2.1.614 Problem 630

Internal problem ID [9786]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 630
Date solved : Wednesday, March 05, 2025 at 07:58:58 AM
CAS classification : [_Gegenbauer]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.297 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{2t^2-3}{{(t^2-1)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =2t^2-3\\ t& ={(t^2-1)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\left(\frac{2t^2-3}{{(t^2-1)}^2}\right)z(t)\end{array}$$

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 2.614: Necessary conditions for each Kovacic case

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

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

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t={(t^2-1)}^2$. There is a pole at $t=1$ of order $2$. There is a pole at $t=-1$ 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. Since pole order is not larger than $2$ and the order at $\mathrm{\infty }$ is $2$ then the necessary conditions for case three are met. Therefore

$$\begin{array}{rl}L& =[1,2,4,6,12]\end{array}$$

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

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

$$r=\frac{5}{4(t-1)}-\frac{1}{4{(t-1)}^2}-\frac{1}{4{(t+1)}^2}-\frac{5}{4(t+1)}$$

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

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

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

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

Since the order of $r$ at $\mathrm{\infty }$ is 2 then $[\sqrt{r}{]}_{\mathrm{\infty }}=0$. Let $b$ be the coefficient of $\frac{1}{t^2}$ in the Laurent series expansion of $r$ at $\mathrm{\infty }$. which can be found by dividing the leading coefficient of $s$ by the leading coefficient of $t$ from

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{2t^2-3}{{(t^2-1)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=2$. Hence

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

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

$$r=\frac{2t^2-3}{{(t^2-1)}^2}$$

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

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$2$	$-1$

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 ${\alpha }_{\mathrm{\infty }}^{+}=2$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})\\ & =2-\left(1\right)\\ & =1\end{array}$$

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

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

Substituting the above values in the above results in

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{t-c_1})+((-)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{-}}{t-c_2})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =\frac{1}{2t-2}+\frac{1}{2t+2}+\left(0\right)\\ & =\frac{1}{2t-2}+\frac{1}{2t+2}\\ & =\frac{t}{t^2-1}\end{array}$$

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

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

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

$$\{a_0=0\}$$

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

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

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

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

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

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

Which simplifies to

$$y_1=\frac{t\sqrt{t^2-1}}{\sqrt{t-1}\sqrt{t+1}}$$

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

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.006 (sec). Leaf size: 25

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

$$y=-\frac{c_2\ln (t+1)t}{2}+\frac{c_2\ln (t-1)t}{2}+c_{1}t+c_2$$

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 
<- Kovacics algorithm successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & (-t^2+1)\left(\frac{d^2}{d{t}^2}y\left(t\right)\right)-2t\left(\frac{d}{dt}y\left(t\right)\right)+2y\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{t}^2}y\left(t\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)=\frac{2y\left(t\right)}{t^2-1}-\frac{2\left(\frac{d}{dt}y\left(t\right)\right)t}{t^2-1}\\ \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^2}{d{t}^2}y\left(t\right)+\frac{2\left(\frac{d}{dt}y\left(t\right)\right)t}{t^2-1}-\frac{2y\left(t\right)}{t^2-1}=0\\ ◻& & \text{Check to see if}{t}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(t\right)=\frac{2t}{t^2-1},P_3\left(t\right)=-\frac{2}{t^2-1}]\\ & \circ & (t+1)\cdot P_2\left(t\right)\text{is analytic at}t=-1\\ & & ((t+1)\cdot P_2\left(t\right))|{}_{t=-1}=1\\ & \circ & {(t+1)}^2\cdot P_3\left(t\right)\text{is analytic at}t=-1\\ & & ({(t+1)}^2\cdot P_3\left(t\right))|{}_{t=-1}=0\\ & \circ & t=-1\text{is a regular singular point}\\ & & \text{Check to see if}{t}_0\text{is a regular singular point}\\ & & t_0=-1\\ \bullet & & \text{Multiply by denominators}\\ & & (t^2-1)\left(\frac{d^2}{d{t}^2}y\left(t\right)\right)+2t\left(\frac{d}{dt}y\left(t\right)\right)-2y\left(t\right)=0\\ \bullet & & \text{Change variables using}t=u-1\text{so that the regular singular point is at}u=0\\ & & (u^2-2u)\left(\frac{d^2}{d{u}^2}y\left(u\right)\right)+(2u-2)\left(\frac{d}{du}y\left(u\right)\right)-2y\left(u\right)=0\\ \bullet & & \text{Assume series solution for}y\left(u\right)\\ & & y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=0..1\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)u^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & u^m\cdot \left(\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)u^{k+r}\\ & \circ & \text{Convert}{u}^m\cdot \left(\frac{d^2}{d{u}^2}y\left(u\right)\right)\text{to series expansion for}m=1..2\\ & & u^m\cdot \left(\frac{d^2}{d{u}^2}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)u^{k+r-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & u^m\cdot \left(\frac{d^2}{d{u}^2}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)u^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & -2a_0{r}^2{u}^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(-2a_{k+1}{(k+1+r)}^2+a_k(k+r+2)(k+r-1))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -2r^2=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & -2a_{k+1}{(k+1)}^2+a_k(k+2)(k-1)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{a_k(k+2)(k-1)}{2{(k+1)}^2}\\ \bullet & & \text{Recursion relation for}r=0\text{; series terminates at}k=1\\ & & a_{k+1}=\frac{a_k(k+2)(k-1)}{2{(k+1)}^2}\\ \bullet & & \text{Apply recursion relation for}k=0\\ & & a_1=-a_0\\ \bullet & & \text{Terminating series solution of the ODE for}r=0\text{. Use reduction of order to find the second linearly independent solution}\\ & & y\left(u\right)=a_0\cdot (-u+1)\\ \bullet & & \text{Revert the change of variables}u=t+1\\ & & [y\left(t\right)=-a_{0}t]\end{array}$$

✓ Mathematica. Time used: 0.02 (sec). Leaf size: 33

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

$$y(t)\to c_{1}t-\frac{1}{2}{c}_2(t\log (1-t)-t\log (t+1)+2)$$

✓ Sympy. Time used: 0.869 (sec). Leaf size: 20

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

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