2.1.612 Problem 628

Internal problem ID [9782]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 628
Date solved : Friday, April 25, 2025 at 06:19:36 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.422 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

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

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =6\\ t& ={(t^2+2t-1)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\left(\frac{6}{{(t^2+2t-1)}^2}\right)z(t)\end{array}$$

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 2.612: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =4-0\\ & =4\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t={(t^2+2t-1)}^2$. There is a pole at $t=\sqrt{2}-1$ of order $2$. There is a pole at $t=-1-\sqrt{2}$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $4$ 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 $4$ 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

$$r=\frac{3}{4{(t-\sqrt{2}+1)}^2}+\frac{3}{4{(t+1+\sqrt{2})}^2}-\frac{3\sqrt{2}}{8(t-\sqrt{2}+1)}+\frac{3\sqrt{2}}{8(t+1+\sqrt{2})}$$

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

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

For the pole at $t=-1-\sqrt{2}$ let $b$ be the coefficient of $\frac{1}{{(t+1+\sqrt{2})}^2}$ 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 $4>2$ then

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

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

$$r=\frac{6}{{(t^2+2t-1)}^2}$$

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

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$4$	$0$	$0$	$1$

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 ${\alpha }_{\mathrm{\infty }}^{-}=1$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})\\ & =1-\left(1\right)\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{t-c_1})+((+)[\sqrt{r}{]}_{c_2}+\frac{{\alpha }_{c_2}^{+}}{t-c_2})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{2(t-\sqrt{2}+1)}+\frac{3}{2(t+1+\sqrt{2})}+(-)\left(0\right)\\ & =-\frac{1}{2(t-\sqrt{2}+1)}+\frac{3}{2(t+1+\sqrt{2})}\\ & =\frac{t+1-2\sqrt{2}}{t^2+2t-1}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(t)$ of degree $d=0$ 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

$$\begin{array}{}\text{(2A)}& p(t)& =1\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{2(t-\sqrt{2}+1)}+\frac{3}{2(t+1+\sqrt{2})})\left(0\right)+((\frac{1}{2{(t-\sqrt{2}+1)}^2}-\frac{3}{2{(t+1+\sqrt{2})}^2})+{(-\frac{1}{2(t-\sqrt{2}+1)}+\frac{3}{2(t+1+\sqrt{2})})}^2-\left(\frac{6}{{(t^2+2t-1)}^2}\right))& =0\\ 0=0\end{array}$$

The equation is satisfied since both sides are zero. Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(t)& =p{e}^{\int \omega dt}\\ & ={\mathrm{e}}^{\int (-\frac{1}{2(t-\sqrt{2}+1)}+\frac{3}{2(t+1+\sqrt{2})})dt}\\ & =\frac{{(t+1+\sqrt{2})}^{3/2}}{\sqrt{t-\sqrt{2}+1}}\end{array}$$

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

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

Which simplifies to

$$y_1=\frac{\sqrt{t^2+2t-1}{(t+1+\sqrt{2})}^{3/2}}{\sqrt{t-\sqrt{2}+1}}$$

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

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

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{\sqrt{t^2+2t-1}{(t+1+\sqrt{2})}^{3/2}}{\sqrt{t-\sqrt{2}+1}}\right)+c_2\left(\frac{\sqrt{t^2+2t-1}{(t+1+\sqrt{2})}^{3/2}}{\sqrt{t-\sqrt{2}+1}}(\frac{\sqrt{2}}{{(t+1+\sqrt{2})}^2}-\frac{1}{t+1+\sqrt{2}})\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.004 (sec). Leaf size: 15

