2.1.742 Problem 762

Internal problem ID [9914]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 762
Date solved : Wednesday, March 05, 2025 at 08:00:51 AM
CAS classification : [_Hermite]

Solve

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

Solved as second order ode using Kovacic algorithm

Time used: 0.229 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{x^2-10}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =x^2-10\\ t& =4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(\frac{x^2}{4}-\frac{5}{2})z(x)\end{array}$$

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 2.742: Necessary conditions for each Kovacic case

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

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

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. 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. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Attempting to find a solution using case $n=1$.

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=-2$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{2}{2}& & =1\end{array}$$

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

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{x}^i\\ \text{(8)}& & =\sum _{i=0}^1{a}_i{x}^i\end{array}$$

Let $a$ be the coefficient of $x^v=x^1$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{x}{2}-\frac{5}{2x}-\frac{25}{4x^3}-\frac{125}{4x^5}-\frac{3125}{16x^7}-\frac{21875}{16x^9}-\frac{328125}{32x^{11}}-\frac{2578125}{32x^{13}}+\dots \end{array}$$

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

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

From Eq. (9) the sum up to $v=1$ gives

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

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^0=1$ 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

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

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

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

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $1$ in $r$ will be the coefficient this term in the quotient. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2-10}{4}\\ & =Q+\frac{R}{4}\\ & =(\frac{x^2}{4}-\frac{5}{2})+\left(0\right)\\ & =\frac{x^2}{4}-\frac{5}{2}\end{array}$$

We see that the coefficient of the term $\frac{1}{x}$ in the quotient is $-\frac{5}{2}$. Now $b$ can be found.

$$\begin{array}{rl}b& =(-\frac{5}{2})-\left(0\right)\\ & =-\frac{5}{2}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{x}{2}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{-\frac{5}{2}}{\frac{1}{2}}-1)& & =-3\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{-\frac{5}{2}}{\frac{1}{2}}-1)& & =2\end{array}$$

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

$$r=\frac{x^2}{4}-\frac{5}{2}$$

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

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 ${\alpha }_{\mathrm{\infty }}^{-}=2$, and since there are no poles then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}\\ & =2\end{array}$$

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

$$\begin{array}{rl}\omega & =(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =0+(-)\left(\frac{x}{2}\right)\\ & =-\frac{x}{2}\\ & =-\frac{x}{2}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=2$ 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

$$\begin{array}{}\text{(2A)}& p(x)& =x^2+a_{1}x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(2\right)+2(-\frac{x}{2})(2x+a_1)+((-\frac{1}{2})+{(-\frac{x}{2})}^2-(\frac{x^2}{4}-\frac{5}{2}))& =0\\ a_{1}x+2a_0+2=0\end{array}$$

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

$$\{a_0=-1,a_1=0\}$$

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

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

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

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

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

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

Which simplifies to

$$y_1=x^2-1$$

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

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

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1(x^2-1)+c_2(x^2-1(\int \frac{{\mathrm{e}}^{\frac{x^2}{2}}}{{(x^2-1)}^2}dx))\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.030 (sec). Leaf size: 42

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

$$y=2c_1{\mathrm{e}}^{\frac{x^2}{2}}x-(x-1)(x+1)(c_1\sqrt{\pi }\mathrm{erfi}\left(\frac{\sqrt{2}x}{2}\right)\sqrt{2}-c_2)$$

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 
   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 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals 
   <- Kovacics algorithm successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)-x\left(\frac{d}{dx}y\left(x\right)\right)+2y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k\\ ◻& & \text{Rewrite DE with series expansions}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_{k}k{x}^k\\ & \circ & \text{Convert}\frac{d^2}{d{x}^2}y\left(x\right)\text{to series expansion}\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=2}}{a}_{k}k(k-1)x^{k-2}\\ & \circ & \text{Shift index using}k->k+2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_{k+2}(k+2)(k+1)x^k\\ & & \text{Rewrite DE with series expansions}\\ & & \stackrel{\mathrm{\infty }}{\sum _{k=0}}(a_{k+2}(k+2)(k+1)-a_k(k-2))x^k=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k^2+3k+2)a_{k+2}-a_k(k-2)=0\\ \bullet & & \text{Recursion relation; series terminates at}k=2\\ & & a_{k+2}=\frac{a_k(k-2)}{k^2+3k+2}\\ \bullet & & \text{Apply recursion relation for}k=0\\ & & a_2=-a_0\\ \bullet & & \text{Terminating series solution of the ODE. Use reduction of order to find the second linearly independent solution}\\ & & y\left(x\right)=A_2{x}^2+A_{1}x-a_0\end{array}$$

✓ Mathematica. Time used: 0.207 (sec). Leaf size: 43

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

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

✗ Sympy

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

False