2.1.749 Problem 771

Internal problem ID [9921]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 771
Date solved : Wednesday, March 05, 2025 at 08:00:57 AM
CAS classification : [_Laguerre]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.395 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{4x^2+4x+21}{16x^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =4x^2+4x+21\\ t& =16x^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =\left(\frac{4x^2+4x+21}{16x^2}\right)z(x)\end{array}$$

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 2.749: 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 $t=16x^2$. There is a pole at $x=0$ 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

$$r=\frac{1}{4}+\frac{1}{4x}+\frac{21}{16x^2}$$

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 $b=\frac{21}{16}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{7}{4}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{3}{4}\end{array}$$

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

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{1}{2}+\frac{1}{4x}+\frac{5}{4x^2}-\frac{5}{8x^3}-\frac{5}{4x^4}+\frac{35}{16x^5}+\frac{105}{64x^6}-\frac{1005}{128x^7}+\dots \end{array}$$

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

$$a=\frac{1}{2}$$

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

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^0{a}_i{x}^i\\ \text{(10)}& & =\frac{1}{2}\end{array}$$

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

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{1}{4}$$

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

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{4x^2+4x+21}{16x^2}\\ & =Q+\frac{R}{16x^2}\\ & =\left(\frac{1}{4}\right)+\left(\frac{4x+21}{16x^2}\right)\\ & =\frac{1}{4}+\frac{4x+21}{16x^2}\end{array}$$

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 $16$ gives $\frac{1}{4}$. Now $b$ can be found.

$$\begin{array}{rl}b& =\left(\frac{1}{4}\right)-\left(0\right)\\ & =\frac{1}{4}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{\frac{1}{4}}{\frac{1}{2}}-0)& & =\frac{1}{4}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{\frac{1}{4}}{\frac{1}{2}}-0)& & =-\frac{1}{4}\end{array}$$

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

$$r=\frac{4x^2+4x+21}{16x^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{7}{4}$	$-\frac{3}{4}$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$0$	$\frac{1}{2}$	$\frac{1}{4}$	$-\frac{1}{4}$

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 ${\alpha }_{\mathrm{\infty }}^{+}=\frac{1}{4}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{-}\right)\\ & =\frac{1}{4}-(-\frac{3}{4})\\ & =1\end{array}$$

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

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

Substituting the above values in the above results in

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+(+)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{3}{4x}+\left(\frac{1}{2}\right)\\ & =-\frac{3}{4x}+\frac{1}{2}\\ & =-\frac{3}{4x}+\frac{1}{2}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=1$ 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

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

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{3}{4x}+\frac{1}{2})\left(1\right)+(\left(\frac{3}{4x^2}\right)+{(-\frac{3}{4x}+\frac{1}{2})}^2-\left(\frac{4x^2+4x+21}{16x^2}\right))& =0\\ \frac{-3-2a_0}{2x}=0\end{array}$$

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

$$\{a_0=-\frac{3}{2}\}$$

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

$$\begin{array}{rl}p(x)& =x-\frac{3}{2}\end{array}$$

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

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

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

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

Which simplifies to

$$y_1=\frac{{\mathrm{e}}^x(-3+2x)}{2}$$

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

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

Substituting gives

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

Therefore the solution is

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

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.041 (sec). Leaf size: 24

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

$$y=c_1\mathrm{hypergeom}(\left[2\right],\left[\frac{7}{2}\right],x)x^{5/2}-\frac{2c_2{\mathrm{e}}^x(x-\frac{3}{2})}{3}$$

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 
   Solution has integrals. Trying a special function solution free of integrals... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      <- Kummer successful 
   <- special function solution successful 
      Solution using Kummer functions still has integrals. Trying a hypergeometric solution. 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
   <- Kovacics algorithm successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & 2\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)x-(2x+3)\left(\frac{d}{dx}y\left(x\right)\right)+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{y\left(x\right)}{2x}+\frac{(2x+3)\left(\frac{d}{dx}y\left(x\right)\right)}{2x}\\ \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{(2x+3)\left(\frac{d}{dx}y\left(x\right)\right)}{2x}+\frac{y\left(x\right)}{2x}=0\\ ◻& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=-\frac{2x+3}{2x},P_3\left(x\right)=\frac{1}{2x}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=-\frac{3}{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=0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & 2\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)x+(-2x-3)\left(\frac{d}{dx}y\left(x\right)\right)+y\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}{x}^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=0..1\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & x^m\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1+m}}{a}_{k+1-m}(k+1-m+r)x^{k+r}\\ & \circ & \text{Convert}x\cdot \left(\frac{d^2}{d{x}^2}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d^2}{d{x}^2}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & x\cdot \left(\frac{d^2}{d{x}^2}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)(k+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(-5+2r)x^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(a_{k+1}(k+1+r)(2k-3+2r)-a_k(2k+2r-1))x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(-5+2r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,\frac{5}{2}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 2(k+1+r)(k+r-\frac{3}{2})a_{k+1}-2a_k(k+r-\frac{1}{2})=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{a_k(2k+2r-1)}{(k+1+r)(2k-3+2r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=\frac{a_k(2k-1)}{(k+1)(2k-3)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+1}=\frac{a_k(2k-1)}{(k+1)(2k-3)}]\\ \bullet & & \text{Recursion relation for}r=\frac{5}{2}\\ & & a_{k+1}=\frac{a_k(2k+4)}{(k+\frac{7}{2})(2k+2)}\\ \bullet & & \text{Solution for}r=\frac{5}{2}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+\frac{5}{2}},a_{k+1}=\frac{a_k(2k+4)}{(k+\frac{7}{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{5}{2}}),a_{k+1}=\frac{a_k(2k-1)}{(k+1)(2k-3)},b_{k+1}=\frac{b_k(2k+4)}{(k+\frac{7}{2})(2k+2)}]\end{array}$$

✓ Mathematica. Time used: 0.225 (sec). Leaf size: 52

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

$$y(x)\to \frac{1}{2}{e}^x(2x-3)(c_2{\int }_1^x\frac{4e^{-K[1]}K[1{]}^{3/2}}{(3-2K[1]{)}^2}dK[1]+c_1)$$

✗ Sympy

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

False