2.2.20 Problem 21
Internal
problem
ID
[9143]
Book
:
Second
order
enumerated
odes
Section
:
section
2
Problem
number
:
21
Date
solved
:
Friday, February 21, 2025 at 09:20:52 PM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
Solve
\begin{align*} y^{\prime \prime }+\cot \left (x \right ) y^{\prime }+4 y \csc \left (x \right )^{2}&=0 \end{align*}
Solved as second order ode using change of variable on x method 2
Time used: 0.496 (sec)
In normal form the ode
\begin{align*} y^{\prime \prime }+\cot \left (x \right ) y^{\prime }+4 y \csc \left (x \right )^{2}&=0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=\cot \left (x \right )\\ q \left (x \right )&=4 \csc \left (x \right )^{2} \end{align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) gives
\begin{align*} \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 \tag {3} \end{align*}
Where \(\tau \) is the new independent variable, and
\begin{align*} 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}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}
Let \(p_{1} = 0\). Eq (4) simplifies to
\begin{align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end{align*}
This ode is solved resulting in
\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int \cot \left (x \right )d x}d x\\ &= \int e^{-\ln \left (\sin \left (x \right )\right )} \,dx\\ &= \int \csc \left (x \right )d x\\ &= -\ln \left (\csc \left (x \right )+\cot \left (x \right )\right )\tag {6} \end{align*}
Using (6) to evaluate \(q_{1}\) from (5) gives
\begin{align*} 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}}\\ &= 4\tag {7} \end{align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in
\begin{align*} \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 )+4 y \left (\tau \right )&=0 \end{align*}
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
\[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \]
Where in the above \(A=1, B=0, C=4\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives
\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }+4 \,{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]
Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives
\[ \lambda ^{2}+4 = 0 \tag {2} \]
Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic
formula
\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]
Substituting \(A=1, B=0, C=4\) into the above gives
\begin{align*} \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 2 i \end{align*}
Hence
\begin{align*} \lambda _1 &= + 2 i\\ \lambda _2 &= - 2 i \end{align*}
Which simplifies to
\begin{align*}
\lambda _1 &= 2 i \\
\lambda _2 &= -2 i \\
\end{align*}
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
\[
y \left (\tau \right ) = e^{\alpha \tau } \left ( c_1 \cos (\beta \tau ) + c_2 \sin (\beta \tau ) \right )
\]
Which becomes
\[
y \left (\tau \right ) = e^{0}\left (c_1 \cos \left (2 \tau \right )+c_2 \sin \left (2 \tau \right )\right )
\]
Or
\[
y \left (\tau \right ) = c_1 \cos \left (2 \tau \right )+c_2 \sin \left (2 \tau \right )
\]
Will add steps showing solving for IC soon.
The above solution is now transformed back to \(y\) using (6) which results in
\[
y = c_1 \cos \left (2 \ln \left (\csc \left (x \right )+\cot \left (x \right )\right )\right )-c_2 \sin \left (2 \ln \left (\csc \left (x \right )+\cot \left (x \right )\right )\right )
\]
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 \cos \left (2 \ln \left (\csc \left (x \right )+\cot \left (x \right )\right )\right )-c_2 \sin \left (2 \ln \left (\csc \left (x \right )+\cot \left (x \right )\right )\right ) \\
\end{align*}
Solved as second order ode using change of variable on x method 1
Time used: 0.362 (sec)
In normal form the ode
\begin{align*} y^{\prime \prime }+\cot \left (x \right ) y^{\prime }+4 y \csc \left (x \right )^{2}&=0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=\cot \left (x \right )\\ q \left (x \right )&=4 \csc \left (x \right )^{2} \end{align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) results
\begin{align*} \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 \tag {3} \end{align*}
Where \(\tau \) is the new independent variable, and
\begin{align*} 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}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end{align*}
Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5)
\begin{align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {2 \sqrt {\csc \left (x \right )^{2}}}{c}\tag {6} \\ \tau '' &= -\frac {2 \csc \left (x \right )^{2} \cot \left (x \right )}{c \sqrt {\csc \left (x \right )^{2}}} \end{align*}
Substituting the above into (4) results in
\begin{align*} 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{align*}
Therefore ode (3) now becomes
\begin{align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end{align*}
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{align*} y \left (\tau \right ) &= c_1 \cos \left (c \tau \right )+c_2 \sin \left (c \tau \right ) \end{align*}
Now from (6)
\begin{align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int 2 \sqrt {\csc \left (x \right )^{2}}d x}{c}\\ &= \frac {2 \ln \left (\csc \left (x \right )-\cot \left (x \right )\right )}{c} \end{align*}
Substituting the above into the solution obtained gives
\[
y = c_1 \cos \left (2 \ln \left (\csc \left (x \right )-\cot \left (x \right )\right )\right )+c_2 \sin \left (2 \ln \left (\csc \left (x \right )-\cot \left (x \right )\right )\right )
\]
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 \cos \left (2 \ln \left (\csc \left (x \right )-\cot \left (x \right )\right )\right )+c_2 \sin \left (2 \ln \left (\csc \left (x \right )-\cot \left (x \right )\right )\right ) \\
\end{align*}
Maple step by step solution
Maple trace
`Methods for second order ODEs:
--- Trying classification methods ---
trying a symmetry of the form [xi=0, eta=F(x)]
<- linear_1 successful`
Maple dsolve solution
Solving time : 0.003 (sec)
Leaf size : 23
dsolve(diff(diff(y(x),x),x)+cot(x)*diff(y(x),x)+4*y(x)*csc(x)^2 = 0,y(x),singsol=all)
\[
y = c_{1} \left (\csc \left (x \right )+\cot \left (x \right )\right )^{-2 i}+c_{2} \left (\csc \left (x \right )+\cot \left (x \right )\right )^{2 i}
\]
✓Mathematica DSolve solution
Solving time : 0.048 (sec)
Leaf size : 25
DSolve[{D[y[x],{x,2}]+Cot[x]*D[y[x],x]+4*y[x]*Csc[x]^2==0,{}},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 solution
Solving time : 0.000 (sec)
Leaf size : 0
Python version: 3.13.1 (main, Dec 4 2024, 18:05:56) [GCC 14.2.1 20240910]
Sympy version 1.13.3
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