2.6.14 Problem 10 (c)

Internal problem ID [19749]
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 33. Problems at page 91
Problem number : 10 (c)
Date solved : Saturday, January 31, 2026 at 01:35:28 AM
CAS classification : [_separable]

2.6.14.1 Solved using first_order_ode_separable

0.388 (sec)

Entering first order ode separable solver

\begin{align*} u \ln \left (u \right ) v^{\prime }+\sin \left (v\right )^{2}&=1 \\ \end{align*}

The ode

\begin{equation} v^{\prime } = -\frac {\left (\sin \left (v\right )-1\right ) \left (\sin \left (v\right )+1\right )}{u \ln \left (u \right )} \end{equation}

is separable as it can be written as

\begin{align*} v^{\prime }&= -\frac {\left (\sin \left (v\right )-1\right ) \left (\sin \left (v\right )+1\right )}{u \ln \left (u \right )}\\ &= f(u) g(v) \end{align*}

Where

\begin{align*} f(u) &= -\frac {1}{u \ln \left (u \right )}\\ g(v) &= \left (\sin \left (v \right )-1\right ) \left (\sin \left (v \right )+1\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(v)} \,dv} &= \int { f(u) \,du} \\ \int { \frac {1}{\left (\sin \left (v \right )-1\right ) \left (\sin \left (v \right )+1\right )}\,dv} &= \int { -\frac {1}{u \ln \left (u \right )} \,du} \\ \end{align*}

\[ \tan \left (v\right )=\ln \left (\ln \left (u \right )\right )-c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(v)\) is zero, since we had to divide by this above. Solving \(g(v)=0\) or

\[ \left (\sin \left (v \right )-1\right ) \left (\sin \left (v \right )+1\right )=0 \]

for \(v\) gives

\begin{align*} v&=-\frac {\pi }{2}\\ v&=\frac {\pi }{2} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \tan \left (v\right ) &= \ln \left (\ln \left (u \right )\right )-c_1 \\ v &= -\frac {\pi }{2} \\ v &= \frac {\pi }{2} \\ \end{align*}

Solving for \(v\) gives

\begin{align*} v &= -\frac {\pi }{2} \\ v &= \frac {\pi }{2} \\ v &= -\arctan \left (-\ln \left (\ln \left (u \right )\right )+c_1 \right ) \\ \end{align*}

Direction field \(u \ln \left (u \right ) v^{\prime }+\sin \left (v\right )^{2} = 1\)	Isoclines for \(u \ln \left (u \right ) v^{\prime }+\sin \left (v\right )^{2} = 1\)

Summary of solutions found

\begin{align*} v &= -\frac {\pi }{2} \\ v &= \frac {\pi }{2} \\ v &= -\arctan \left (-\ln \left (\ln \left (u \right )\right )+c_1 \right ) \\ \end{align*}

2.6.14.2 Solved using first_order_ode_isobaric

0.325 (sec)

Entering first order ode isobaric solver

\begin{align*} u \ln \left (u \right ) v^{\prime }+\sin \left (v\right )^{2}&=1 \\ \end{align*}

Solving for \(v'\) gives

\[ v' = -\frac {\sin \left (v\right )^{2}-1}{u \ln \left (u \right )} \]

An ode \(v^{\prime }=f(u,v)\) is isobaric if

\[ f(t u, t^m v) = t^{m-1} f(u,v)\tag {1} \]

Where here

\[ f(u,v) = -\frac {\sin \left (v\right )^{2}-1}{u \ln \left (u \right )}\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = 0 \]

Since the ode is isobaric of order \(m=0\), then the substitution

\begin{align*} v&=u u^m \\ &=u \end{align*}

Converts the ODE to a separable in \(u \left (u \right )\). Performing this substitution gives

\[ u^{\prime }\left (u \right ) = -\frac {\sin \left (u \left (u \right )\right )^{2}-1}{u \ln \left (u \right )} \]

Entering first order ode separable solverThe ode

\begin{equation} u^{\prime }\left (u \right ) = -\frac {\left (\sin \left (u \left (u \right )\right )-1\right ) \left (\sin \left (u \left (u \right )\right )+1\right )}{u \ln \left (u \right )} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (u \right )&= -\frac {\left (\sin \left (u \left (u \right )\right )-1\right ) \left (\sin \left (u \left (u \right )\right )+1\right )}{u \ln \left (u \right )}\\ &= f(u) g(u) \end{align*}

Where

\begin{align*} f(u) &= -\frac {1}{u \ln \left (u \right )}\\ g(u) &= \left (\sin \left (u \right )-1\right ) \left (\sin \left (u \right )+1\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(u) \,du} \\ \int { \frac {1}{\left (\sin \left (u \right )-1\right ) \left (\sin \left (u \right )+1\right )}\,du} &= \int { -\frac {1}{u \ln \left (u \right )} \,du} \\ \end{align*}

\[ \tan \left (u \left (u \right )\right )=\ln \left (\ln \left (u \right )\right )-c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \left (\sin \left (u \right )-1\right ) \left (\sin \left (u \right )+1\right )=0 \]

for \(u \left (u \right )\) gives

\begin{align*} u \left (u \right )&=-\frac {\pi }{2}\\ u \left (u \right )&=\frac {\pi }{2} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \tan \left (u \left (u \right )\right ) &= \ln \left (\ln \left (u \right )\right )-c_1 \\ u \left (u \right ) &= -\frac {\pi }{2} \\ u \left (u \right ) &= \frac {\pi }{2} \\ \end{align*}

