2.5.3 Problem 3

Internal problem ID [8963]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 3
Date solved : Friday, April 25, 2025 at 05:27:00 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order solved by an integrating factor

Time used: 0.070 (sec)

Solve

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

The ode satisfies this form

$$y^{\prime \prime }+p\left(x\right)y^{\prime }+\frac{(p^{\prime }\left(x\right)+p{\left(x\right)}^2)y}{2}=f\left(x\right)$$

Where $p(x)=2\cot \left(x\right)$. Therefore, there is an integrating factor given by

$$\begin{array}{rl}M(x)& =e^{\frac{1}{2}\int pdx}\\ & =e^{\int 2\cot \left(x\right)dx}\\ & =\sin \left(x\right)\end{array}$$

Multiplying both sides of the ODE by the integrating factor $M(x)$ makes the left side of the ODE a complete differential

$$\begin{array}{rl}{(M(x)y)}^{″}& =0\\ {(\sin \left(x\right)y)}^{″}& =0\end{array}$$

Integrating once gives

$${(\sin \left(x\right)y)}^{\prime }=c_1$$

Integrating again gives

$$(\sin \left(x\right)y)=c_{1}x+c_2$$

Hence the solution is

$$\begin{array}{rl}y& =\frac{c_{1}x+c_2}{\sin \left(x\right)}\end{array}$$

Or

$$y=\frac{c_{1}x}{\sin \left(x\right)}+\frac{c_2}{\sin \left(x\right)}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{c_{1}x}{\sin \left(x\right)}+\frac{c_2}{\sin \left(x\right)}\end{array}$$

Solved as second order ode using change of variable on y method 1

Time used: 0.156 (sec)

Solve

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

In normal form the given ode is written as

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =2\cot \left(x\right)\\ q\left(x\right)& =-1\end{array}$$

Calculating the Liouville ode invariant $Q$ given by

$$\begin{array}{rl}Q& =q-\frac{p^{\prime }}{2}-\frac{p^2}{4}\\ & =-1-\frac{{(2\cot \left(x\right))}^{\prime }}{2}-\frac{{(2\cot \left(x\right))}^2}{4}\\ & =-1-\frac{(-2-2\cot {\left(x\right)}^2)}{2}-\frac{(4\cot {\left(x\right)}^2)}{4}\\ & =-1-(-1-\cot {\left(x\right)}^2)-\cot {\left(x\right)}^2\\ & =0\end{array}$$

Since the Liouville ode invariant does not depend on the independent variable $x$ then the transformation

$$\begin{array}{}\text{(3)}& y=v\left(x\right)z\left(x\right)\end{array}$$

is used to change the original ode to a constant coefficients ode in $v$. In (3) the term $z\left(x\right)$ is given by

$$\begin{array}{rl}z\left(x\right)& ={\mathrm{e}}^{-\int \frac{p\left(x\right)}{2}dx}\\ & =e^{-\int \frac{2\cot \left(x\right)}{2}}\\ \text{(5)}& & =\csc \left(x\right)\end{array}$$

Hence (3) becomes

$$\begin{array}{}\text{(4)}& y=v\left(x\right)\csc \left(x\right)\end{array}$$

Applying this change of variable to the original ode results in

$$\begin{array}{r}{v}^{\prime \prime }\left(x\right)\csc \left(x\right)=0\end{array}$$

Which is now solved for $v\left(x\right)$.

The above ode can be simplified to

$$\begin{array}{r}{v}^{\prime \prime }\left(x\right)=0\end{array}$$

Integrating twice gives the solution

$$v=c_{1}x+c_2$$

Will add steps showing solving for IC soon.

Now that $v$ is known, then

$$\begin{array}{rl}y\left(x\right)& =vz\left(x\right)\\ \text{(7)}& & =(c_{1}x+c_2)\left(z\left(x\right)\right)\end{array}$$

But from (5)

$$\begin{array}{rl}z\left(x\right)& =\csc \left(x\right)\end{array}$$

Hence (7) becomes

$$\begin{array}{r}y\left(x\right)=(c_{1}x+c_2)\csc \left(x\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y\left(x\right)& =(c_{1}x+c_2)\csc \left(x\right)\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.080 (sec)

Solve

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

Writing the ode as

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

Comparing (1) and (2) shows that

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

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

$$\begin{array}{}\text{(6)}& r& =\frac{0}{1}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =0\\ t& =1\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =0\end{array}$$

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 2.116: Necessary conditions for each Kovacic case

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

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

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

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

Since $r=0$ 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

$$z_1(x)=1$$

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

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

Which simplifies to

$$y_1=\csc \left(x\right)$$

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

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

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1(\csc \left(x\right))+c_2(\csc \left(x\right)\left(x\right))\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\csc \left(x\right)+c_{2}x\csc \left(x\right)\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 12

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

$$y=\csc \left(x\right)(c_{2}x+c_1)$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
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) 
<- Kovacics algorithm successful
 

✓ Mathematica. Time used: 0.033 (sec). Leaf size: 15

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

$$y(x)\to (c_{2}x+c_1)\csc (x)$$

✗ Sympy

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

NotImplementedError : The given ODE -(y(x) - Derivative(y(x), (x, 2)))*tan(x)/2 + Derivative(y(x), x) cannot be solved by the factorable group method