2.1.788 Problem 810

Internal problem ID [9960]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 810
Date solved : Sunday, March 30, 2025 at 02:50:26 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.111 (sec)

Writing the ode as

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

Comparing (1) and (2) shows that

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

Applying the Liouville transformation on the dependent variable gives

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{1}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =1\\ t& =4\end{array}$$

Therefore eq. (4) becomes

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

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

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

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 2.788: Necessary conditions for each Kovacic case

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

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

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 $0$ then the necessary conditions for case one are met. Therefore

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

Since $r=\frac{1}{4}$ is not a function of $x$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(x)={\mathrm{e}}^{-\frac{x}{2}}$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $u$ is found from

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

Which simplifies to

$$u_1={\mathrm{e}}^{\frac{x^2}{2}}$$

The second solution $u_2$ to the original ode is found using reduction of order

$$u_2=u_1\int \frac{e^{\int -\frac{B}{A}dx}}{u_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{u}_2& =u_1\int \frac{e^{\int -\frac{-2x-1}{1}dx}}{{\left(u_1\right)}^2}dx\\ & =u_1\int \frac{e^{x^2+x}}{{\left(u_1\right)}^2}dx\\ & =u_1\left({\mathrm{e}}^{x^2+x}{\mathrm{e}}^{-x^2}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}u& =c_1{u}_1+c_2{u}_2\\ & =c_1\left({\mathrm{e}}^{\frac{x^2}{2}}\right)+c_2\left({\mathrm{e}}^{\frac{x^2}{2}}\left({\mathrm{e}}^{x^2+x}{\mathrm{e}}^{-x^2}\right)\right)\end{array}$$

Will add steps showing solving for IC soon.

✓ Maple. Time used: 0.003 (sec). Leaf size: 17

ode:=diff(diff(u(x),x),x)-(2*x+1)*diff(u(x),x)+(x^2+x-1)*u(x) = 0; 
dsolve(ode,u(x), singsol=all);
 

$$u={\mathrm{e}}^{\frac{x^2}{2}}(c_1+c_2{\mathrm{e}}^x)$$

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) 
   Reducible group (found another exponential solution) 
<- Kovacics algorithm successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}u\left(x\right)-(2x+1)\left(\frac{d}{dx}u\left(x\right)\right)+(x^2+x-1)u\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}u\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}u\left(x\right)=(-x^2-x+1)u\left(x\right)+(2x+1)\left(\frac{d}{dx}u\left(x\right)\right)\\ \bullet & & \text{Group terms with}u\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}u\left(x\right)+(-2x-1)\left(\frac{d}{dx}u\left(x\right)\right)+(x^2+x-1)u\left(x\right)=0\\ \bullet & & \text{Assume series solution for}u\left(x\right)\\ & & u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{x}^m\cdot u\left(x\right)\text{to series expansion for}m=0..2\\ & & x^m\cdot u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)}}{a}_k{x}^{k+m}\\ & \circ & \text{Shift index using}k->k-m\\ & & x^m\cdot u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,-m)+m}}{a}_{k-m}{x}^k\\ & \circ & \text{Convert}{x}^m\cdot \left(\frac{d}{dx}u\left(x\right)\right)\text{to series expansion for}m=0..1\\ & & x^m\cdot \left(\frac{d}{dx}u\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,1-m)}}{a}_{k}k{x}^{k-1+m}\\ & \circ & \text{Shift index using}k->k+1-m\\ & & x^m\cdot \left(\frac{d}{dx}u\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=max(0,1-m)+m-1}}{a}_{k+1-m}(k+1-m)x^k\\ & \circ & \text{Convert}\frac{d}{dx}\frac{d}{dx}u\left(x\right)\text{to series expansion}\\ & & \frac{d}{dx}\frac{d}{dx}u\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}u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_{k+2}(k+2)(k+1)x^k\\ & & \text{Rewrite ODE with series expansions}\\ & & 2a_2-a_1-a_0+(6a_3-2a_2-3a_1+a_0)x+(\stackrel{\mathrm{\infty }}{\sum _{k=2}}(a_{k+2}(k+2)(k+1)-a_{k+1}(k+1)-a_k(2k+1)+a_{k-1}+a_{k-2})x^k)=0\\ \bullet & & \text{The coefficients of each power of}x\text{must be 0}\\ & & [2a_2-a_1-a_0=0,6a_3-2a_2-3a_1+a_0=0]\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & \{a_2=\frac{a_1}{2}+\frac{a_0}{2},a_3=\frac{2a_1}{3}\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & k^2{a}_{k+2}+(-2a_k-a_{k+1}+3a_{k+2})k-a_k+a_{k-2}+a_{k-1}-a_{k+1}+2a_{k+2}=0\\ \bullet & & \text{Shift index using}k->k+2\\ & & {(k+2)}^2{a}_{k+4}+(-2a_{k+2}-a_{k+3}+3a_{k+4})(k+2)-a_{k+2}+a_k+a_{k+1}-a_{k+3}+2a_{k+4}=0\\ \bullet & & \text{Recursion relation that defines the series solution to the ODE}\\ & & [u\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+4}=\frac{2k{a}_{k+2}+k{a}_{k+3}-a_k-a_{k+1}+5a_{k+2}+3a_{k+3}}{k^2+7k+12},a_2=\frac{a_1}{2}+\frac{a_0}{2},a_3=\frac{2a_1}{3}]\end{array}$$

✓ Mathematica. Time used: 0.026 (sec). Leaf size: 24

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

$$u(x)\to e^{\frac{x^2}{2}}(c_2{e}^x+c_1)$$

✗ Sympy

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

False