2.6.14 Problem 10 (c)

Internal problem ID [18578]
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 : Tuesday, January 28, 2025 at 12:00:55 PM
CAS classification : [_separable]

Solve

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

Solved as first order separable ode

Time used: 0.183 (sec)

The ode

\begin{equation} v^{\prime } = -\frac {\sin \left (v\right )^{2}-1}{\ln \left (u \right ) u} \end{equation}

is separable as it can be written as

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

Where

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

Integrating gives

\begin{align*} \int { \frac {1}{g(v)} \,dv} &= \int { f(u) \,du} \\ \int { \frac {1}{-\sin \left (v \right )^{2}+1}\,dv} &= \int { \frac {1}{\ln \left (u \right ) u} \,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

\[ -\sin \left (v \right )^{2}+1=0 \]

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*}

Figure 2.80: Slope field plot
\(u \ln \left (u \right ) v^{\prime }+\sin \left (v\right )^{2} = 1\)

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*}

Maple step by step solution

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

Maple trace

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

Maple dsolve solution

Solving time : 0.004 (sec)
Leaf size : 10

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

\[ v = \arctan \left (\ln \left (\ln \left (u \right )\right )+c_{1} \right ) \]

Mathematica DSolve solution

Solving time : 0.359 (sec)
Leaf size : 52

DSolve[{u*Log[u]*D[v[u],u]+Sin[v[u]]==0,{}},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*}