2.1.262 Problem 265

Internal problem ID [9432]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 265
Date solved : Friday, April 25, 2025 at 06:08:10 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

$$\begin{array}{rl}{x}^2(1-4x)y^{\prime \prime }-\frac{x{y}^{\prime }}{2}-\frac{3xy}{4}& =0\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.407 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{-48x^2-20x+5}{16{(4x^2-x)}^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-48x^2-20x+5\\ t& =16{(4x^2-x)}^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{-48x^2-20x+5}{16{(4x^2-x)}^2}\right)z(x)\end{array}$$

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 2.262: 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 $t=16{(4x^2-x)}^2$. There is a pole at $x=0$ of order $2$. There is a pole at $x=\frac{1}{4}$ 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 $n=1$.

Unable to find solution using case one

Attempting to find a solution using case $n=2$.

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

$$r=-\frac{3}{16{(x-\frac{1}{4})}^2}-\frac{5}{4(x-\frac{1}{4})}+\frac{5}{4x}+\frac{5}{16x^2}$$

For the pole at $x=0$ let $b$ be the coefficient of $\frac{1}{x^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=\frac{5}{16}$. Hence

$$\begin{array}{rl}{E}_c& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{-1,2,5\}\end{array}$$

For the pole at $x=\frac{1}{4}$ let $b$ be the coefficient of $\frac{1}{{(x-\frac{1}{4})}^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=-\frac{3}{16}$. Hence

$$\begin{array}{rl}{E}_c& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{1,2,3\}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is 2 then let $b$ be the coefficient of $\frac{1}{x^2}$ 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

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{-48x^2-20x+5}{16{(4x^2-x)}^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=-\frac{3}{16}$. Hence

$$\begin{array}{rl}{E}_{\mathrm{\infty }}& =\{2,2+2\sqrt{1+4b},2-2\sqrt{1+4b}\}\\ & =\{1,2,3\}\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ for case 2 of Kovacic algorithm.

pole $c$ location	pole order	$E_c$
$0$	$2$	$\{-1,2,5\}$
$\frac{1}{4}$	$2$	$\{1,2,3\}$

Order of $r$ at $\mathrm{\infty }$	$E_{\mathrm{\infty }}$
$2$	$\{1,2,3\}$

Using the family $\{e_1,e_2,\dots ,e_{\mathrm{\infty }}\}$ given by

$$e_1=2,e_2=1,e_{\mathrm{\infty }}=3$$

Gives a non negative integer $d$ (the degree of the polynomial $p(x)$), which is generated using

$$\begin{array}{rl}d& =\frac{1}{2}(e_{\mathrm{\infty }}-\sum _{c\in \mathrm{\Gamma }}{e}_c)\\ & =\frac{1}{2}(3-(2+\left(1\right)))\\ & =0\end{array}$$

We now form the following rational function

$$\begin{array}{rl}\theta & =\frac{1}{2}\sum _{c\in \mathrm{\Gamma }}\frac{e_c}{x-c}\\ & =\frac{1}{2}(\frac{2}{(x-\left(0\right))}+\frac{1}{(x-\left(\frac{1}{4}\right))})\\ & =\frac{1}{x}+\frac{1}{2x-\frac{1}{2}}\end{array}$$

Now we search for a monic polynomial $p(x)$ of degree $d=0$ such that

$$\begin{array}{}\text{(1A)}& p^{‴}+3\theta p^{″}+(3{\theta }^2+3{\theta }^{\prime }-4r)p^{\prime }+({\theta }^{″}+3\theta {\theta }^{\prime }+{\theta }^3-4r\theta -2r^{\prime })p=0\end{array}$$

Since $d=0$, then letting

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

Substituting $p$ and $\theta$ into Eq. (1A) gives

$$0=0$$

And solving for $p$ gives

$$p=1$$

Now that $p(x)$ is found let

$$\begin{array}{rl}\varphi & =\theta +\frac{p^{\prime }}{p}\\ & =\frac{1}{x}+\frac{1}{2x-\frac{1}{2}}\end{array}$$

Let $\omega$ be the solution of

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

Substituting the values for $\varphi$ and $r$ into the above equation gives

$$w^2-(\frac{1}{x}+\frac{1}{2x-\frac{1}{2}})w+\frac{144x^2-12x-5}{16x^2{(-1+4x)}^2}=0$$

Solving for $\omega$ gives

$$\begin{array}{rl}\omega & =\frac{12x-2+3\sqrt{1-4x}}{4x(-1+4x)}\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =e^{\int \omega dx}\\ & ={\mathrm{e}}^{\int \frac{12x-2+3\sqrt{1-4x}}{4x(-1+4x)}dx}\\ & =\frac{{(-1+4x)}^{1/4}\sqrt{x}\sqrt{2}{\left(\frac{\sqrt{1-4x}+1}{\sqrt{x}}\right)}^{3/2}}{4}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-\frac{x}{2}}{-4x^3+x^2}dx}\\ & =z_1{e}^{-\frac{\ln (-1+4x)}{4}+\frac{\ln \left(x\right)}{4}}\\ & =z_1\left(\frac{x^{1/4}}{{(-1+4x)}^{1/4}}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{x^{1/4}\sqrt{2}(\sqrt{1-4x}+1)\sqrt{\frac{\sqrt{1-4x}+1}{\sqrt{x}}}}{4}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-\frac{x}{2}}{-4x^3+x^2}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-\frac{\ln (-1+4x)}{2}+\frac{\ln \left(x\right)}{2}}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(\frac{2{\mathrm{e}}^{-\frac{\ln (-1+4x)}{2}+\frac{\ln (1-4x)}{2}}{(-{(\sqrt{1-4x}+1)}^2+2\sqrt{1-4x}+2)}^{3/2}}{3{(\sqrt{1-4x}+1)}^3}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{x^{1/4}\sqrt{2}(\sqrt{1-4x}+1)\sqrt{\frac{\sqrt{1-4x}+1}{\sqrt{x}}}}{4}\right)+c_2\left(\frac{x^{1/4}\sqrt{2}(\sqrt{1-4x}+1)\sqrt{\frac{\sqrt{1-4x}+1}{\sqrt{x}}}}{4}\left(\frac{2{\mathrm{e}}^{-\frac{\ln (-1+4x)}{2}+\frac{\ln (1-4x)}{2}}{(-{(\sqrt{1-4x}+1)}^2+2\sqrt{1-4x}+2)}^{3/2}}{3{(\sqrt{1-4x}+1)}^3}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.016 (sec). Leaf size: 46

