2.1.12 Problem 12

Internal problem ID [9182]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 12
Date solved : Friday, April 25, 2025 at 05:59:14 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}t{y}^{\prime \prime }+(t^2-1)y^{\prime }+t^{2}y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.526 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& t{y}^{\prime \prime }+(t^2-1)y^{\prime }+t^{2}y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =t\\ \text{(3)}& B& =t^2-1C& =t^2\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(t)& =y{e}^{\int \frac{B}{2A}dt}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(t)=rz(t)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{t^4-4t^3+3}{4t^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =t^4-4t^3+3\\ t& =4t^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\left(\frac{t^4-4t^3+3}{4t^2}\right)z(t)\end{array}$$

Equation (7) is now solved. After finding $z(t)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(t\right)e^{-\int \frac{B}{2A}dt}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.12: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-4\\ & =-2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=4t^2$. There is a pole at $t=0$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $-2$ then the necessary conditions for case one are met. Since there is a pole of order $2$ then necessary conditions for case two are met. Therefore

$$\begin{array}{rl}L& =[1,2]\end{array}$$

Attempting to find a solution using case $n=1$.

Looking at poles of order 2. The partial fractions decomposition of $r$ is

$$r=\frac{t^2}{4}-t+\frac{3}{4t^2}$$

For the pole at $t=0$ let $b$ be the coefficient of $\frac{1}{t^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{3}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{3}{2}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{2}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=-2$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{2}{2}& & =1\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $t^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{t}^i\\ \text{(8)}& & =\sum _{i=0}^1{a}_i{t}^i\end{array}$$

Let $a$ be the coefficient of $t^v=t^1$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{t}{2}-1-\frac{1}{t}-\frac{2}{t^2}-\frac{17}{4t^3}-\frac{25}{2t^4}-\frac{75}{2t^5}-\frac{117}{t^6}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=\frac{1}{2}$$

From Eq. (9) the sum up to $v=1$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^1{a}_i{t}^i\\ \text{(10)}& & =\frac{t}{2}-1\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $t^{v-1}=t^0=1$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{1}{4}{t}^2-t+1$$

This shows that the coefficient of $1$ in the above is $1$. Now we need to find the coefficient of $1$ in $r$. How this is done depends on if $v=0$ or not. Since $v=1$ which is not zero, then starting $r=\frac{s}{t}$, we do long division and write this in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $1$ in $r$ will be the coefficient this term in the quotient. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{t^4-4t^3+3}{4t^2}\\ & =Q+\frac{R}{4t^2}\\ & =(\frac{1}{4}{t}^2-t)+\left(\frac{3}{4t^2}\right)\\ & =\frac{t^2}{4}-t+\frac{3}{4t^2}\end{array}$$

We see that the coefficient of the term $t$ in the quotient is $0$. Now $b$ can be found.

$$\begin{array}{rl}b& =\left(0\right)-\left(1\right)\\ & =-1\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{t}{2}-1\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{-1}{\frac{1}{2}}-1)& & =-\frac{3}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{-1}{\frac{1}{2}}-1)& & =\frac{1}{2}\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ where $r$ is

$$r=\frac{t^4-4t^3+3}{4t^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{3}{2}$	$-\frac{1}{2}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$-2$	$\frac{t}{2}-1$	$-\frac{3}{2}$	$\frac{1}{2}$

Now that the all $[\sqrt{r}{]}_c$ and its associated ${\alpha }_c^{\pm }$ have been determined for all the poles in the set $\mathrm{\Gamma }$ and $[\sqrt{r}{]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$ have also been found, the next step is to determine possible non negative integer $d$ from these using

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{s(\mathrm{\infty })}-\sum _{c\in \mathrm{\Gamma }}{\alpha }_c^{s(c)}\end{array}$$

Where $s(c)$ is either $+$ or $-$ and $s(\mathrm{\infty })$ is the sign of ${\alpha }_{\mathrm{\infty }}^{\pm }$. This is done by trial over all set of families $s=(s(c){)}_{c\in \mathrm{\Gamma }\cup \mathrm{\infty }}$ until such $d$ is found to work in finding candidate $\omega$. Trying ${\alpha }_{\mathrm{\infty }}^{-}=\frac{1}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)\\ & =\frac{1}{2}-(-\frac{1}{2})\\ & =1\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{t-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

