2.1.33 Problem 34

Internal problem ID [9205]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 34
Date solved : Wednesday, March 05, 2025 at 07:37:49 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}(2x+1)y^{\prime \prime }-2y^{\prime }-(2x+3)y& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.273 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& (2x+1)y^{\prime \prime }-2y^{\prime }+(-2x-3)y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

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

Applying the Liouville transformation on the dependent variable gives

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

Then (2) becomes

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{4x^2+8x+6}{{(2x+1)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =4x^2+8x+6\\ t& ={(2x+1)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{4x^2+8x+6}{{(2x+1)}^2}\right)z(x)\end{array}$$

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

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

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 2.33: 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 $t={(2x+1)}^2$. There is a pole at $x=-\frac{1}{2}$ 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

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

For the pole at $x=-\frac{1}{2}$ let $b$ be the coefficient of $\frac{1}{{(x+\frac{1}{2})}^2}$ 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 $x^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

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

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

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx 1+\frac{1}{2x}-\frac{1}{4x^3}+\frac{11}{32x^4}-\frac{21}{64x^5}+\frac{15}{64x^6}-\frac{3}{32x^7}-\frac{117}{2048x^8}+\dots \end{array}$$

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

$$a=1$$

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

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

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^{-1}=\frac{1}{x}$ 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 $\frac{1}{x}$ in the above is $0$. Now we need to find the coefficient of $\frac{1}{x}$ 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 $\frac{1}{x}$ in $r$ will be the coefficient in $R$ of the term in $x$ of degree of $t$ minus one, divided by the leading coefficient in $t$. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{4x^2+8x+6}{4x^2+4x+1}\\ & =Q+\frac{R}{4x^2+4x+1}\\ & =\left(1\right)+\left(\frac{4x+5}{4x^2+4x+1}\right)\\ & =1+\frac{4x+5}{4x^2+4x+1}\end{array}$$

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

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

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

$$r=\frac{4x^2+8x+6}{{(2x+1)}^2}$$

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

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

The above gives

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

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=0$ to solve the ode. The polynomial $p(x)$ needs to satisfy the equation

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

Let

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

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{2(x+\frac{1}{2})}-1)\left(0\right)+(\left(\frac{1}{2{(x+\frac{1}{2})}^2}\right)+{(-\frac{1}{2(x+\frac{1}{2})}-1)}^2-\left(\frac{4x^2+8x+6}{{(2x+1)}^2}\right))& =0\\ 0=0\end{array}$$

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

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

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

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

Which simplifies to

$$y_1={\mathrm{e}}^{-x}$$

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

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

Substituting gives

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.006 (sec). Leaf size: 16

ode:=(2*x+1)*diff(diff(y(x),x),x)-2*diff(y(x),x)-(2*x+3)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1{\mathrm{e}}^{-x}+c_2{\mathrm{e}}^{x}x$$

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}\\ & & (2x+1)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)-2\frac{d}{dx}y\left(x\right)-(2x+3)y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=\frac{(2x+3)y\left(x\right)}{2x+1}+\frac{2\left(\frac{d}{dx}y\left(x\right)\right)}{2x+1}\\ \bullet & & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)-\frac{2\left(\frac{d}{dx}y\left(x\right)\right)}{2x+1}-\frac{(2x+3)y\left(x\right)}{2x+1}=0\\ ◻& & \text{Check to see if}{x}_0=-\frac{1}{2}\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=-\frac{2}{2x+1},P_3\left(x\right)=-\frac{2x+3}{2x+1}]\\ & \circ & (x+\frac{1}{2})\cdot P_2\left(x\right)\text{is analytic at}x=-\frac{1}{2}\\ & & ((x+\frac{1}{2})\cdot P_2\left(x\right))|{}_{x=-\frac{1}{2}}=-1\\ & \circ & {(x+\frac{1}{2})}^2\cdot P_3\left(x\right)\text{is analytic at}x=-\frac{1}{2}\\ & & ({(x+\frac{1}{2})}^2\cdot P_3\left(x\right))|{}_{x=-\frac{1}{2}}=0\\ & \circ & x=-\frac{1}{2}\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=-\frac{1}{2}\text{is a regular singular point}\\ & & x_0=-\frac{1}{2}\\ \bullet & & \text{Multiply by denominators}\\ & & (2x+1)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)-2\frac{d}{dx}y\left(x\right)+(-2x-3)y\left(x\right)=0\\ \bullet & & \text{Change variables using}x=u-\frac{1}{2}\text{so that the regular singular point is at}u=0\\ & & 2u\left(\frac{d^2}{d{u}^2}y\left(u\right)\right)-2\frac{d}{du}y\left(u\right)+(-2u-2)y\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 y\left(u\right)\text{to series expansion for}m=0..1\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+r+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & u^m\cdot y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=m}}{a}_{k-m}{u}^{k+r}\\ & \circ & \text{Convert}\frac{d}{du}y\left(u\right)\text{to series expansion}\\ & & \frac{d}{du}y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)u^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & \frac{d}{du}y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)u^{k+r}\\ & \circ & \text{Convert}u\cdot \left(\frac{d^2}{d{u}^2}y\left(u\right)\right)\text{to series expansion}\\ & & u\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-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & u\cdot \left(\frac{d^2}{d{u}^2}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}+(2a_1(1+r)(-1+r)-2a_0)u^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(2a_{k+1}(k+1+r)(k+r-1)-2a_k-2a_{k-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 must be 0}\\ & & 2a_1(1+r)(-1+r)-2a_0=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 2a_{k+1}(k+1+r)(k+r-1)-2a_k-2a_{k-1}=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & 2a_{k+2}(k+2+r)(k+r)-2a_{k+1}-2a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+2}=\frac{a_{k+1}+a_k}{(k+2+r)(k+r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+2}=\frac{a_{k+1}+a_k}{(k+2)k}\\ \bullet & & \text{Series not valid for}r=0\text{, division by}0\text{in the recursion relation at}k=0\\ & & a_{k+2}=\frac{a_{k+1}+a_k}{(k+2)k}\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+2}=\frac{a_{k+1}+a_k}{(k+4)(k+2)}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(u\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{u}^{k+2},a_{k+2}=\frac{a_{k+1}+a_k}{(k+4)(k+2)},6a_1-2a_0=0]\\ \bullet & & \text{Revert the change of variables}u=x+\frac{1}{2}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{(x+\frac{1}{2})}^{k+2},a_{k+2}=\frac{a_{k+1}+a_k}{(k+4)(k+2)},6a_1-2a_0=0]\end{array}$$

✓ Mathematica. Time used: 0.358 (sec). Leaf size: 69

ode=(2*x+1)*D[y[x],{x,2}]-2*D[y[x],x]-(2*x+3)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((2*x + 1)*Derivative(y(x), (x, 2)) - (2*x + 3)*y(x) - 2*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False