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

$$y=c_2{t}^2+c_{1}t+c_1+c_2$$

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) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)-\frac{2(t+1)\left(\frac{d}{dt}y\left(t\right)\right)}{t^2+2t-1}+\frac{2y\left(t\right)}{t^2+2t-1}=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)\\ ◻& & \text{Check to see if}{t}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(t\right)=-\frac{2(t+1)}{t^2+2t-1},P_3\left(t\right)=\frac{2}{t^2+2t-1}]\\ & \circ & (t+1+\sqrt{2})\cdot P_2\left(t\right)\text{is analytic at}t=-1-\sqrt{2}\\ & & ((t+1+\sqrt{2})\cdot P_2\left(t\right))|{}_{t=-1-\sqrt{2}}=0\\ & \circ & {(t+1+\sqrt{2})}^2\cdot P_3\left(t\right)\text{is analytic at}t=-1-\sqrt{2}\\ & & ({(t+1+\sqrt{2})}^2\cdot P_3\left(t\right))|{}_{t=-1-\sqrt{2}}=0\\ & \circ & t=-1-\sqrt{2}\text{is a regular singular point}\\ & & \text{Check to see if}{t}_0\text{is a regular singular point}\\ & & t_0=-1-\sqrt{2}\\ \bullet & & \text{Multiply by denominators}\\ & & (t^2+2t-1)\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+(-2t-2)\left(\frac{d}{dt}y\left(t\right)\right)+2y\left(t\right)=0\\ \bullet & & \text{Change variables using}t=u-1-\sqrt{2}\text{so that the regular singular point is at}u=0\\ & & (u^2-2u\sqrt{2})\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(-2u+2\sqrt{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}{du}\frac{d}{du}y\left(u\right)\right)\text{to series expansion for}m=1..2\\ & & u^m\cdot \left(\frac{d}{du}\frac{d}{du}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}{du}\frac{d}{du}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}\\ & & -2\sqrt{2}(r-2)r{a}_0{u}^{r-1}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(-2\sqrt{2}(k+r-1)(k+1+r)a_{k+1}+a_k(k+r-1)(k+r-2))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -2\sqrt{2}(r-2)r=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,2\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k+r-1)(-2a_{k+1}(k+1+r)\sqrt{2}+a_k(k+r-2))=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{a_k(k+r-2)\sqrt{2}}{4(k+1+r)}\\ \bullet & & \text{Recursion relation for}r=0\text{; series terminates at}k=2\\ & & a_{k+1}=\frac{a_k(k-2)\sqrt{2}}{4(k+1)}\\ \bullet & & \text{Apply recursion relation for}k=0\\ & & a_1=-\frac{a_0\sqrt{2}}{2}\\ \bullet & & \text{Apply recursion relation for}k=1\\ & & a_2=-\frac{a_1\sqrt{2}}{8}\\ \bullet & & \text{Express in terms of}{a}_0\\ & & a_2=\frac{a_0}{8}\\ \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 (1-\frac{u\sqrt{2}}{2}+\frac{u^2}{8})\\ \bullet & & \text{Revert the change of variables}u=t+1+\sqrt{2}\\ & & [y\left(t\right)=a_0(\frac{(-2t-2)\sqrt{2}}{8}+\frac{t^2}{8}+\frac{t}{4}+\frac{3}{8})]\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+1}=\frac{a_{k}k\sqrt{2}}{4(k+3)}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+2},a_{k+1}=\frac{a_{k}k\sqrt{2}}{4(k+3)}]\\ \bullet & & \text{Revert the change of variables}u=t+1+\sqrt{2}\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(t+1+\sqrt{2})}^{k+2},a_{k+1}=\frac{a_{k}k\sqrt{2}}{4(k+3)}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(t\right)=a_0(\frac{(-2t-2)\sqrt{2}}{8}+\frac{t^2}{8}+\frac{t}{4}+\frac{3}{8})+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(t+1+\sqrt{2})}^{k+2}),b_{k+1}=\frac{b_{k}k\sqrt{2}}{4(k+3)}]\end{array}$$

✓ Mathematica. Time used: 0.447 (sec). Leaf size: 98

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

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

✓ Sympy. Time used: 0.986 (sec). Leaf size: 19

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

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