Internal problem ID [1821]
Internal file name [OUTPUT/1822_Sunday_June_05_2022_02_34_02_AM_74330320/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 2.8.3, The method of Frobenius. Equal roots, and roots differering by an integer.
Page 223
Problem number: 2.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {t^{2} y^{\prime \prime }-t \left (t +1\right ) y^{\prime }+y=0} \] 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. \[ t^{2} y^{\prime \prime }+\left (-t^{2}-t \right ) y^{\prime }+y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(t) y^{\prime } + q(t) y &=0 \end {align*}
Where \begin {align*} p(t) &= -\frac {t +1}{t}\\ q(t) &= \frac {1}{t^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ t^{2} y^{\prime \prime }+\left (-t^{2}-t \right ) y^{\prime }+y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] 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^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+\left (-t^{2}-t \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 }}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} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{1+n +r} a_{n} \left (n +r \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 }}a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{1+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}\right ) \\ \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 \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{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 ) t^{n +r}\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 }}a_{n} t^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-t^{n +r} a_{n} \left (n +r \right )+a_{n} t^{n +r} = 0 \] When \(n = 0\) the above becomes \[ t^{r} a_{0} r \left (-1+r \right )-t^{r} a_{0} r +a_{0} t^{r} = 0 \] Or \[ \left (t^{r} r \left (-1+r \right )-t^{r} r +t^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-1+r \right )^{2} t^{r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ \left (-1+r \right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-1+r \right )^{2} t^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 1]\).
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. Using the value of the indicial root found earlier, \(r = 1\), Eqs (1A,1B) become \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{1+n}\\ y_{2}\left (t \right ) &= y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{1+n}\right ) \end {align*}
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 )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1}}{n +r -1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {a_{n -1}}{n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {1}{r} \] Which for the root \(r = 1\) becomes \[ a_{1}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{r \left (1+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r}\) | \(1\) |
| \(a_{2}\) | \(\frac {1}{r \left (1+r \right )}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {1}{r \left (1+r \right ) \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r}\) | \(1\) |
| \(a_{2}\) | \(\frac {1}{r \left (1+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {1}{r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{6}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (3+r \right ) r \left (1+r \right ) \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r}\) | \(1\) |
| \(a_{2}\) | \(\frac {1}{r \left (1+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {1}{r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{6}\) |
| \(a_{4}\) | \(\frac {1}{\left (3+r \right ) r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1}{\left (3+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{5}={\frac {1}{120}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r}\) | \(1\) |
| \(a_{2}\) | \(\frac {1}{r \left (1+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {1}{r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{6}\) |
| \(a_{4}\) | \(\frac {1}{\left (3+r \right ) r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{24}\) |
| \(a_{5}\) | \(\frac {1}{\left (3+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{120}\) |
Using the above table, then the first 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+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\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 \(r = 1\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =1\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {1}{r}\) | \(1\) | \(-\frac {1}{r^{2}}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{r \left (1+r \right )}\) | \(\frac {1}{2}\) | \(\frac {-1-2 r}{r^{2} \left (1+r \right )^{2}}\) | \(-{\frac {3}{4}}\) |
| \(b_{3}\) | \(\frac {1}{r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{6}\) | \(\frac {-3 r^{2}-6 r -2}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(-{\frac {11}{36}}\) |
| \(b_{4}\) | \(\frac {1}{\left (3+r \right ) r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{24}\) | \(\frac {-4 r^{3}-18 r^{2}-22 r -6}{\left (3+r \right )^{2} r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(-{\frac {25}{288}}\) |
| \(b_{5}\) | \(\frac {1}{\left (3+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{120}\) | \(\frac {-5 r^{4}-40 r^{3}-105 r^{2}-100 r -24}{\left (3+r \right )^{2} r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {137}{7200}}\) |
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 \\ &= t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+t \left (-t -\frac {3 t^{2}}{4}-\frac {11 t^{3}}{36}-\frac {25 t^{4}}{288}-\frac {137 t^{5}}{7200}+O\left (t^{6}\right )\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} t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) + c_{2} \left (t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+t \left (-t -\frac {3 t^{2}}{4}-\frac {11 t^{3}}{36}-\frac {25 t^{4}}{288}-\frac {137 t^{5}}{7200}+O\left (t^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right )+c_{2} \left (t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+t \left (-t -\frac {3 t^{2}}{4}-\frac {11 t^{3}}{36}-\frac {25 t^{4}}{288}-\frac {137 t^{5}}{7200}+O\left (t^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right )+c_{2} \left (t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+t \left (-t -\frac {3 t^{2}}{4}-\frac {11 t^{3}}{36}-\frac {25 t^{4}}{288}-\frac {137 t^{5}}{7200}+O\left (t^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right )+c_{2} \left (t \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+t \left (-t -\frac {3 t^{2}}{4}-\frac {11 t^{3}}{36}-\frac {25 t^{4}}{288}-\frac {137 t^{5}}{7200}+O\left (t^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} y^{\prime \prime }+\left (-t^{2}-t \right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {y}{t^{2}}+\frac {\left (t +1\right ) y^{\prime }}{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 {y}{t^{2}}-\frac {\left (t +1\right ) y^{\prime }}{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 {1}{t^{2}}\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}}}=1 \\ {} & \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^{2} y^{\prime \prime }-t \left (t +1\right ) 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^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & 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^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+r \right )^{2} t^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r -1\right )^{2}-a_{k -1} \left (k +r -1\right )\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =1 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -1\right ) \left (a_{k} \left (k +r -1\right )-a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r \right ) \left (a_{k +1} \left (k +r \right )-a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +1}, a_{k +1}=\frac {a_{k}}{k +1}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 63
Order:=6; dsolve(t^2*diff(y(t),t$2)-t*(1+t)*diff(y(t),t)+y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = \left (\left (c_{2} \ln \left (t \right )+c_{1} \right ) \left (1+t +\frac {1}{2} t^{2}+\frac {1}{6} t^{3}+\frac {1}{24} t^{4}+\frac {1}{120} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+\left (-t -\frac {3}{4} t^{2}-\frac {11}{36} t^{3}-\frac {25}{288} t^{4}-\frac {137}{7200} t^{5}+\operatorname {O}\left (t^{6}\right )\right ) c_{2} \right ) t \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 112
AsymptoticDSolveValue[t^2*y''[t]-t*(1+t)*y'[t]+y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 t \left (\frac {t^5}{120}+\frac {t^4}{24}+\frac {t^3}{6}+\frac {t^2}{2}+t+1\right )+c_2 \left (t \left (-\frac {137 t^5}{7200}-\frac {25 t^4}{288}-\frac {11 t^3}{36}-\frac {3 t^2}{4}-t\right )+t \left (\frac {t^5}{120}+\frac {t^4}{24}+\frac {t^3}{6}+\frac {t^2}{2}+t+1\right ) \log (t)\right ) \]