Internal problem ID [1793]
Internal file name [OUTPUT/1794_Sunday_June_05_2022_02_31_32_AM_67098885/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: 1.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {t \left (t -2\right )^{2} y^{\prime \prime }+t y^{\prime }+y=0} \] 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. \[ \left (t^{3}-4 t^{2}+4 t \right ) y^{\prime \prime }+t y^{\prime }+y = 0 \] 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 {1}{\left (t -2\right )^{2}}\\ q(t) &= \frac {1}{t \left (t -2\right )^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \([2]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ t \left (t^{2}-4 t +4\right ) y^{\prime \prime }+t y^{\prime }+y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] 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 (t^{2}-4 t +4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+t \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 }}a_{n} t^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) t^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) t^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ 4 t^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ 4 t^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 4 t^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ 4 r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ 4 t^{-1+r} r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 0]\).
Since \(r_1 - r_2 = 1\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (t \right ) &= t^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\\ y_{2}\left (t \right ) &= C y_{1}\left (t \right ) \ln \left (t \right )+t^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= t \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\\ y_{2}\left (t \right ) &= C y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{1+n}\\ y_{2}\left (t \right ) &= C y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -2}-4 n^{2} a_{n -1}+2 n r a_{n -2}-8 n r a_{n -1}+r^{2} a_{n -2}-4 r^{2} a_{n -1}-5 n a_{n -2}+13 n a_{n -1}-5 r a_{n -2}+13 r a_{n -1}+6 a_{n -2}-8 a_{n -1}}{4 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (-a_{n -2}+4 a_{n -1}\right ) n^{2}+\left (3 a_{n -2}-5 a_{n -1}\right ) n -2 a_{n -2}-a_{n -1}}{4 n \left (1+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )}\) | \(-{\frac {1}{4}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {12 r^{4}-8 r^{3}-23 r^{2}+7 r +2}{16 r \left (1+r \right )^{2} \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{2}=-{\frac {5}{96}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )}\) | \(-{\frac {1}{4}}\) |
| \(a_{2}\) | \(\frac {12 r^{4}-8 r^{3}-23 r^{2}+7 r +2}{16 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(-{\frac {5}{96}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {32 r^{6}+56 r^{5}-116 r^{4}-181 r^{3}+30 r^{2}+65 r +10}{64 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {13}{1152}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )}\) | \(-{\frac {1}{4}}\) |
| \(a_{2}\) | \(\frac {12 r^{4}-8 r^{3}-23 r^{2}+7 r +2}{16 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(-{\frac {5}{96}}\) |
| \(a_{3}\) | \(\frac {32 r^{6}+56 r^{5}-116 r^{4}-181 r^{3}+30 r^{2}+65 r +10}{64 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {13}{1152}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {80 r^{8}+480 r^{7}+484 r^{6}-1708 r^{5}-3435 r^{4}-538 r^{3}+2011 r^{2}+930 r +104}{256 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{4}=-{\frac {199}{92160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )}\) | \(-{\frac {1}{4}}\) |
| \(a_{2}\) | \(\frac {12 r^{4}-8 r^{3}-23 r^{2}+7 r +2}{16 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(-{\frac {5}{96}}\) |
| \(a_{3}\) | \(\frac {32 r^{6}+56 r^{5}-116 r^{4}-181 r^{3}+30 r^{2}+65 r +10}{64 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {13}{1152}}\) |
| \(a_{4}\) | \(\frac {80 r^{8}+480 r^{7}+484 r^{6}-1708 r^{5}-3435 r^{4}-538 r^{3}+2011 r^{2}+930 r +104}{256 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {199}{92160}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {192 r^{10}+2320 r^{9}+9328 r^{8}+8240 r^{7}-37948 r^{6}-100469 r^{5}-56451 r^{4}+60511 r^{3}+74719 r^{2}+19998 r +1592}{1024 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {1123}{5529600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )}\) | \(-{\frac {1}{4}}\) |
| \(a_{2}\) | \(\frac {12 r^{4}-8 r^{3}-23 r^{2}+7 r +2}{16 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(-{\frac {5}{96}}\) |
| \(a_{3}\) | \(\frac {32 r^{6}+56 r^{5}-116 r^{4}-181 r^{3}+30 r^{2}+65 r +10}{64 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(-{\frac {13}{1152}}\) |