The above gives

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{t-c_1})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{2t}+(-)(\frac{t}{2}-1)\\ & =-\frac{1}{2t}-\frac{t}{2}+1\\ & =-\frac{{(t-1)}^2}{2t}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(t)$ of degree $d=1$ to solve the ode. The polynomial $p(t)$ needs to satisfy the equation

$$\begin{array}{}\text{(1A)}& p^{″}+2\omega p^{\prime }+({\omega }^{\prime }+{\omega }^2-r)p=0\end{array}$$

Let

$$\begin{array}{}\text{(2A)}& p(t)& =t+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{2t}-\frac{t}{2}+1)\left(1\right)+((\frac{1}{2t^2}-\frac{1}{2})+{(-\frac{1}{2t}-\frac{t}{2}+1)}^2-\left(\frac{t^4-4t^3+3}{4t^2}\right))& =0\\ \frac{(a_0+1)(t-1)}{t}=0\end{array}$$

Solving for the coefficients $a_i$ in the above using method of undetermined coefficients gives

$$\{a_0=-1\}$$

Substituting these coefficients in $p(t)$ in eq. (2A) results in

$$\begin{array}{rl}p(t)& =t-1\end{array}$$

Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(t)& =p{e}^{\int \omega dt}\\ & =(t-1){\mathrm{e}}^{\int (-\frac{1}{2t}-\frac{t}{2}+1)dt}\\ & =(t-1){\mathrm{e}}^{-\frac{t^2}{4}+t-\frac{\ln \left(t\right)}{2}}\\ & =\frac{(t-1){\mathrm{e}}^{-\frac{t(t-4)}{4}}}{\sqrt{t}}\end{array}$$

The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{t^2-1}{t}dt}\\ & =z_1{e}^{-\frac{t^2}{4}+\frac{\ln \left(t\right)}{2}}\\ & =z_1\left(\sqrt{t}{\mathrm{e}}^{-\frac{t^2}{4}}\right)\end{array}$$

Which simplifies to

$$y_1=(t-1){\mathrm{e}}^{-\frac{t(t-2)}{2}}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dt}}{y_1^2}dt$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{t^2-1}{t}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{-\frac{t^2}{2}+\ln \left(t\right)}}{{\left(y_1\right)}^2}dt\\ & =y_1(\int \frac{{\mathrm{e}}^{-\frac{t^2}{2}+\ln \left(t\right)}{\mathrm{e}}^{t(t-2)}}{{(t-1)}^2}dt)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left((t-1){\mathrm{e}}^{-\frac{t(t-2)}{2}}\right)+c_2\left((t-1){\mathrm{e}}^{-\frac{t(t-2)}{2}}(\int \frac{{\mathrm{e}}^{-\frac{t^2}{2}+\ln \left(t\right)}{\mathrm{e}}^{t(t-2)}}{{(t-1)}^2}dt)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.035 (sec). Leaf size: 82

ode:=diff(diff(y(t),t),t)*t+(t^2-1)*diff(y(t),t)+t^2*y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 

$$y=-\frac{\sqrt{2}{\mathrm{e}}^{-\frac{t(t-2)}{2}}(-c_2\sqrt{\pi }(-1+t)(t-2)\mathrm{erf}\left(\frac{\sqrt{2}\sqrt{-{(t-2)}^2}}{2}\right)+\sqrt{2}(c_2{\mathrm{e}}^{\frac{{(t-2)}^2}{2}}-t{c}_1+c_1)\sqrt{-{(t-2)}^2})}{2\sqrt{-{(t-2)}^2}}$$