Solving for \(u \left (u \right )\) gives

\begin{align*} u \left (u \right ) &= -\frac {\pi }{2} \\ u \left (u \right ) &= \frac {\pi }{2} \\ u \left (u \right ) &= -\arctan \left (-\ln \left (\ln \left (u \right )\right )+c_1 \right ) \\ \end{align*}

Converting \(u \left (u \right ) = -\frac {\pi }{2}\) back to \(v\) gives

\begin{align*} v = -\frac {\pi }{2} \end{align*}

Converting \(u \left (u \right ) = \frac {\pi }{2}\) back to \(v\) gives

\begin{align*} v = \frac {\pi }{2} \end{align*}

Converting \(u \left (u \right ) = -\arctan \left (-\ln \left (\ln \left (u \right )\right )+c_1 \right )\) back to \(v\) gives

\begin{align*} v = -\arctan \left (-\ln \left (\ln \left (u \right )\right )+c_1 \right ) \end{align*}

Direction field \(u \ln \left (u \right ) v^{\prime }+\sin \left (v\right )^{2} = 1\)	Isoclines for \(u \ln \left (u \right ) v^{\prime }+\sin \left (v\right )^{2} = 1\)

Summary of solutions found

\begin{align*} v &= -\frac {\pi }{2} \\ v &= \frac {\pi }{2} \\ v &= -\arctan \left (-\ln \left (\ln \left (u \right )\right )+c_1 \right ) \\ \end{align*}

2.6.14.3 ✓ Maple. Time used: 0.004 (sec). Leaf size: 10

ode:=u*ln(u)*diff(v(u),u)+sin(v(u))^2 = 1; 
dsolve(ode,v(u), singsol=all);
 

\[ v = \arctan \left (\ln \left (\ln \left (u \right )\right )+c_1 \right ) \]

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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & u \ln \left (u \right ) \left (\frac {d}{d u}v \left (u \right )\right )+\sin \left (v \left (u \right )\right )^{2}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d u}v \left (u \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d u}v \left (u \right )=\frac {-\sin \left (v \left (u \right )\right )^{2}+1}{u \ln \left (u \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d u}v \left (u \right )}{-\sin \left (v \left (u \right )\right )^{2}+1}=\frac {1}{u \ln \left (u \right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} u \\ {} & {} & \int \frac {\frac {d}{d u}v \left (u \right )}{-\sin \left (v \left (u \right )\right )^{2}+1}d u =\int \frac {1}{u \ln \left (u \right )}d u +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \tan \left (v \left (u \right )\right )=\ln \left (\ln \left (u \right )\right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} v \left (u \right ) \\ {} & {} & v \left (u \right )=\arctan \left (\ln \left (\ln \left (u \right )\right )+\mathit {C1} \right ) \end {array} \]

2.6.14.4 ✓ Mathematica. Time used: 0.288 (sec). Leaf size: 52

ode=u*Log[u]*D[v[u],u]+Sin[v[u]]==0; 
ic={}; 
DSolve[{ode,ic},v[u],u,IncludeSingularSolutions->True]
 

\begin{align*} v(u)&\to -\arccos (-\tanh (-\log (\log (u))+c_1))\\ v(u)&\to \arccos (-\tanh (-\log (\log (u))+c_1))\\ v(u)&\to 0\\ v(u)&\to -\pi \\ v(u)&\to \pi \end{align*}

2.6.14.5 ✓ Sympy. Time used: 1.250 (sec). Leaf size: 76

from sympy import * 
u = symbols("u") 
v = Function("v") 
ode = Eq(u*log(u)*Derivative(v(u), u) + sin(v(u))**2 - 1,0) 
ics = {} 
dsolve(ode,func=v(u),ics=ics)
 

\[ \left [ v{\left (u \right )} = 2 \operatorname {atan}{\left (\frac {\sqrt {C_{1}^{2} + 2 C_{1} \log {\left (\log {\left (u \right )} \right )} + \log {\left (\log {\left (u \right )} \right )}^{2} + 1} - 1}{C_{1} + \log {\left (\log {\left (u \right )} \right )}} \right )}, \ v{\left (u \right )} = - 2 \operatorname {atan}{\left (\frac {\sqrt {C_{1}^{2} + 2 C_{1} \log {\left (\log {\left (u \right )} \right )} + \log {\left (\log {\left (u \right )} \right )}^{2} + 1} + 1}{C_{1} + \log {\left (\log {\left (u \right )} \right )}} \right )}\right ] \]

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=v(u)) 
 
('factorable', 'separable', 'lie_group', 'separable_Integral')