| \(a_{4}\) | \(\frac {80 r^{8}+480 r^{7}+484 r^{6}-1708 r^{5}-3435 r^{4}-538 r^{3}+2011 r^{2}+930 r +104}{256 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {199}{92160}}\) |
| \(a_{5}\) | \(\frac {192 r^{10}+2320 r^{9}+9328 r^{8}+8240 r^{7}-37948 r^{6}-100469 r^{5}-56451 r^{4}+60511 r^{3}+74719 r^{2}+19998 r +1592}{1024 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(-{\frac {1123}{5529600}}\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \begin {align*} y_{1}\left (t \right )&= t \left (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 \right ) \\ &= t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (t \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{1} \\ &= \frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )}&= \lim _{r\rightarrow 0}\frac {4 r^{2}-5 r -1}{4 r \left (1+r \right )}\\ &= \textit {undefined} \end {align*}
Since the limit does not exist then the log term is needed. Therefore the second solution has the form \[ y_{2}\left (t \right ) = C y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) \] Therefore \begin{align*} \frac {d}{d t}y_{2}\left (t \right ) &= C y_{1}^{\prime }\left (t \right ) \ln \left (t \right )+\frac {C y_{1}\left (t \right )}{t}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right ) \\ &= C y_{1}^{\prime }\left (t \right ) \ln \left (t \right )+\frac {C y_{1}\left (t \right )}{t}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d t^{2}}y_{2}\left (t \right ) &= C y_{1}^{\prime \prime }\left (t \right ) \ln \left (t \right )+\frac {2 C y_{1}^{\prime }\left (t \right )}{t}-\frac {C y_{1}\left (t \right )}{t^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (t \right ) \ln \left (t \right )+\frac {2 C y_{1}^{\prime }\left (t \right )}{t}-\frac {C y_{1}\left (t \right )}{t^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*} Substituting these back into the given ode \(t \left (t^{2}-4 t +4\right ) y^{\prime \prime }+t y^{\prime }+y = 0\) gives \[ t \left (t^{2}-4 t +4\right ) \left (C y_{1}^{\prime \prime }\left (t \right ) \ln \left (t \right )+\frac {2 C y_{1}^{\prime }\left (t \right )}{t}-\frac {C y_{1}\left (t \right )}{t^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right )\right )+t \left (C y_{1}^{\prime }\left (t \right ) \ln \left (t \right )+\frac {C y_{1}\left (t \right )}{t}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )\right )+C y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (t \left (t^{2}-4 t +4\right ) y_{1}^{\prime \prime }\left (t \right )+y_{1}^{\prime }\left (t \right ) t +y_{1}\left (t \right )\right ) \ln \left (t \right )+t \left (t^{2}-4 t +4\right ) \left (\frac {2 y_{1}^{\prime }\left (t \right )}{t}-\frac {y_{1}\left (t \right )}{t^{2}}\right )+y_{1}\left (t \right )\right ) C +t \left (t^{2}-4 t +4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right )\right )+t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (t \right )\) is a solution to the ode, then \[ t \left (t^{2}-4 t +4\right ) y_{1}^{\prime \prime }\left (t \right )+y_{1}^{\prime }\left (t \right ) t +y_{1}\left (t \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (t \left (t^{2}-4 t +4\right ) \left (\frac {2 y_{1}^{\prime }\left (t \right )}{t}-\frac {y_{1}\left (t \right )}{t^{2}}\right )+y_{1}\left (t \right )\right ) C +t \left (t^{2}-4 t +4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right )\right )+t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \frac {\left (2 t \left (t -2\right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )-\left (t -1\right ) \left (t -4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r_{1}}\right )\right ) C}{t}+\frac {t^{2} \left (t -2\right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) t^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) t}{t} = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 t \left (t -2\right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n} a_{n} \left (1+n \right )\right )-\left (t -1\right ) \left (t -4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{1+n}\right )\right ) C}{t}+\frac {t^{2} \left (t -2\right )^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n -2} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n -1} b_{n} n \right ) t^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) t}{t} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,t^{n +2} a_{n} \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 C \,t^{1+n} a_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 C \,t^{n} a_{n} \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,t^{n +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 C \,t^{1+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C