2.1.617 Problem 633

Internal problem ID [9789]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 633
Date solved : Sunday, March 30, 2025 at 02:46:40 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.261 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (2t+1)y^{\prime \prime }+(-4t-4)y^{\prime }+4y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =2t+1\\ \text{(3)}& B& =-4t-4C& =4\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{4t^2+2}{{(2t+1)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =4t^2+2\\ t& ={(2t+1)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\left(\frac{4t^2+2}{{(2t+1)}^2}\right)z(t)\end{array}$$

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 2.617: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-2\\ & =0\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t={(2t+1)}^2$. There is a pole at $t=-\frac{1}{2}$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ 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

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

For the pole at $t=-\frac{1}{2}$ let $b$ be the coefficient of $\frac{1}{{(t+\frac{1}{2})}^2}$ 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 $O_r(\mathrm{\infty })=0$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{0}{2}& & =0\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ 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

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

Let $a$ be the coefficient of $t^v=t^0$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx 1-\frac{1}{2t}+\frac{1}{2t^2}-\frac{1}{4t^3}+\frac{3}{32t^4}-\frac{3}{64t^5}+\frac{1}{32t^6}-\frac{1}{64t^7}+\dots \end{array}$$

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

$$a=1$$

From Eq. (9) the sum up to $v=0$ gives

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

Now we need to find $b$, where $b$ be the coefficient of $t^{v-1}=t^{-1}=\frac{1}{t}$ 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

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=1$$

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

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

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $\frac{1}{t}$ in $r$ will be the coefficient in $R$ of the term in $t$ of degree of $t$ minus one, divided by the leading coefficient in $t$. Doing long division gives

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

Since the degree of $t$ is $2$, then we see that the coefficient of the term $t$ in the remainder $R$ is $-4$. Dividing this by leading coefficient in $t$ which is $4$ gives $-1$. Now $b$ can be found.

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

Hence

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

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

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

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

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

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

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{-}\right)\\ & =-\frac{1}{2}-(-\frac{1}{2})\\ & =0\end{array}$$

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

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

Substituting the above values in the above results in

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

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

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

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

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

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

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

Which simplifies to

$$y_1={\mathrm{e}}^{2t}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-4t-4}{2t+1}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{2t+\ln (2t+1)}}{{\left(y_1\right)}^2}dt\\ & =y_1(-\frac{(t+1){\mathrm{e}}^{2t+\ln (2t+1)}{\mathrm{e}}^{-4t}}{2t+1})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left({\mathrm{e}}^{2t}\right)+c_2\left({\mathrm{e}}^{2t}(-\frac{(t+1){\mathrm{e}}^{2t+\ln (2t+1)}{\mathrm{e}}^{-4t}}{2t+1})\right)\end{array}$$

Will add steps showing solving for IC soon.

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

ode:=(2*t+1)*diff(diff(y(t),t),t)-4*(1+t)*diff(y(t),t)+4*y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 

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

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) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & (2t+1)\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)-4(t+1)\left(\frac{d}{dt}y\left(t\right)\right)+4y\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)=-\frac{4y\left(t\right)}{2t+1}+\frac{4(t+1)\left(\frac{d}{dt}y\left(t\right)\right)}{2t+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}{dt}\frac{d}{dt}y\left(t\right)-\frac{4(t+1)\left(\frac{d}{dt}y\left(t\right)\right)}{2t+1}+\frac{4y\left(t\right)}{2t+1}=0\\ ◻& & \text{Check to see if}{t}_0=-\frac{1}{2}\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(t\right)=-\frac{4(t+1)}{2t+1},P_3\left(t\right)=\frac{4}{2t+1}]\\ & \circ & (t+\frac{1}{2})\cdot P_2\left(t\right)\text{is analytic at}t=-\frac{1}{2}\\ & & ((t+\frac{1}{2})\cdot P_2\left(t\right))|{}_{t=-\frac{1}{2}}=-1\\ & \circ & {(t+\frac{1}{2})}^2\cdot P_3\left(t\right)\text{is analytic at}t=-\frac{1}{2}\\ & & ({(t+\frac{1}{2})}^2\cdot P_3\left(t\right))|{}_{t=-\frac{1}{2}}=0\\ & \circ & t=-\frac{1}{2}\text{is a regular singular point}\\ & & \text{Check to see if}{t}_0=-\frac{1}{2}\text{is a regular singular point}\\ & & t_0=-\frac{1}{2}\\ \bullet & & \text{Multiply by denominators}\\ & & (2t+1)\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+(-4t-4)\left(\frac{d}{dt}y\left(t\right)\right)+4y\left(t\right)=0\\ \bullet & & \text{Change variables using}t=u-\frac{1}{2}\text{so that the regular singular point is at}u=0\\ & & 2u\left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)+(-4u-2)\left(\frac{d}{du}y\left(u\right)\right)+4y\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\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)\text{to series expansion}\\ & & u\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-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & u\cdot \left(\frac{d}{du}\frac{d}{du}y\left(u\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)(k+r)u^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & 2a_{0}r(-2+r)u^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(2a_{k+1}(k+1+r)(k+r-1)-4a_k(k+r-1))u^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & 2r(-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}\\ & & 2(k+r-1)(a_{k+1}(k+1+r)-2a_k)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{2a_k}{k+1+r}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=\frac{2a_k}{k+1}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^k,a_{k+1}=\frac{2a_k}{k+1}]\\ \bullet & & \text{Revert the change of variables}u=t+\frac{1}{2}\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(t+\frac{1}{2})}^k,a_{k+1}=\frac{2a_k}{k+1}]\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+1}=\frac{2a_k}{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{2a_k}{k+3}]\\ \bullet & & \text{Revert the change of variables}u=t+\frac{1}{2}\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(t+\frac{1}{2})}^{k+2},a_{k+1}=\frac{2a_k}{k+3}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(t\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(t+\frac{1}{2})}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{(t+\frac{1}{2})}^{k+2}),a_{k+1}=\frac{2a_k}{k+1},b_{k+1}=\frac{2b_k}{k+3}]\end{array}$$

✓ Mathematica. Time used: 0.169 (sec). Leaf size: 88

ode=(2*t+1)*D[y[t],{t,2}]-4*(t+1)*D[y[t],t]+4*y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to \exp ({\int }_1^t\frac{2K[1]}{2K[1]+1}dK[1]-\frac{1}{2}{\int }_1^t(-2-\frac{2}{2K[2]+1})dK[2])(c_2{\int }_1^t\exp (-2{\int }_1^{K[3]}\frac{2K[1]}{2K[1]+1}dK[1])dK[3]+c_1)$$

✓ Sympy. Time used: 1.140 (sec). Leaf size: 63

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq((2*t + 1)*Derivative(y(t), (t, 2)) - (4*t + 4)*Derivative(y(t), t) + 4*y(t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=-t^{4}r\left(3\right)+\frac{17t^{5}r\left(3\right)}{10}+C_2(\frac{52t^5}{15}-\frac{11t^4}{3}-4t^2+1)+C_{1}t(-2t^4+2t^3+2t+1)+O\left(t^6\right)$$