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 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> 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 
               <- hyper3 successful: indirect Equivalence to 0F1 under ``^ @ Mo\ 
ebius`` is resolved 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form could result into a too large expression - returning s\ 
pecial function form of solution, free of uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & t\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+(t^2-1)\left(\frac{d}{dt}y\left(t\right)\right)+t^{2}y\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)=-ty\left(t\right)-\frac{(t^2-1)\left(\frac{d}{dt}y\left(t\right)\right)}{t}\\ \bullet & & \text{Group terms with}y\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)+\frac{(t^2-1)\left(\frac{d}{dt}y\left(t\right)\right)}{t}+ty\left(t\right)=0\\ ◻& & \text{Check to see if}{t}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(t\right)=\frac{t^2-1}{t},P_3\left(t\right)=t]\\ & \circ & t\cdot P_2\left(t\right)\text{is analytic at}t=0\\ & & (t\cdot P_2\left(t\right))|{}_{t=0}=-1\\ & \circ & t^2\cdot P_3\left(t\right)\text{is analytic at}t=0\\ & & (t^2\cdot P_3\left(t\right))|{}_{t=0}=0\\ & \circ & t=0\text{is a regular singular point}\\ & & \text{Check to see if}{t}_0=0\text{is a regular singular point}\\ & & t_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & t\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+(t^2-1)\left(\frac{d}{dt}y\left(t\right)\right)+t^{2}y\left(t\right)=0\\ \bullet & & \text{Assume series solution for}y\left(t\right)\\ & & y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{t}^2\cdot y\left(t\right)\text{to series expansion}\\ & & t^2\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+r+2}\\ & \circ & \text{Shift index using}k->k-2\\ & & t^2\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=2}}{a}_{k-2}{t}^{k+r}\\ & \circ & \text{Convert}{t}^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)\text{to series expansion for}m=0..2\\ & & t^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)t^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & t^m\cdot \left(\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)t^{k+r}\\ & \circ & \text{Convert}t\cdot \left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)\text{to series expansion}\\ & & t\cdot \left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)t^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & t\cdot \left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)(k+r)t^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(-2+r)t^{-1+r}+a_1(1+r)(-1+r)t^r+(a_2(2+r)r+a_{0}r)t^{1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_{k+1}(k+1+r)(k+r-1)+a_{k-1}(k+r-1)+a_{k-2})t^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(-2+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,2\}\\ \bullet & & \text{The coefficients of each power of}t\text{must be 0}\\ & & [a_1(1+r)(-1+r)=0,a_2(2+r)r+a_{0}r=0]\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & \{a_1=0,a_2=-\frac{a_0}{2+r}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+1+r)(k+r-1)+a_{k-1}(k+r-1)+a_{k-2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & a_{k+3}(k+3+r)(k+1+r)+a_{k+1}(k+1+r)+a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+3}=-\frac{k{a}_{k+1}+r{a}_{k+1}+a_k+a_{k+1}}{(k+3+r)(k+1+r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+3}=-\frac{k{a}_{k+1}+a_k+a_{k+1}}{(k+3)(k+1)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^k,a_{k+3}=-\frac{k{a}_{k+1}+a_k+a_{k+1}}{(k+3)(k+1)},a_1=0,a_2=-\frac{a_0}{2}]\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+3}=-\frac{k{a}_{k+1}+a_k+3a_{k+1}}{(k+5)(k+3)}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+2},a_{k+3}=-\frac{k{a}_{k+1}+a_k+3a_{k+1}}{(k+5)(k+3)},a_1=0,a_2=-\frac{a_0}{4}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(t\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{t}^{k+2}),a_{k+3}=-\frac{k{a}_{k+1}+a_k+a_{k+1}}{(k+3)(k+1)},a_1=0,a_2=-\frac{a_0}{2},b_{k+3}=-\frac{k{b}_{k+1}+b_k+3b_{k+1}}{(k+5)(k+3)},b_1=0,b_2=-\frac{b_0}{4}]\end{array}$$

✓ Mathematica. Time used: 0.557 (sec). Leaf size: 54

ode=t*D[y[t],{t,2}]+(t^2-1)*D[y[t],t]+t^2*y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to e^{t-\frac{t^2}{2}}(t-1)(c_2{\int }_1^t\frac{e^{\frac{1}{2}(K[1]-4)K[1]}K[1]}{(K[1]-1{)}^2}dK[1]+c_1)$$

✓ Sympy. Time used: 0.826 (sec). Leaf size: 36

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t**2*y(t) + t*Derivative(y(t), (t, 2)) + (t**2 - 1)*Derivative(y(t), t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=C_2(\frac{t^5}{15}-\frac{t^3}{3}+1)+C_1{t}^2(-\frac{t^3}{15}-\frac{t^2}{4}+1)+O\left(t^6\right)$$