2.4.72 Problem 69

Internal problem ID [10235]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 69
Date solved : Monday, January 26, 2026 at 09:24:58 PM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

2.4.72.1 second order change of variable on x method 2

1.066 (sec)

\begin{align*} y^{\prime \prime } \sin \left (2 x \right )^{2}+y^{\prime } \sin \left (4 x \right )-4 y&=0 \\ \end{align*}

Entering second order change of variable on \(x\) method 2 solverIn normal form the ode

\begin{align*} y^{\prime \prime } \sin \left (2 x \right )^{2}+y^{\prime } \sin \left (4 x \right )-4 y = 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 )&=2 \cot \left (2 x \right )\\ q \left (x \right )&=-4 \csc \left (2 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 2 \cot \left (2 x \right )d x}d x\\ &= \int e^{\frac {\ln \left (\csc \left (2 x \right )^{2}\right )}{2}} \,dx\\ &= \int \sqrt {\csc \left (2 x \right )^{2}}d x\\ &= \frac {\ln \left (\csc \left (2 x \right )-\cot \left (2 x \right )\right ) \sin \left (2 x \right ) \sqrt {\csc \left (2 x \right )^{2}}}{2}\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 (2 x \right )^{2}}{\csc \left (2 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 \(y \left (\tau \right )\).Entering second order linear constant coefficient ode solver

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 the roots are

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

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 \end{align*}

Hence

\begin{align*} \lambda _1 &= + 2 \\ \lambda _2 &= - 2 \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= 2 \\ \lambda _2 &= -2 \\ \end{align*}

Since the roots are distinct, the solution is

\begin{align*} y \left (\tau \right ) &= c_1 e^{\lambda _1 \tau } + c_2 e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_1 e^{\left (2\right )\tau } +c_2 e^{\left (-2\right )\tau } \\ \end{align*}

Or

\[ y \left (\tau \right ) =c_1 \,{\mathrm e}^{2 \tau }+c_2 \,{\mathrm e}^{-2 \tau } \]

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

\[ y = c_1 \,{\mathrm e}^{\ln \left (\csc \left (2 x \right )-\cot \left (2 x \right )\right ) \sin \left (2 x \right ) \sqrt {\csc \left (2 x \right )^{2}}}+c_2 \,{\mathrm e}^{-\ln \left (\csc \left (2 x \right )-\cot \left (2 x \right )\right ) \sin \left (2 x \right ) \sqrt {\csc \left (2 x \right )^{2}}} \]

Summary of solutions found

\begin{align*} y &= c_1 \,{\mathrm e}^{\ln \left (\csc \left (2 x \right )-\cot \left (2 x \right )\right ) \sin \left (2 x \right ) \sqrt {\csc \left (2 x \right )^{2}}}+c_2 \,{\mathrm e}^{-\ln \left (\csc \left (2 x \right )-\cot \left (2 x \right )\right ) \sin \left (2 x \right ) \sqrt {\csc \left (2 x \right )^{2}}} \\ \end{align*}

2.4.72.2 ✓ Maple. Time used: 0.102 (sec). Leaf size: 17

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

\[ y = c_1 \csc \left (2 x \right )+c_2 \cot \left (2 x \right ) \]

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 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \ 
@ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x),\ 
 dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
   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)] 
   <- linear_1 successful 
   Change of variables used: 
      [x = 1/4*arccos(t)] 
   Linear ODE actually solved: 
      -u(t)+(3*t^2-2*t-1)*diff(u(t),t)+(2*t^3-2*t^2-2*t+2)*diff(diff(u(t),t),t)\ 
 = 0 
<- change of variables successful
 

2.4.72.3 ✓ Mathematica. Time used: 0.032 (sec). Leaf size: 29

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

\begin{align*} y(x)&\to \frac {c_1-i c_2 \cos (2 x)}{\sqrt {\sin ^2(2 x)}} \end{align*}

2.4.72.4 ✗ Sympy

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

NotImplementedError : The given ODE -(4*y(x) - sin(2*x)**2*Derivative(y(x), (x, 2)))/sin(4*x) + Deri
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('factorable',)