2.1.403 Problem 415

Internal problem ID [9575]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 415
Date solved : Sunday, March 30, 2025 at 02:37:51 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.362 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& 5y^{\prime \prime }-2x{y}^{\prime }+10y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =5\\ \text{(3)}& B& =-2xC& =10\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{x^2-55}{25}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =x^2-55\\ t& =25\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(\frac{x^2}{25}-\frac{11}{5})z(x)\end{array}$$

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 2.403: 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

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{x}{5}-\frac{11}{2x}-\frac{605}{8x^3}-\frac{33275}{16x^5}-\frac{9150625}{128x^7}-\frac{704598125}{256x^9}-\frac{116258690625}{1024x^{11}}-\frac{10048072546875}{2048x^{13}}+\dots \end{array}$$

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

$$a=\frac{1}{5}$$

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

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^1{a}_i{x}^i\\ \text{(10)}& & =\frac{x}{5}\end{array}$$

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

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{x^2}{25}$$

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

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2-55}{25}\\ & =Q+\frac{R}{25}\\ & =(\frac{x^2}{25}-\frac{11}{5})+\left(0\right)\\ & =\frac{x^2}{25}-\frac{11}{5}\end{array}$$

We see that the coefficient of the term $\frac{1}{x}$ in the quotient is $-\frac{11}{5}$. Now $b$ can be found.

$$\begin{array}{rl}b& =(-\frac{11}{5})-\left(0\right)\\ & =-\frac{11}{5}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{x}{5}\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{-\frac{11}{5}}{\frac{1}{5}}-1)& & =-6\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{-\frac{11}{5}}{\frac{1}{5}}-1)& & =5\end{array}$$

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

$$r=\frac{x^2}{25}-\frac{11}{5}$$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$-2$	$\frac{x}{5}$	$-6$	$5$

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 ${\alpha }_{\mathrm{\infty }}^{-}=5$, and since there are no poles then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}\\ & =5\end{array}$$

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

$$\begin{array}{rl}\omega & =(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =0+(-)\left(\frac{x}{5}\right)\\ & =-\frac{x}{5}\\ & =-\frac{x}{5}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=5$ 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

$$\begin{array}{}\text{(2A)}& p(x)& =x^5+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}(20x^3+12x^2{a}_4+6x{a}_3+2a_2)+2(-\frac{x}{5})(5x^4+4x^3{a}_4+3x^2{a}_3+2x{a}_2+a_1)+((-\frac{1}{5})+{(-\frac{x}{5})}^2-(\frac{x^2}{25}-\frac{11}{5}))& =0\\ \frac{2a_4{x}^4}{5}+\frac{4(25+a_3)x^3}{5}+\frac{6(a_2+10a_4)x^2}{5}+\frac{2(4a_1+15a_3)x}{5}+2a_0+2a_2=0\end{array}$$

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

$$\{a_0=0,a_1=\frac{375}{4},a_2=0,a_3=-25,a_4=0\}$$

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

$$\begin{array}{rl}p(x)& =x^5-25x^3+\frac{375}{4}x\end{array}$$

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

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(x^5-25x^3+\frac{375}{4}x){\mathrm{e}}^{\int -\frac{x}{5}dx}\\ & =(x^5-25x^3+\frac{375}{4}x){\mathrm{e}}^{-\frac{x^2}{10}}\\ & =\frac{(4x^5-100x^3+375x){\mathrm{e}}^{-\frac{x^2}{10}}}{4}\end{array}$$

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

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-2x}{5}dx}\\ & =z_1{e}^{\frac{x^2}{10}}\\ & =z_1\left({\mathrm{e}}^{\frac{x^2}{10}}\right)\end{array}$$

Which simplifies to

$$y_1=x^5-25x^3+\frac{375}{4}x$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-2x}{5}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{\frac{x^2}{5}}}{{\left(y_1\right)}^2}dx\\ & =y_1(\int \frac{{\mathrm{e}}^{\frac{x^2}{5}}}{{(x^5-25x^3+\frac{375}{4}x)}^2}dx)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1(x^5-25x^3+\frac{375}{4}x)+c_2(x^5-25x^3+\frac{375}{4}x(\int \frac{{\mathrm{e}}^{\frac{x^2}{5}}}{{(x^5-25x^3+\frac{375}{4}x)}^2}dx))\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.008 (sec). Leaf size: 31

ode:=5*diff(diff(y(x),x),x)-2*diff(y(x),x)*x+10*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_2\mathrm{hypergeom}([-\frac{5}{2}],\left[\frac{1}{2}\right],\frac{x^2}{5})+\frac{4(x^4-25x^2+\frac{375}{4})x{c}_1}{375}$$

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 hypergeomet\ 
ric 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}\\ & & 5\frac{d}{dx}\frac{d}{dx}y\left(x\right)-2x\left(\frac{d}{dx}y\left(x\right)\right)+10y\left(x\right)=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{2x\left(\frac{d}{dx}y\left(x\right)\right)}{5}-2y\left(x\right)\\ \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{2x\left(\frac{d}{dx}y\left(x\right)\right)}{5}+2y\left(x\right)=0\\ \bullet & & \text{Multiply by denominators}\\ & & 5\frac{d}{dx}\frac{d}{dx}y\left(x\right)-2x\left(\frac{d}{dx}y\left(x\right)\right)+10y\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\\ ◻& & \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}{dx}\frac{d}{dx}y\left(x\right)\text{to series expansion}\\ & & \frac{d}{dx}\frac{d}{dx}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}{dx}\frac{d}{dx}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}}(5a_{k+2}(k+2)(k+1)-2a_k(k-5))x^k=0\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & 5(k^2+3k+2)a_{k+2}-2a_k(k-5)=0\\ \bullet & & \text{Recursion relation that defines the series solution to the ODE}\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+2}=\frac{2a_k(k-5)}{5(k^2+3k+2)}]\end{array}$$

✓ Mathematica. Time used: 0.115 (sec). Leaf size: 138

ode=5*D[y[x],{x,2}]-2*x*D[y[x],x]+10*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to -\frac{1}{200}\sqrt{\frac{\pi }{5}}{c}_2\sqrt{x^2}(4x^4-100x^2+375)\text{erfi}\left(\frac{\sqrt{x^2}}{\sqrt{5}}\right)+\frac{32c_1{x}^5}{25\sqrt{5}}-\frac{32c_1{x}^3}{\sqrt{5}}-\frac{9}{20}{c}_2{e}^{\frac{x^2}{5}}{x}^2+c_2{e}^{\frac{x^2}{5}}+\frac{1}{50}{c}_2{e}^{\frac{x^2}{5}}{x}^4+24\sqrt{5}{c}_{1}x$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-2*x*Derivative(y(x), x) + 10*y(x) + 5*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False