2.20 problem 21

Internal problem ID [8109]
Book : Second order enumerated odes
Section : section 2
Problem number : 21
Date solved : Monday, October 21, 2024 at 04:52:34 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*}

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

Time used: 0.461 (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}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \cot \left (x \right )d x \right )}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.

2.20.2 Maple step by step solution

2.20.3 Maple trace

Methods for second order ODEs:
 

2.20.4 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)
 

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

2.20.5 Mathematica DSolve solution

Solving time : 0.071 (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]
 

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