2.2.20 Problem 21

Internal problem ID [9142]
Book : Second order enumerated odes
Section : section 2
Problem number : 21
Date solved : Friday, April 25, 2025 at 05:57:25 PM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

Solved as second order ode using change of variable on x method 2

Time used: 0.461 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+\cot \left(x\right)y^{\prime }+4y\csc {\left(x\right)}^2& =0\end{array}$$

In normal form the ode

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+\cot \left(x\right)y^{\prime }+4y\csc {\left(x\right)}^2& =0\end{array}$$

Becomes

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

Where

$$\begin{array}{rl}p\left(x\right)& =\cot \left(x\right)\\ q\left(x\right)& =4\csc {\left(x\right)}^2\end{array}$$

Applying change of variables $\tau =g\left(x\right)$ to (2) gives

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $p_1=0$. Eq (4) simplifies to

$$\begin{array}{rl}{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)& =0\end{array}$$

This ode is solved resulting in

$$\begin{array}{rl}\tau & =\int {\mathrm{e}}^{-\int p\left(x\right)dx}dx\\ & =\int {\mathrm{e}}^{-\int \cot \left(x\right)dx}dx\\ & =\int e^{-\ln (\sin \left(x\right))}dx\\ & =\int \csc \left(x\right)dx\\ \text{(6)}& & =-\ln (\csc \left(x\right)+\cot \left(x\right))\end{array}$$

Using (6) to evaluate $q_1$ from (5) gives

$$\begin{array}{rl}{q}_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{4\csc {\left(x\right)}^2}{\csc {\left(x\right)}^2}\\ \text{(7)}& & =4\end{array}$$

Substituting the above in (3) and noting that now $p_1=0$ results in

$$\begin{array}{rl}\frac{d^2}{d{\tau }^2}y\left(\tau \right)+q_{1}y\left(\tau \right)& =0\\ \frac{d^2}{d{\tau }^2}y\left(\tau \right)+4y\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $y\left(\tau \right)$.This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(\tau )+B{y}^{\prime }(\tau )+Cy(\tau )=0$$

Where in the above $A=1,B=0,C=4$. Let the solution be $y=e^{\lambda \tau }$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{\tau \lambda }+4{\mathrm{e}}^{\tau \lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda \tau }$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2+4=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=4$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{0^2-(4)\left(1\right)\left(4\right)}\\ & =\pm 2i\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =+2i\\ {\lambda }_2& =-2i\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =2i\\ {\lambda }_2& =-2i\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =0$ and $\beta =2$. Therefore the final solution, when using Euler relation, can be written as

$$y=e^{\alpha \tau }(c_1\cos (\beta \tau )+c_2\sin (\beta \tau ))$$

Which becomes

$$y=e^0(c_1\cos \left(2\tau \right)+c_2\sin \left(2\tau \right))$$

Or

$$y=c_1\cos \left(2\tau \right)+c_2\sin \left(2\tau \right)$$

Will add steps showing solving for IC soon.

The above solution is now transformed back to $y\left(x\right)$ using (6) which results in

$$y\left(x\right)=c_1\cos (2\ln (\csc \left(x\right)+\cot \left(x\right)))-c_2\sin (2\ln (\csc \left(x\right)+\cot \left(x\right)))$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y\left(x\right)& =c_1\cos (2\ln (\csc \left(x\right)+\cot \left(x\right)))-c_2\sin (2\ln (\csc \left(x\right)+\cot \left(x\right)))\end{array}$$

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

Time used: 0.450 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+\cot \left(x\right)y^{\prime }+4y\csc {\left(x\right)}^2& =0\end{array}$$

In normal form the ode

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+\cot \left(x\right)y^{\prime }+4y\csc {\left(x\right)}^2& =0\end{array}$$

Becomes

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

Where

$$\begin{array}{rl}p\left(x\right)& =\cot \left(x\right)\\ q\left(x\right)& =4\csc {\left(x\right)}^2\end{array}$$

Applying change of variables $\tau =g\left(x\right)$ to (2) results

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $q_1=c^2$ where $c$ is some constant. Therefore from (5)

$$\begin{array}{rl}{\tau }^{\prime }& =\frac{1}{c}\sqrt{q}\\ \text{(6)}& & =\frac{2\sqrt{\csc {\left(x\right)}^2}}{c}{\tau }^{″}& =-\frac{2\csc {\left(x\right)}^2\cot \left(x\right)}{c\sqrt{\csc {\left(x\right)}^2}}\end{array}$$

Substituting the above into (4) results in

$$\begin{array}{rl}{p}_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{-\frac{2\csc {\left(x\right)}^2\cot \left(x\right)}{c\sqrt{\csc {\left(x\right)}^2}}+\cot \left(x\right)\frac{2\sqrt{\csc {\left(x\right)}^2}}{c}}{{\left(\frac{2\sqrt{\csc {\left(x\right)}^2}}{c}\right)}^2}\\ & =0\end{array}$$

Therefore ode (3) now becomes

$$\begin{array}{rl}y{\left(\tau \right)}^{″}+p_{1}y{\left(\tau \right)}^{\prime }+q_{1}y\left(\tau \right)& =0\\ \text{(7)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+c^{2}y\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $y\left(\tau \right)$. Since the ode is now constant coefficients, it can be easily solved to give

$$\begin{array}{rl}y\left(\tau \right)& =c_1\cos \left(c\tau \right)+c_2\sin \left(c\tau \right)\end{array}$$

Now from (6)

$$\begin{array}{rl}\tau & =\int \frac{1}{c}\sqrt{q}dx\\ & =\frac{\int 2\sqrt{\csc {\left(x\right)}^2}dx}{c}\\ & =\frac{2\ln (\csc \left(x\right)-\cot \left(x\right))}{c}\end{array}$$

Substituting the above into the solution obtained gives

$$y=c_1\cos (2\ln (\csc \left(x\right)-\cot \left(x\right)))+c_2\sin (2\ln (\csc \left(x\right)-\cot \left(x\right)))$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\cos (2\ln (\csc \left(x\right)-\cot \left(x\right)))+c_2\sin (2\ln (\csc \left(x\right)-\cot \left(x\right)))\end{array}$$

✓ Maple. Time used: 0.002 (sec). Leaf size: 23

ode:=diff(diff(y(x),x),x)+cot(x)*diff(y(x),x)+4*y(x)*csc(x)^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1{(\csc \left(x\right)+\cot \left(x\right))}^{-2i}+c_2{(\csc \left(x\right)+\cot \left(x\right))}^{2i}$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
<- linear_1 successful
 

✓ Mathematica. Time used: 0.048 (sec). Leaf size: 25

ode=D[y[x],{x,2}]+Cot[x]*D[y[x],x]+4*y[x]*Csc[x]^2==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to c_1\cos (2\text{arctanh}(\cos (x)))-c_2\sin (2\text{arctanh}(\cos (x)))$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(4*y(x)/sin(x)**2 + Derivative(y(x), (x, 2)) + Derivative(y(x), x)/tan(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE 4*y(x)*tan(x)/sin(x)**2 + tan(x)*Derivative(y(x), (x, 2)) + Derivative(y(x), x) cannot be solved by the factorable group method