2.3.4 Problem 4

Internal problem ID [19719]
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 : Wednesday, January 28, 2026 at 10:56:21 AM
CAS classification : [_separable]

2.3.4.1 Solved using first_order_ode_separable

1.230 (sec)

Entering first order ode separable solver

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

The ode

\begin{equation} y^{\prime } = -\frac {\sec \left (y\right )^{2} \tan \left (x \right )}{\sec \left (x \right )^{2} \tan \left (y\right )} \end{equation}

is separable as it can be written as

\begin{align*} 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{align*}

Where

\begin{align*} 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{align*}

Integrating gives

\begin{align*} \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{align*}

\[ -\frac {\cos \left (y\right )^{2}}{2}=\frac {\cos \left (x \right )^{2}}{2}+c_1 \]

Solving for \(y\) gives

\begin{align*} y &= \pi -\arccos \left (\sqrt {-\cos \left (x \right )^{2}-2 c_1}\right ) \\ y &= \arccos \left (\sqrt {-\cos \left (x \right )^{2}-2 c_1}\right ) \\ \end{align*}

Direction field \(\sec \left (x \right )^{2} \tan \left (y\right ) y^{\prime }+\sec \left (y\right )^{2} \tan \left (x \right ) = 0\)	Isoclines for \(\sec \left (x \right )^{2} \tan \left (y\right ) y^{\prime }+\sec \left (y\right )^{2} \tan \left (x \right ) = 0\)

Summary of solutions found

\begin{align*} y &= \pi -\arccos \left (\sqrt {-\cos \left (x \right )^{2}-2 c_1}\right ) \\ y &= \arccos \left (\sqrt {-\cos \left (x \right )^{2}-2 c_1}\right ) \\ \end{align*}

2.3.4.2 ✓ 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);
 

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

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} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \sec \left (x \right )^{2} \tan \left (y \left (x \right )\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\sec \left (y \left (x \right )\right )^{2} \tan \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}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 & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d x}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 & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\left (\frac {d}{d x}y \left (x \right )\right ) \tan \left (y \left (x \right )\right )}{\sec \left (y \left (x \right )\right )^{2}}d x =\int -\frac {\tan \left (x \right )}{\sec \left (x \right )^{2}}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 \sec \left (y \left (x \right )\right )^{2}}=\frac {1}{2 \sec \left (x \right )^{2}}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \left \{y \left (x \right )=\pi -\mathrm {arcsec}\left (\frac {1}{\sqrt {-\cos \left (x \right )^{2}-2 \mathit {C1}}}\right ), y \left (x \right )=\mathrm {arcsec}\left (\frac {1}{\sqrt {-\cos \left (x \right )^{2}-2 \mathit {C1}}}\right )\right \} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=\pi -\mathrm {arcsec}\left (\frac {1}{\sqrt {-\cos \left (x \right )^{2}+\mathit {C1}}}\right ), y \left (x \right )=\mathrm {arcsec}\left (\frac {1}{\sqrt {-\cos \left (x \right )^{2}+\mathit {C1}}}\right )\right \} \end {array} \]

2.3.4.3 ✓ Mathematica. Time used: 0.488 (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{align*} y(x)&\to -\frac {1}{2} \arccos (-\cos (2 x)-2 c_1)\\ y(x)&\to \frac {1}{2} \arccos (-\cos (2 x)-2 c_1) \end{align*}

2.3.4.4 ✓ Sympy. Time used: 2.198 (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)
 

\[ \left [ y{\left (x \right )} = \pi - \frac {\operatorname {acos}{\left (C_{1} - \cos {\left (2 x \right )} \right )}}{2}, \ y{\left (x \right )} = \frac {\operatorname {acos}{\left (C_{1} - \cos {\left (2 x \right )} \right )}}{2}\right ] \]

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', 'separable', '1st_power_series', 'lie_group', 'separable_Integral')