}+\left (1160 \lambda ^{3}-52900 \lambda ^{2}+120980 \lambda +1676150\right ) r^{7}+\left (-40 \lambda ^{4}+9780 \lambda ^{3}-203520 \lambda ^{2}+137910 \lambda +3163500\right ) r^{6}+\left (-630 \lambda ^{4}+47400 \lambda ^{3}-539850 \lambda ^{2}-21000 \lambda +4244280\right ) r^{5}+\left (10 \lambda ^{5}-4100 \lambda ^{4}+145650 \lambda ^{3}-1011600 \lambda ^{2}-313620 \lambda +3942800\right ) r^{4}+\left (120 \lambda ^{5}-14350 \lambda ^{4}+293040 \lambda ^{3}-1343250 \lambda ^{2}-444640 \lambda +2404800\right ) r^{3}+\left (510 \lambda ^{5}-28800 \lambda ^{4}+380430 \lambda ^{3}-1232700 \lambda ^{2}-260640 \lambda +864000\right ) r^{2}+\left (900 \lambda ^{5}-31280 \lambda ^{4}+292900 \lambda ^{3}-718040 \lambda ^{2}-24000 \lambda +138240\right ) r +548 \lambda ^{5}-14040 \lambda ^{4}+101604 \lambda ^{3}-202160 \lambda ^{2}+27648 \lambda }{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(-\frac {\left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{1800}-\frac {\left (-\lambda +16\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{2400}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +1\right ) \lambda }{3600}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \lambda }{7200}+\frac {137 \left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{432000}\) |
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 \\ &= \left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right ) \ln \left (z \right )+2 \lambda z +\left (-\frac {\lambda }{2}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) z^{2}+\left (-\frac {\left (-\lambda +1\right ) \lambda }{9}-\frac {\left (-\lambda +4\right ) \lambda }{18}+\frac {11 \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{108}\right ) z^{3}+\left (-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{96}-\frac {\left (\lambda -9\right ) \left (\lambda -1\right ) \lambda }{144}-\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \lambda }{288}-\frac {25 \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) z^{4}+\left (-\frac {\left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{1800}-\frac {\left (-\lambda +16\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{2400}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +1\right ) \lambda }{3600}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \lambda }{7200}+\frac {137 \left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{432000}\right ) z^{5}+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} \left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right ) + c_{2} \left (\left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right ) \ln \left (z \right )+2 \lambda z +\left (-\frac {\lambda }{2}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) z^{2}+\left (-\frac {\left (-\lambda +1\right ) \lambda }{9}-\frac {\left (-\lambda +4\right ) \lambda }{18}+\frac {11 \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{108}\right ) z^{3}+\left (-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{96}-\frac {\left (\lambda -9\right ) \left (\lambda -1\right ) \lambda }{144}-\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \lambda }{288}-\frac {25 \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) z^{4}+\left (-\frac {\left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{1800}-\frac {\left (-\lambda +16\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{2400}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +1\right ) \lambda }{3600}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \lambda }{7200}+\frac {137 \left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{432000}\right ) z^{5}+O\left (z^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right )+c_{2} \left (\left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right ) \ln \left (z \right )+2 \lambda z +\left (-\frac {\lambda }{2}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) z^{2}+\left (-\frac {\left (-\lambda +1\right ) \lambda }{9}-\frac {\left (-\lambda +4\right ) \lambda }{18}+\frac {11 \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{108}\right ) z^{3}+\left (-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{96}-\frac {\left (\lambda -9\right ) \left (\lambda -1\right ) \lambda }{144}-\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \lambda }{288}-\frac {25 \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) z^{4}+\left (-\frac {\left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{1800}-\frac {\left (-\lambda +16\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{2400}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +1\right ) \lambda }{3600}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \lambda }{7200}+\frac {137 \left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{432000}\right ) z^{5}+O\left (z^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right )+c_{2} \left (\left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right ) \ln \left (z \right )+2 \lambda z +\left (-\frac {\lambda }{2}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) z^{2}+\left (-\frac {\left (-\lambda +1\right ) \lambda }{9}-\frac {\left (-\lambda +4\right ) \lambda }{18}+\frac {11 \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{108}\right ) z^{3}+\left (-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{96}-\frac {\left (\lambda -9\right ) \left (\lambda -1\right ) \lambda }{144}-\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \lambda }{288}-\frac {25 \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) z^{4}+\left (-\frac {\left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{1800}-\frac {\left (-\lambda +16\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{2400}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +1\right ) \lambda }{3600}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \lambda }{7200}+\frac {137 \left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{432000}\right ) z^{5}+O\left (z^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right )+c_{2} \left (\left (-\lambda z +1+\frac {\left (\lambda -1\right ) \lambda \,z^{2}}{4}-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{3}}{36}+\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{4}}{576}-\frac {\left (\lambda -16\right ) \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda \,z^{5}}{14400}+O\left (z^{6}\right )\right ) \ln \left (z \right )+2 \lambda z +\left (-\frac {\lambda }{2}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) z^{2}+\left (-\frac {\left (-\lambda +1\right ) \lambda }{9}-\frac {\left (-\lambda +4\right ) \lambda }{18}+\frac {11 \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{108}\right ) z^{3}+\left (-\frac {\left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{96}-\frac {\left (\lambda -9\right ) \left (\lambda -1\right ) \lambda }{144}-\frac {\left (\lambda -9\right ) \left (\lambda -4\right ) \lambda }{288}-\frac {25 \left (\lambda -9\right ) \left (\lambda -4\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) z^{4}+\left (-\frac {\left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{1800}-\frac {\left (-\lambda +16\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{2400}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +1\right ) \lambda }{3600}-\frac {\left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \lambda }{7200}+\frac {137 \left (-\lambda +16\right ) \left (-\lambda +9\right ) \left (-\lambda +4\right ) \left (-\lambda +1\right ) \lambda }{432000}\right ) z^{5}+O\left (z^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{\prime \prime } z \left (z -1\right )+\left (1-z \right ) y^{\prime }+\lambda 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 {\lambda y}{z \left (z -1\right )}-\frac {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 {y^{\prime }}{z}-\frac {\lambda y}{z \left (z -1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} z_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (z \right )=\frac {1}{z}, P_{3}\left (z \right )=-\frac {\lambda }{z \left (z -1\right )}\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}}}=1 \\ {} & \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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & z_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } z \left (z -1\right )+\left (z -1\right ) y^{\prime }-\lambda 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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & z^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) z^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & z^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) z^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r^{2} z^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (k^{2}+2 k r +r^{2}-\lambda \right )\right ) z^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +1} \left (k +1\right )^{2}+a_{k} \left (k^{2}-\lambda \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}-\lambda \right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}-\lambda \right )}{\left (k +1\right )^{2}} \\ \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} \left (k^{2}-\lambda \right )}{\left (k +1\right )^{2}}\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 -> Whittaker -> 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: 261
Order:=6; dsolve(z*(1-z)*diff(y(z),z$2)+(1-z)*diff(y(z),z)+lambda*y(z)=0,y(z),type='series',z=0);
\[ y \left (z \right ) = \left (2 \lambda z +\left (\frac {1}{4} \lambda -\frac {3}{4} \lambda ^{2}\right ) z^{2}+\left (-\frac {37}{108} \lambda ^{2}+\frac {2}{27} \lambda +\frac {11}{108} \lambda ^{3}\right ) z^{3}+\left (\frac {139}{1728} \lambda ^{3}-\frac {649}{3456} \lambda ^{2}+\frac {1}{32} \lambda -\frac {25}{3456} \lambda ^{4}\right ) z^{4}+\left (-\frac {13}{1600} \lambda ^{4}+\frac {8467}{144000} \lambda ^{3}-\frac {2527}{21600} \lambda ^{2}+\frac {2}{125} \lambda +\frac {137}{432000} \lambda ^{5}\right ) z^{5}+\operatorname {O}\left (z^{6}\right )\right ) c_{2} +\left (1-\lambda z +\frac {1}{4} \left (-1+\lambda \right ) \lambda z^{2}-\frac {1}{36} \lambda \left (\lambda ^{2}-5 \lambda +4\right ) z^{3}+\frac {1}{576} \lambda \left (\lambda ^{3}-14 \lambda ^{2}+49 \lambda -36\right ) z^{4}-\frac {1}{14400} \lambda \left (-1+\lambda \right ) \left (\lambda -4\right ) \left (\lambda -16\right ) \left (\lambda -9\right ) z^{5}+\operatorname {O}\left (z^{6}\right )\right ) \left (c_{2} \ln \left (z \right )+c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 940
AsymptoticDSolveValue[z*(1-z)*y''[z]+(1-z)*y'[z]+\[Lambda]*y[z]==0,y[z],{z,0,5}]
\[ y(z)\to \left (\frac {1}{25} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\frac {1}{16} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) z^5+\frac {1}{16} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) z^4+\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) z^3+\frac {1}{4} \left (\lambda ^2-\lambda \right ) z^2-\lambda z+1\right ) c_1+c_2 \left (-\frac {2}{125} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\frac {1}{16} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) z^5+\frac {1}{25} \left (\frac {\lambda ^3}{2}-2 \lambda ^2+\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda +\frac {2}{27} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\frac {1}{9} \left (\frac {\lambda ^3}{2}-2 \lambda ^2+\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda \right ) \lambda +\frac {1}{32} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) \lambda -\frac {1}{16} \left (\frac {\lambda ^3}{2}-2 \lambda ^2+\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda +\frac {2}{27} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\frac {1}{9} \left (\frac {\lambda ^3}{2}-2 \lambda ^2+\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda \right ) \lambda \right ) \lambda \right ) z^5-\frac {1}{32} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) z^4+\frac {1}{16} \left (\frac {\lambda ^3}{2}-2 \lambda ^2+\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda +\frac {2}{27} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\frac {1}{9} \left (\frac {\lambda ^3}{2}-2 \lambda ^2+\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda \right ) \lambda \right ) z^4-\frac {2}{27} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) z^3+\frac {1}{9} \left (\frac {\lambda ^3}{2}-2 \lambda ^2+\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda \right ) z^3-\frac {\lambda ^2 z^2}{2}-\frac {1}{4} \left (\lambda ^2-\lambda \right ) z^2+2 \lambda z+\left (\frac {1}{25} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\frac {1}{16} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) z^5+\frac {1}{16} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) \lambda -\lambda \right ) z^4+\frac {1}{9} \left (\lambda ^2-\frac {1}{4} \left (\lambda ^2-\lambda \right ) \lambda -\lambda \right ) z^3+\frac {1}{4} \left (\lambda ^2-\lambda \right ) z^2-\lambda z+1\right ) \log (z)\right ) \]