a_{n} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,t^{1+n} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 t^{n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 n \,t^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,t^{n +2} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (n -2\right ) t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 C \,t^{1+n} a_{n} \left (1+n \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-8 C a_{n -2} \left (n -1\right ) t^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}8 C \,t^{n} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}8 C a_{n -1} n \,t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,t^{n +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{n -3} t^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 C \,t^{1+n} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}5 C a_{n -2} t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C a_{n} t^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 C a_{n -1} t^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,t^{1+n} b_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (n -2\right ) b_{n -2} \left (n -3\right ) t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 t^{n} b_{n} n \left (n -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 \left (n -1\right ) b_{n -1} \left (n -2\right ) t^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}t^{n} b_{n} n &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} t^{n -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 \(n -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (n -2\right ) t^{n -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-8 C a_{n -2} \left (n -1\right ) t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}8 C a_{n -1} n \,t^{n -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{n -3} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}5 C a_{n -2} t^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 C a_{n -1} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\left (n -2\right ) b_{n -2} \left (n -3\right ) t^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 \left (n -1\right ) b_{n -1} \left (n -2\right ) t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 n \,t^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} t^{n -1}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the difference between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives \[ 4 C +1 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {1}{4}} \] For \(n=2\), Eq (2B) gives \[ \left (-3 a_{0}+12 a_{1}\right ) C +2 b_{1}+8 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {3}{2}+8 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {3}{16}} \] For \(n=3\), Eq (2B) gives \[ \left (a_{0}-11 a_{1}+20 a_{2}\right ) C -5 b_{2}+24 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {25}{96}+24 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-{\frac {25}{2304}} \] For \(n=4\), Eq (2B) gives \[ \left (3 a_{1}-19 a_{2}+28 a_{3}\right ) C +2 b_{2}-20 b_{3}+48 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {5}{36}+48 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {5}{1728}} \] For \(n=5\), Eq (2B) gives \[ \left (5 a_{2}-27 a_{3}+36 a_{4}\right ) C +6 b_{3}-43 b_{4}+80 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {50087}{276480}+80 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {50087}{22118400}} \] Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from \[ y_{2}\left (t \right ) = C y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) \] Using the above value found for \(C=-{\frac {1}{4}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (t \right )= -\frac {1}{4}\eslowast \left (t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right )\right ) \ln \left (t \right )+1-\frac {3 t^{2}}{16}-\frac {25 t^{3}}{2304}+\frac {5 t^{4}}{1728}+\frac {50087 t^{5}}{22118400}+O\left (t^{6}\right ) \] 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} t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right ) + c_{2} \left (-\frac {1}{4}\eslowast \left (t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right )\right ) \ln \left (t \right )+1-\frac {3 t^{2}}{16}-\frac {25 t^{3}}{2304}+\frac {5 t^{4}}{1728}+\frac {50087 t^{5}}{22118400}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right )+c_{2} \left (-\frac {t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right ) \ln \left (t \right )}{4}+1-\frac {3 t^{2}}{16}-\frac {25 t^{3}}{2304}+\frac {5 t^{4}}{1728}+\frac {50087 t^{5}}{22118400}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right )+c_{2} \left (-\frac {t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right ) \ln \left (t \right )}{4}+1-\frac {3 t^{2}}{16}-\frac {25 t^{3}}{2304}+\frac {5 t^{4}}{1728}+\frac {50087 t^{5}}{22118400}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right )+c_{2} \left (-\frac {t \left (1-\frac {t}{4}-\frac {5 t^{2}}{96}-\frac {13 t^{3}}{1152}-\frac {199 t^{4}}{92160}-\frac {1123 t^{5}}{5529600}+O\left (t^{6}\right )\right ) \ln \left (t \right )}{4}+1-\frac {3 t^{2}}{16}-\frac {25 t^{3}}{2304}+\frac {5 t^{4}}{1728}+\frac {50087 t^{5}}{22118400}+O\left (t^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t \left (t^{2}-4 t +4\right ) y^{\prime \prime }+t y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {y}{t \left (t^{2}-4 t +4\right )}-\frac {y^{\prime }}{t^{2}-4 t +4} \\ \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 }}{t^{2}-4 t +4}+\frac {y}{t \left (t^{2}-4 t +4\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (t \right )=\frac {1}{t^{2}-4 t +4}, P_{3}\left (t \right )=\frac {1}{t \left (t^{2}-4 t +4\right )}\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}}}=0 \\ {} & \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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & t \left (t^{2}-4 t +4\right ) y^{\prime \prime }+t y^{\prime }+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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & t^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & t^{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 ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 4 a_{0} r \left (-1+r \right ) t^{-1+r}+\left (4 a_{1} \left (1+r \right ) r -a_{0} \left (4 r^{2}-5 r -1\right )\right ) t^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +1+r \right ) \left (k +r \right )-a_{k} \left (4 k^{2}+8 k r +4 r^{2}-5 k -5 r -1\right )+a_{k -1} \left (k +r -1\right ) \left (k -2+r \right )\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) r -a_{0} \left (4 r^{2}-5 r -1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) k^{2}+\left (\left (-8 a_{k}+2 a_{k -1}+8 a_{k +1}\right ) r +5 a_{k}-3 a_{k -1}+4 a_{k +1}\right ) k +\left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r^{2}+\left (5 a_{k}-3 a_{k -1}+4 a_{k +1}\right ) r +a_{k}+2 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) \left (k +1\right )^{2}+\left (\left (-8 a_{k +1}+2 a_{k}+8 a_{k +2}\right ) r +5 a_{k +1}-3 a_{k}+4 a_{k +2}\right ) \left (k +1\right )+\left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r^{2}+\left (5 a_{k +1}-3 a_{k}+4 a_{k +2}\right ) r +a_{k +1}+2 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+2 k r a_{k}-8 k r a_{k +1}+r^{2} a_{k}-4 r^{2} a_{k +1}-k a_{k}-3 k a_{k +1}-r a_{k}-3 r a_{k +1}+2 a_{k +1}}{4 \left (k^{2}+2 k r +r^{2}+3 k +3 r +2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}-k a_{k}-3 k a_{k +1}+2 a_{k +1}}{4 \left (k^{2}+3 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}-k a_{k}-3 k a_{k +1}+2 a_{k +1}}{4 \left (k^{2}+3 k +2\right )}, a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+k a_{k}-11 k a_{k +1}-5 a_{k +1}}{4 \left (k^{2}+5 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +1}, a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+k a_{k}-11 k a_{k +1}-5 a_{k +1}}{4 \left (k^{2}+5 k +6\right )}, 8 a_{1}+2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{1+k}\right ), a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{1+k}-k a_{k}-3 k a_{1+k}+2 a_{1+k}}{4 \left (k^{2}+3 k +2\right )}, a_{0}=0, b_{k +2}=-\frac {k^{2} b_{k}-4 k^{2} b_{1+k}+k b_{k}-11 k b_{1+k}-5 b_{1+k}}{4 \left (k^{2}+5 k +6\right )}, 8 b_{1}+2 b_{0}=0\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 -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a = 0, e <> 0, c <> 0 `
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 60
Order:=6; dsolve(t*(t-2)^2*diff(y(t),t$2)+t*diff(y(t),t)+y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = c_{1} t \left (1-\frac {1}{4} t -\frac {5}{96} t^{2}-\frac {13}{1152} t^{3}-\frac {199}{92160} t^{4}-\frac {1123}{5529600} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+c_{2} \left (\ln \left (t \right ) \left (-\frac {1}{4} t +\frac {1}{16} t^{2}+\frac {5}{384} t^{3}+\frac {13}{4608} t^{4}+\frac {199}{368640} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+\left (1-\frac {1}{4} t -\frac {1}{8} t^{2}+\frac {5}{2304} t^{3}+\frac {79}{13824} t^{4}+\frac {62027}{22118400} t^{5}+\operatorname {O}\left (t^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.067 (sec). Leaf size: 87
AsymptoticDSolveValue[t*(t-2)^2*y''[t]+t*y'[t]+y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 \left (\frac {t \left (13 t^3+60 t^2+288 t-1152\right ) \log (t)}{4608}+\frac {98 t^4+285 t^3+432 t^2-6912 t+6912}{6912}\right )+c_2 \left (-\frac {199 t^5}{92160}-\frac {13 t^4}{1152}-\frac {5 t^3}{96}-\frac {t^2}{4}+t\right ) \]