2.1.638 Problem 655

Internal problem ID [9810]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 655
Date solved : Wednesday, March 05, 2025 at 07:59:19 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}z{y}^{\prime \prime }+(2z-3)y^{\prime }+\frac{4y}{z}& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.322 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& z{y}^{\prime \prime }+(2z-3)y^{\prime }+\frac{4y}{z}& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

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

Applying the Liouville transformation on the dependent variable gives

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

Then (2) becomes

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{4z^2-12z-1}{4z^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =4z^2-12z-1\\ t& =4z^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(z)& =\left(\frac{4z^2-12z-1}{4z^2}\right)z(z)\end{array}$$

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

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

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 2.638: 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 $t=4z^2$. There is a pole at $z=0$ 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

$$r=1-\frac{1}{4z^2}-\frac{3}{z}$$

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

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

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

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx 1-\frac{3}{2z}-\frac{5}{4z^2}-\frac{15}{8z^3}-\frac{115}{32z^4}-\frac{495}{64z^5}-\frac{2285}{128z^6}-\frac{11055}{256z^7}+\dots \end{array}$$

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

$$a=1$$

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

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

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

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{4z^2-12z-1}{4z^2}\\ & =Q+\frac{R}{4z^2}\\ & =\left(1\right)+\left(\frac{-12z-1}{4z^2}\right)\\ & =1+\frac{-12z-1}{4z^2}\end{array}$$

Since the degree of $t$ is $2$, then we see that the coefficient of the term $z$ in the remainder $R$ is $-12$. Dividing this by leading coefficient in $t$ which is $4$ gives $-3$. Now $b$ can be found.

$$\begin{array}{rl}b& =(-3)-\left(0\right)\\ & =-3\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =1\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{-3}{1}-0)& & =-\frac{3}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{-3}{1}-0)& & =\frac{3}{2}\end{array}$$

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

$$r=\frac{4z^2-12z-1}{4z^2}$$

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

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$0$	$1$	$-\frac{3}{2}$	$\frac{3}{2}$

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 ${\alpha }_{\mathrm{\infty }}^{-}=\frac{3}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{+}\right)\\ & =\frac{3}{2}-\left(\frac{1}{2}\right)\\ & =1\end{array}$$

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

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

The above gives

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

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

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

Let

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

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(\frac{1}{2z}-1)\left(1\right)+((-\frac{1}{2z^2})+{(\frac{1}{2z}-1)}^2-\left(\frac{4z^2-12z-1}{4z^2}\right))& =0\\ \frac{1+2a_0}{z}=0\end{array}$$

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

$$\{a_0=-\frac{1}{2}\}$$

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

$$\begin{array}{rl}p(z)& =z-\frac{1}{2}\end{array}$$

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

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

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dz}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{2z-3}{z}dz}\\ & =z_1{e}^{-z+\frac{3\ln \left(z\right)}{2}}\\ & =z_1\left(z^{3/2}{\mathrm{e}}^{-z}\right)\end{array}$$

Which simplifies to

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

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

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

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{2z-3}{z}dz}}{{\left(y_1\right)}^2}dz\\ & =y_1\int \frac{e^{-2z+3\ln \left(z\right)}}{{\left(y_1\right)}^2}dz\\ & =y_1(-4{\mathrm{Ei}}_1(-2z)-\frac{4{\mathrm{e}}^{2z}}{-1+2z})\end{array}$$

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.008 (sec). Leaf size: 36

ode:=z*diff(diff(y(z),z),z)+(2*z-3)*diff(y(z),z)+4/z*y(z) = 0; 
dsolve(ode,y(z), singsol=all);
 

$$y\left(z\right)=2z^2({\mathrm{e}}^{-2z}{c}_2(-\frac{1}{2}+z){\mathrm{Ei}}_1(-2z)+c_1(-\frac{1}{2}+z){\mathrm{e}}^{-2z}+\frac{c_2}{2})$$

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 
<- Kovacics algorithm successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & z\left(\frac{d^2}{d{z}^2}y\left(z\right)\right)+(-3+2z)\left(\frac{d}{dz}y\left(z\right)\right)+\frac{4y\left(z\right)}{z}=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{z}^2}y\left(z\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{z}^2}y\left(z\right)=-\frac{4y\left(z\right)}{z^2}-\frac{(-3+2z)\left(\frac{d}{dz}y\left(z\right)\right)}{z}\\ \bullet & & \text{Group terms with}y\left(z\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d^2}{d{z}^2}y\left(z\right)+\frac{(-3+2z)\left(\frac{d}{dz}y\left(z\right)\right)}{z}+\frac{4y\left(z\right)}{z^2}=0\\ ◻& & \text{Check to see if}{z}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(z\right)=\frac{-3+2z}{z},P_3\left(z\right)=\frac{4}{z^2}]\\ & \circ & z\cdot P_2\left(z\right)\text{is analytic at}z=0\\ & & (z\cdot P_2\left(z\right))|{}_{z=0}=-3\\ & \circ & z^2\cdot P_3\left(z\right)\text{is analytic at}z=0\\ & & (z^2\cdot P_3\left(z\right))|{}_{z=0}=4\\ & \circ & z=0\text{is a regular singular point}\\ & & \text{Check to see if}{z}_0=0\text{is a regular singular point}\\ & & z_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & \left(\frac{d^2}{d{z}^2}y\left(z\right)\right)z^2+z(-3+2z)\left(\frac{d}{dz}y\left(z\right)\right)+4y\left(z\right)=0\\ \bullet & & \text{Assume series solution for}y\left(z\right)\\ & & y\left(z\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{z}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{z}^m\cdot \left(\frac{d}{dz}y\left(z\right)\right)\text{to series expansion for}m=1..2\\ & & z^m\cdot \left(\frac{d}{dz}y\left(z\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)z^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & z^m\cdot \left(\frac{d}{dz}y\left(z\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)z^{k+r}\\ & \circ & \text{Convert}{z}^2\cdot \left(\frac{d^2}{d{z}^2}y\left(z\right)\right)\text{to series expansion}\\ & & z^2\cdot \left(\frac{d^2}{d{z}^2}y\left(z\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)z^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_0{(-2+r)}^2{z}^r+(\stackrel{\mathrm{\infty }}{\sum _{k=1}}(a_k{(k+r-2)}^2+2a_{k-1}(k+r-1))z^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & {(-2+r)}^2=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r=2\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_k{(k+r-2)}^2+2a_{k-1}(k+r-1)=0\\ \bullet & & \text{Shift index using}k->k+1\\ & & a_{k+1}{(k+r-1)}^2+2a_k(k+r)=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=-\frac{2a_k(k+r)}{{(k+r-1)}^2}\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+1}=-\frac{2a_k(k+2)}{{(k+1)}^2}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(z\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{z}^{k+2},a_{k+1}=-\frac{2a_k(k+2)}{{(k+1)}^2}]\end{array}$$

✓ Mathematica. Time used: 0.225 (sec). Leaf size: 55

ode=z*D[y[z],{z,2}]+(2*z-3)*D[y[z],z]+4/z*y[z]==0; 
ic={}; 
DSolve[{ode,ic},y[z],z,IncludeSingularSolutions->True]
 

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

✓ Sympy. Time used: 1.741 (sec). Leaf size: 416

from sympy import * 
z = symbols("z") 
y = Function("y") 
ode = Eq(z*Derivative(y(z), (z, 2)) + (2*z - 3)*Derivative(y(z), z) + 4*y(z)/z,0) 
ics = {} 
dsolve(ode,func=y(z),ics=ics)
 

$$\text{Solution too large to show}$$