Internal problem ID [1817]
Internal file name [OUTPUT/1818_Sunday_June_05_2022_02_33_41_AM_2944121/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 2.8.2, Regular singular points, the method of Frobenius. Page 214
Problem number: 25.
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
[_Laguerre]
\[ \boxed {t y^{\prime \prime }+\left (1-t \right ) y^{\prime }+\lambda y=0} \] With the expansion point for the power series method at \(t = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ t y^{\prime \prime }+\left (1-t \right ) y^{\prime }+\lambda y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(t) y^{\prime } + q(t) y &=0 \end {align*}
Where \begin {align*} p(t) &= -\frac {t -1}{t}\\ q(t) &= \frac {\lambda }{t}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ t y^{\prime \prime }+\left (1-t \right ) y^{\prime }+\lambda y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+\left (1-t \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\lambda \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\lambda a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\lambda a_{n} t^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}\lambda a_{n -1} t^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{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 ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\lambda a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ t^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} t^{-1+r} = 0 \] Or \[ \left (t^{-1+r} r \left (-1+r \right )+r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ t^{-1+r} r^{2} = 0 \] Since the above is true for all \(t\) 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 \[ t^{-1+r} r^{2} = 0 \] 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 (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (t \right ) &= y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \] 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 \(y_{1}\left (t \right )\). 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 )-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 \[ a_{n} = -\frac {a_{n -1} \left (\lambda -n -r +1\right )}{n^{2}+2 n r +r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {a_{n -1} \left (-\lambda +n -1\right )}{n^{2}}\tag {5} \] 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 \[ a_{1}=\frac {r -\lambda }{\left (r +1\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{1}=-\lambda \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}=\frac {\left (\lambda -1\right ) \lambda }{4} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
| \(a_{2}\) | \(\frac {\left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {\left (r +2-\lambda \right ) \left (r +1-\lambda \right ) \left (r -\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{3}=-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{36} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
| \(a_{2}\) | \(\frac {\left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) |
| \(a_{3}\) | \(\frac {\left (r +2-\lambda \right ) \left (r +1-\lambda \right ) \left (r -\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{36}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (\lambda -3-r \right ) \left (\lambda -2-r \right ) \left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{4}=\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
| \(a_{2}\) | \(\frac {\left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) |
| \(a_{3}\) | \(\frac {\left (r +2-\lambda \right ) \left (r +1-\lambda \right ) \left (r -\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{36}\) |
| \(a_{4}\) | \(\frac {\left (\lambda -3-r \right ) \left (\lambda -2-r \right ) \left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {\left (r +4-\lambda \right ) \left (r +3-\lambda \right ) \left (r +2-\lambda \right ) \left (r +1-\lambda \right ) \left (r -\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{5}=-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{14400} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r -\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) |
| \(a_{2}\) | \(\frac {\left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) |
| \(a_{3}\) | \(\frac {\left (r +2-\lambda \right ) \left (r +1-\lambda \right ) \left (r -\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{36}\) |
| \(a_{4}\) | \(\frac {\left (\lambda -3-r \right ) \left (\lambda -2-r \right ) \left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}\) |
| \(a_{5}\) | \(\frac {\left (r +4-\lambda \right ) \left (r +3-\lambda \right ) \left (r +2-\lambda \right ) \left (r +1-\lambda \right ) \left (r -\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{14400}\) |