ode:=x^2*(1-4*x)*diff(diff(y(x),x),x)-1/2*x*diff(y(x),x)-3/4*x*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=-\frac{\sqrt{2}(c_1(x-1)\sqrt{1-4x}-2c_2{x}^{3/2}+c_1(3x-1))}{{(1+\sqrt{1-4x})}^{3/2}}$$

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 
   Group is reducible or imprimitive 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^2(1-4x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)-\frac{x\left(\frac{d}{dx}y\left(x\right)\right)}{2}-\frac{3xy\left(x\right)}{4}=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=-\frac{3y\left(x\right)}{4x(-1+4x)}-\frac{\frac{d}{dx}y\left(x\right)}{2x(-1+4x)}\\ \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}{dx}\frac{d}{dx}y\left(x\right)+\frac{\frac{d}{dx}y\left(x\right)}{2x(-1+4x)}+\frac{3y\left(x\right)}{4x(-1+4x)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{1}{2x(-1+4x)},P_3\left(x\right)=\frac{3}{4x(-1+4x)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=-\frac{1}{2}\\ & \circ & x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & (x^2\cdot P_3\left(x\right))|{}_{x=0}=0\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & 4x(-1+4x)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+2\frac{d}{dx}y\left(x\right)+3y\left(x\right)=0\\ \bullet & & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}\frac{d}{dx}y\left(x\right)\text{to series expansion}\\ & & \frac{d}{dx}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & \frac{d}{dx}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)x^{k+r}\\ & \circ & \text{Convert}{x}^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=1..2\\ & & x^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & x^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & -2a_{0}r(-3+2r)x^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(-2a_{k+1}(k+1+r)(2k-1+2r)+a_k(4k+4r-1)(4k+4r-3))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -2r(-3+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,\frac{3}{2}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & -4(k+1+r)(k-\frac{1}{2}+r)a_{k+1}+16(k+r-\frac{3}{4})(k+r-\frac{1}{4})a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{(4k+4r-3)(4k+4r-1)a_k}{2(k+1+r)(2k-1+2r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=\frac{(4k-3)(4k-1)a_k}{2(k+1)(2k-1)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+1}=\frac{(4k-3)(4k-1)a_k}{2(k+1)(2k-1)}]\\ \bullet & & \text{Recursion relation for}r=\frac{3}{2}\\ & & a_{k+1}=\frac{(4k+3)(4k+5)a_k}{2(k+\frac{5}{2})(2k+2)}\\ \bullet & & \text{Solution for}r=\frac{3}{2}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{3}{2}},a_{k+1}=\frac{(4k+3)(4k+5)a_k}{2(k+\frac{5}{2})(2k+2)}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k+\frac{3}{2}}),a_{k+1}=\frac{(4k-3)(4k-1)a_k}{2(k+1)(2k-1)},b_{k+1}=\frac{(4k+3)(4k+5)b_k}{2(k+\frac{5}{2})(2k+2)}]\end{array}$$

✓ Mathematica. Time used: 4.075 (sec). Leaf size: 125

ode=x^2*(1-4*x)*D[y[x],{x,2}]+((1-(3/2))*x-(6-4*(3/2))*x^2)*D[y[x],x]+(3/2)*(1-(3/2))*x*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{\sqrt[4]{4x-1}(6c_1{(\sqrt{4x-1}-i)}^{3/2}+i{c}_2{(\sqrt{4x-1}+i)}^{3/2})\exp (-\frac{1}{2}{\int }_1^x-\frac{1}{2K[1]-8K[1{]}^2}dK[1])}{6\sqrt[4]{\sqrt{4x-1}-i}\sqrt[4]{\sqrt{4x-1}+i}}$$

✓ Sympy. Time used: 0.433 (sec). Leaf size: 60

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(1 - 4*x)*Derivative(y(x), (x, 2)) - 3*x*y(x)/4 - x*Derivative(y(x), x)/2,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)={(4x-1)}^{\frac{3}{4}}(C_1{\left(\frac{x}{4x-1}\right)}^{\frac{3}{2}}{}_2{F}_1\left(\begin{array}{c}\frac{3}{4},\frac{5}{4}\\ \frac{5}{2}\end{array}\right|\frac{4x}{4x-1})+C_2{}_2{F}_1\left(\begin{array}{c}-\frac{3}{4},-\frac{1}{4}\\ -\frac{1}{2}\end{array}\right|\frac{4x}{4x-1}))$$