2.2.29 Problem 29

Internal problem ID [9151]
Book : Second order enumerated odes
Section : section 2
Problem number : 29
Date solved : Friday, April 25, 2025 at 05:58:10 PM
CAS classification : [_Lienard]

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

Time used: 0.350 (sec)

Solve

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

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

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

Calculating the Liouville ode invariant $Q$ given by

$$\begin{array}{rl}Q& =q-\frac{p^{\prime }}{2}-\frac{p^2}{4}\\ & =1-\frac{{(-\frac{\sin \left(2x\right)}{\cos {\left(x\right)}^2})}^{\prime }}{2}-\frac{{(-\frac{\sin \left(2x\right)}{\cos {\left(x\right)}^2})}^2}{4}\\ & =1-\frac{(-\frac{2\cos \left(2x\right)}{\cos {\left(x\right)}^2}-\frac{2\sin \left(2x\right)\sin \left(x\right)}{\cos {\left(x\right)}^3})}{2}-\frac{\left(\frac{\sin {\left(2x\right)}^2}{\cos {\left(x\right)}^4}\right)}{4}\\ & =1-(-\frac{\cos \left(2x\right)}{\cos {\left(x\right)}^2}-\frac{\sin \left(2x\right)\sin \left(x\right)}{\cos {\left(x\right)}^3})-\frac{\sin {\left(2x\right)}^2}{4\cos {\left(x\right)}^4}\\ & =2\end{array}$$

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

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

Hence (3) becomes

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

Applying this change of variable to the original ode results in

$$\begin{array}{r}\cos \left(x\right)(2v\left(x\right)+v^{\prime \prime }\left(x\right))=0\end{array}$$

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

The above ode can be simplified to

$$\begin{array}{r}2v\left(x\right)+v^{\prime \prime }\left(x\right)=0\end{array}$$

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{v}^{″}(x)+B{v}^{\prime }(x)+Cv(x)=0$$

Where in the above $A=1,B=0,C=2$. Let the solution be $v=e^{\lambda x}$. Substituting this into the ODE gives

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

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

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

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 $A=1,B=0,C=2$ into the above gives

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

Hence

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

Which simplifies to

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

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

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

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

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

Which becomes

$$v=e^0(c_1\cos \left(\sqrt{2}x\right)+c_2\sin \left(\sqrt{2}x\right))$$

Or

$$v=c_1\cos \left(\sqrt{2}x\right)+c_2\sin \left(\sqrt{2}x\right)$$

Will add steps showing solving for IC soon.

Now that $v$ is known, then

$$\begin{array}{rl}y\left(x\right)& =vz\left(x\right)\\ \text{(7)}& & =(c_1\cos \left(\sqrt{2}x\right)+c_2\sin \left(\sqrt{2}x\right))\left(z\left(x\right)\right)\end{array}$$

But from (5)

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

Hence (7) becomes

$$\begin{array}{r}y\left(x\right)=(c_1\cos \left(\sqrt{2}x\right)+c_2\sin \left(\sqrt{2}x\right))\sec \left(x\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y\left(x\right)& =(c_1\cos \left(\sqrt{2}x\right)+c_2\sin \left(\sqrt{2}x\right))\sec \left(x\right)\end{array}$$

✓ Maple. Time used: 0.003 (sec). Leaf size: 24

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

$$y=\sec \left(x\right)(c_1\sin \left(\sqrt{2}x\right)+c_2\cos \left(\sqrt{2}x\right))$$

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 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Group is reducible or imprimitive 
<- Kovacics algorithm successful
 

✓ Mathematica. Time used: 0.062 (sec). Leaf size: 51

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

$$y(x)\to \frac{1}{4}{e}^{-i\sqrt{2}x}(4c_1-i\sqrt{2}{c}_2{e}^{2i\sqrt{2}x})\sec (x)$$

✗ Sympy

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

False