Using the above table, then the first solution \(y_{1}\left (t \right )\) becomes \begin{align*} y_{1}\left (t \right )&= a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}\dots \\ &= -\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (t \right ) = y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right ) \] 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 |
| \(b_{1}\) | \(\frac {r -\lambda }{\left (r +1\right )^{2}}\) | \(-\lambda \) | \(\frac {-r +1+2 \lambda }{\left (r +1\right )^{3}}\) | \(1+2 \lambda \) |
| \(b_{2}\) | \(\frac {\left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2}}\) | \(\frac {\left (\lambda -1\right ) \lambda }{4}\) | \(\frac {-2 r^{3}+\left (6 \lambda -3\right ) r^{2}+\left (-4 \lambda ^{2}+10 \lambda +1\right ) r -6 \lambda ^{2}+2 \lambda +2}{\left (r +1\right )^{3} \left (2+r \right )^{3}}\) | \(-\frac {\lambda }{2}+\frac {1}{4}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\) |
| \(b_{3}\) | \(\frac {\left (r +2-\lambda \right ) \left (r +1-\lambda \right ) \left (r -\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{36}\) | \(\frac {12-3 r^{5}+6 \left (-3+2 \lambda \right ) r^{4}+\left (-15 \lambda ^{2}+66 \lambda -35\right ) r^{3}+6 \left (\lambda ^{3}-12 \lambda ^{2}+20 \lambda -3\right ) r^{2}+\left (24 \lambda ^{3}-105 \lambda ^{2}+78 \lambda +14\right ) r +22 \lambda ^{3}-48 \lambda ^{2}+8 \lambda }{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3}}\) | \(-\frac {\left (1-\lambda \right ) \lambda }{36}-\frac {\left (-\lambda +2\right ) \lambda }{36}+\frac {\left (-\lambda +2\right ) \left (1-\lambda \right )}{36}+\frac {11 \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{108}\) |
| \(b_{4}\) | \(\frac {\left (\lambda -3-r \right ) \left (\lambda -2-r \right ) \left (\lambda -1-r \right ) \left (\lambda -r \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}\) | \(\frac {-4 r^{7}+\left (20 \lambda -50\right ) r^{6}+\left (-36 \lambda ^{2}+228 \lambda -246\right ) r^{5}+\left (28 \lambda ^{3}-366 \lambda ^{2}+1014 \lambda -592\right ) r^{4}+\left (-8 \lambda ^{4}+248 \lambda ^{3}-1408 \lambda ^{2}+2208 \lambda -674\right ) r^{3}+\left (-60 \lambda ^{4}+780 \lambda ^{3}-2550 \lambda ^{2}+2382 \lambda -198\right ) r^{2}+\left (-140 \lambda ^{4}+1040 \lambda ^{3}-2152 \lambda ^{2}+1076 \lambda +228\right ) r -100 \lambda ^{4}+504 \lambda ^{3}-668 \lambda ^{2}+72 \lambda +144}{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right )}{576}-\frac {25 \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{3456}\) |
| \(b_{5}\) | \(\frac {\left (r +4-\lambda \right ) \left (r +3-\lambda \right ) \left (r +2-\lambda \right ) \left (r +1-\lambda \right ) \left (r -\lambda \right )}{\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{14400}\) | \(\frac {-5 r^{9}+\left (30 \lambda -105\right ) r^{8}+\left (-70 \lambda ^{2}+580 \lambda -930\right ) r^{7}+\left (80 \lambda ^{3}-1230 \lambda ^{2}+4690 \lambda -4500\right ) r^{6}+\left (-45 \lambda ^{4}+1260 \lambda ^{3}-8895 \lambda ^{2}+20550 \lambda -12819\right ) r^{5}+\left (10 \lambda ^{5}-625 \lambda ^{4}+7950 \lambda ^{3}-34175 \lambda ^{2}+52700 \lambda -21315\right ) r^{4}+\left (120 \lambda ^{5}-3325 \lambda ^{4}+25700 \lambda ^{3}-74885 \lambda ^{2}+79270 \lambda -18310\right ) r^{3}+\left (510 \lambda ^{5}-8475 \lambda ^{4}+44850 \lambda ^{3}-92775 \lambda ^{2}+65340 \lambda -3600\right ) r^{2}+\left (900 \lambda ^{5}-10370 \lambda ^{4}+40060 \lambda ^{3}-59370 \lambda ^{2}+23800 \lambda +5424\right ) r +548 \lambda ^{5}-4880 \lambda ^{4}+14380 \lambda ^{3}-14800 \lambda ^{2}+1152 \lambda +2880}{\left (r +1\right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(-\frac {\left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \lambda }{14400}+\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right )}{14400}+\frac {137 \left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{432000}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (t \right )&=y_{1}\left (t \right ) \ln \left (t \right )+b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}\dots \\ &= \left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\left (1+2 \lambda \right ) t +\left (-\frac {\lambda }{2}+\frac {1}{4}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) t^{2}+\left (-\frac {\left (1-\lambda \right ) \lambda }{36}-\frac {\left (-\lambda +2\right ) \lambda }{36}+\frac {\left (-\lambda +2\right ) \left (1-\lambda \right )}{36}+\frac {11 \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{108}\right ) t^{3}+\left (-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right )}{576}-\frac {25 \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) t^{4}+\left (-\frac {\left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \lambda }{14400}+\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right )}{14400}+\frac {137 \left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{432000}\right ) t^{5}+O\left (t^{6}\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} \left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right ) + c_{2} \left (\left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\left (1+2 \lambda \right ) t +\left (-\frac {\lambda }{2}+\frac {1}{4}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) t^{2}+\left (-\frac {\left (1-\lambda \right ) \lambda }{36}-\frac {\left (-\lambda +2\right ) \lambda }{36}+\frac {\left (-\lambda +2\right ) \left (1-\lambda \right )}{36}+\frac {11 \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{108}\right ) t^{3}+\left (-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right )}{576}-\frac {25 \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) t^{4}+\left (-\frac {\left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \lambda }{14400}+\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right )}{14400}+\frac {137 \left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{432000}\right ) t^{5}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right )+c_{2} \left (\left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\left (1+2 \lambda \right ) t +\left (-\frac {\lambda }{2}+\frac {1}{4}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) t^{2}+\left (-\frac {\left (1-\lambda \right ) \lambda }{36}-\frac {\left (-\lambda +2\right ) \lambda }{36}+\frac {\left (-\lambda +2\right ) \left (1-\lambda \right )}{36}+\frac {11 \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{108}\right ) t^{3}+\left (-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right )}{576}-\frac {25 \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) t^{4}+\left (-\frac {\left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \lambda }{14400}+\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right )}{14400}+\frac {137 \left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{432000}\right ) t^{5}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right )+c_{2} \left (\left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\left (1+2 \lambda \right ) t +\left (-\frac {\lambda }{2}+\frac {1}{4}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) t^{2}+\left (-\frac {\left (1-\lambda \right ) \lambda }{36}-\frac {\left (-\lambda +2\right ) \lambda }{36}+\frac {\left (-\lambda +2\right ) \left (1-\lambda \right )}{36}+\frac {11 \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{108}\right ) t^{3}+\left (-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right )}{576}-\frac {25 \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) t^{4}+\left (-\frac {\left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \lambda }{14400}+\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right )}{14400}+\frac {137 \left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{432000}\right ) t^{5}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right )+c_{2} \left (\left (-\lambda t +1+\frac {\left (\lambda -1\right ) \lambda \,t^{2}}{4}-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{3}}{36}+\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{4}}{576}-\frac {\left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda \,t^{5}}{14400}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\left (1+2 \lambda \right ) t +\left (-\frac {\lambda }{2}+\frac {1}{4}-\frac {3 \left (\lambda -1\right ) \lambda }{4}\right ) t^{2}+\left (-\frac {\left (1-\lambda \right ) \lambda }{36}-\frac {\left (-\lambda +2\right ) \lambda }{36}+\frac {\left (-\lambda +2\right ) \left (1-\lambda \right )}{36}+\frac {11 \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{108}\right ) t^{3}+\left (-\frac {\left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -1\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \lambda }{576}-\frac {\left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right )}{576}-\frac {25 \left (\lambda -3\right ) \left (\lambda -2\right ) \left (\lambda -1\right ) \lambda }{3456}\right ) t^{4}+\left (-\frac {\left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (1-\lambda \right ) \lambda }{14400}-\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \lambda }{14400}+\frac {\left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right )}{14400}+\frac {137 \left (4-\lambda \right ) \left (3-\lambda \right ) \left (-\lambda +2\right ) \left (1-\lambda \right ) \lambda }{432000}\right ) t^{5}+O\left (t^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t y^{\prime \prime }+\left (1-t \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 {\left (t -1\right ) y^{\prime }}{t}-\frac {\lambda y}{t} \\ \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 (t -1\right ) y^{\prime }}{t}+\frac {\lambda y}{t}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (t \right )=-\frac {t -1}{t}, P_{3}\left (t \right )=\frac {\lambda }{t}\right ] \\ {} & \circ & t \cdot P_{2}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t \cdot P_{2}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=1 \\ {} & \circ & t^{2}\cdot P_{3}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t^{2}\cdot P_{3}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=0 \\ {} & \circ & t =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & t y^{\prime \prime }+\left (1-t \right ) y^{\prime }+\lambda y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & t^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & t^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & t \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r^{2} t^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right )^{2}-a_{k} \left (-\lambda +k +r \right )\right ) t^{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 (-\lambda +k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (-\lambda +k \right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (-\lambda +k \right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +1}=\frac {a_{k} \left (-\lambda +k \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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 309
Order:=6; dsolve(t*diff(y(t),t$2)+(1-t)*diff(y(t),t)+lambda*y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = \left (\left (2 \lambda +1\right ) t +\left (\frac {1}{4} \lambda +\frac {1}{4}-\frac {3}{4} \lambda ^{2}\right ) t^{2}+\left (-\frac {2}{9} \lambda ^{2}+\frac {1}{27} \lambda +\frac {1}{18}+\frac {11}{108} \lambda ^{3}\right ) t^{3}+\left (\frac {7}{192} \lambda ^{3}-\frac {167}{3456} \lambda ^{2}+\frac {1}{192} \lambda +\frac {1}{96}-\frac {25}{3456} \lambda ^{4}\right ) t^{4}+\left (\frac {719}{86400} \lambda ^{3}-\frac {61}{21600} \lambda ^{4}+\frac {137}{432000} \lambda ^{5}+\frac {1}{1500} \lambda -\frac {37}{4320} \lambda ^{2}+\frac {1}{600}\right ) t^{5}+\operatorname {O}\left (t^{6}\right )\right ) c_{2} +\left (1-\lambda t +\frac {1}{4} \left (-1+\lambda \right ) \lambda t^{2}-\frac {1}{36} \left (\lambda -2\right ) \left (-1+\lambda \right ) \lambda t^{3}+\frac {1}{576} \left (\lambda -3\right ) \left (\lambda -2\right ) \left (-1+\lambda \right ) \lambda t^{4}-\frac {1}{14400} \left (\lambda -4\right ) \left (\lambda -3\right ) \left (\lambda -2\right ) \left (-1+\lambda \right ) \lambda t^{5}+\operatorname {O}\left (t^{6}\right )\right ) \left (c_{2} \ln \left (t \right )+c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 415
AsymptoticDSolveValue[t*y''[t]+(1-t)*y'[t]+\[Lambda]*y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 \left (-\frac {(\lambda -4) (\lambda -3) (\lambda -2) (\lambda -1) \lambda t^5}{14400}+\frac {1}{576} (\lambda -3) (\lambda -2) (\lambda -1) \lambda t^4-\frac {1}{36} (\lambda -2) (\lambda -1) \lambda t^3+\frac {1}{4} (\lambda -1) \lambda t^2-\lambda t+1\right )+c_2 \left (\frac {(\lambda -4) (\lambda -3) (\lambda -2) (\lambda -1) t^5}{14400}+\frac {(\lambda -4) (\lambda -3) (\lambda -2) \lambda t^5}{14400}+\frac {(\lambda -4) (\lambda -3) (\lambda -1) \lambda t^5}{14400}+\frac {(\lambda -4) (\lambda -2) (\lambda -1) \lambda t^5}{14400}+\frac {137 (\lambda -4) (\lambda -3) (\lambda -2) (\lambda -1) \lambda t^5}{432000}+\frac {(\lambda -3) (\lambda -2) (\lambda -1) \lambda t^5}{14400}-\frac {1}{576} (\lambda -3) (\lambda -2) (\lambda -1) t^4-\frac {1}{576} (\lambda -3) (\lambda -2) \lambda t^4-\frac {1}{576} (\lambda -3) (\lambda -1) \lambda t^4-\frac {25 (\lambda -3) (\lambda -2) (\lambda -1) \lambda t^4}{3456}-\frac {1}{576} (\lambda -2) (\lambda -1) \lambda t^4+\frac {1}{36} (\lambda -2) (\lambda -1) t^3+\frac {1}{36} (\lambda -2) \lambda t^3+\frac {11}{108} (\lambda -2) (\lambda -1) \lambda t^3+\frac {1}{36} (\lambda -1) \lambda t^3-\frac {1}{4} (\lambda -1) t^2-\frac {3}{4} (\lambda -1) \lambda t^2-\frac {\lambda t^2}{4}+\left (-\frac {(\lambda -4) (\lambda -3) (\lambda -2) (\lambda -1) \lambda t^5}{14400}+\frac {1}{576} (\lambda -3) (\lambda -2) (\lambda -1) \lambda t^4-\frac {1}{36} (\lambda -2) (\lambda -1) \lambda t^3+\frac {1}{4} (\lambda -1) \lambda t^2-\lambda t+1\right ) \log (t)+2 \lambda t+t\right ) \]