2.3.4 Problem 4

Internal problem ID [18477]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter IV. Methods of solution: First order equations. section 29. Problems at page 81
Problem number : 4
Date solved : Monday, March 31, 2025 at 05:35:20 PM
CAS classification : [_separable]

Solved using first_order_ode_separable

Time used: 0.159 (sec)

Solve

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

The ode

$$\begin{array}{}\text{(1)}& y^{\prime }=-\frac{\sec {\left(y\right)}^2\tan \left(x\right)}{\sec {\left(x\right)}^2\tan \left(y\right)}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{y}^{\prime }& =-\frac{\sec {\left(y\right)}^2\tan \left(x\right)}{\sec {\left(x\right)}^2\tan \left(y\right)}\\ & =f(x)g(y)\end{array}$$

Where

$$\begin{array}{rl}f(x)& =-\frac{\tan \left(x\right)}{\sec {\left(x\right)}^2}\\ g(y)& =\frac{\sec {\left(y\right)}^2}{\tan \left(y\right)}\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(y)}dy& =\int f(x)dx\\ \int \frac{\tan \left(y\right)}{\sec {\left(y\right)}^2}dy& =\int -\frac{\tan \left(x\right)}{\sec {\left(x\right)}^2}dx\end{array}$$

$$-\frac{1}{2\sec {\left(y\right)}^2}=\frac{1}{2\sec {\left(x\right)}^2}+c_1$$

Solving for $y$ gives

$$\begin{array}{rl}y& =\pi -\mathrm{arcsec}\left(\frac{1}{\sqrt{-\cos {\left(x\right)}^2-2c_1}}\right)\\ y& =\mathrm{arcsec}\left(\frac{1}{\sqrt{-\cos {\left(x\right)}^2-2c_1}}\right)\end{array}$$

Figure 2.30: Slope field $\sec {\left(x\right)}^2\tan \left(y\right)y^{\prime }+\sec {\left(y\right)}^2\tan \left(x\right)=0$

Summary of solutions found

$$\begin{array}{rl}y& =\pi -\mathrm{arcsec}\left(\frac{1}{\sqrt{-\cos {\left(x\right)}^2-2c_1}}\right)\\ y& =\mathrm{arcsec}\left(\frac{1}{\sqrt{-\cos {\left(x\right)}^2-2c_1}}\right)\end{array}$$

✓ Maple. Time used: 0.003 (sec). Leaf size: 17

ode:=sec(x)^2*tan(y(x))*diff(y(x),x)+sec(y(x))^2*tan(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{\arccos (-\cos \left(2x\right)+4c_1)}{2}$$

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \sec {\left(x\right)}^2\tan \left(y\left(x\right)\right)\left(\frac{d}{dx}y\left(x\right)\right)+\sec {\left(y\left(x\right)\right)}^2\tan \left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=-\frac{\sec {\left(y\left(x\right)\right)}^2\tan \left(x\right)}{\sec {\left(x\right)}^2\tan \left(y\left(x\right)\right)}\\ \bullet & & \text{Separate variables}\\ & & \frac{\left(\frac{d}{dx}y\left(x\right)\right)\tan \left(y\left(x\right)\right)}{\sec {\left(y\left(x\right)\right)}^2}=-\frac{\tan \left(x\right)}{\sec {\left(x\right)}^2}\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\left(\frac{d}{dx}y\left(x\right)\right)\tan \left(y\left(x\right)\right)}{\sec {\left(y\left(x\right)\right)}^2}dx=\int -\frac{\tan \left(x\right)}{\sec {\left(x\right)}^2}dx+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & -\frac{1}{2\sec {\left(y\left(x\right)\right)}^2}=\frac{1}{2\sec {\left(x\right)}^2}+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(x\right)\\ & & \{y\left(x\right)=\pi -\mathrm{arcsec}\left(\frac{1}{\sqrt{-\cos {\left(x\right)}^2-2\mathit{C}\mathit{1}}}\right),y\left(x\right)=\mathrm{arcsec}\left(\frac{1}{\sqrt{-\cos {\left(x\right)}^2-2\mathit{C}\mathit{1}}}\right)\}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & \{y\left(x\right)=\pi -\mathrm{arcsec}\left(\frac{1}{\sqrt{-\cos {\left(x\right)}^2+\mathit{C}\mathit{1}}}\right),y\left(x\right)=\mathrm{arcsec}\left(\frac{1}{\sqrt{-\cos {\left(x\right)}^2+\mathit{C}\mathit{1}}}\right)\}\end{array}$$

✓ Mathematica. Time used: 0.504 (sec). Leaf size: 41

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

$$\begin{array}{r}y(x)\to -\frac{1}{2}\arccos (-\cos (2x)-2c_1)\\ y(x)\to \frac{1}{2}\arccos (-\cos (2x)-2c_1)\end{array}$$

✓ Sympy. Time used: 0.635 (sec). Leaf size: 26

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

$$[y\left(x\right)=\pi -\frac{\mathrm{acos}(C_1-\cos \left(2x\right))}{2},y\left(x\right)=\frac{\mathrm{acos}(C_1-\cos \left(2x\right))}{2}]$$