Internal problem ID [12940]
Internal file name [OUTPUT/11593_Tuesday_November_07_2023_11_27_36_PM_36577844/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Exercises section 1.4 page 61
Problem number: 17.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {\theta ^{\prime }+\frac {11 \cos \left (\theta \right )}{10}={\frac {9}{10}}} \] With initial conditions \begin {align*} [\theta \left (0\right ) = 1] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} \theta ^{\prime } &= f(t,\theta )\\ &= \frac {9}{10}-\frac {11 \cos \left (\theta \right )}{10} \end {align*}
The \(\theta \) domain of \(f(t,\theta )\) when \(t=0\) is \[ \{-\infty <\theta <\infty \} \] And the point \(\theta _0 = 1\) is inside this domain. Now we will look at the continuity of \begin {align*} \frac {\partial f}{\partial \theta } &= \frac {\partial }{\partial \theta }\left (\frac {9}{10}-\frac {11 \cos \left (\theta \right )}{10}\right ) \\ &= \frac {11 \sin \left (\theta \right )}{10} \end {align*}
The \(\theta \) domain of \(\frac {\partial f}{\partial \theta }\) when \(t=0\) is \[ \{-\infty <\theta <\infty \} \] And the point \(\theta _0 = 1\) is inside this domain. Therefore solution exists and is unique.
Integrating both sides gives \begin {align*} \int \frac {1}{\frac {9}{10}-\frac {11 \cos \left (\theta \right )}{10}}d \theta &= \int {dt}\\ -\sqrt {10}\, \operatorname {arctanh}\left (\tan \left (\frac {\theta }{2}\right ) \sqrt {10}\right )&= t +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(\theta =1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\sqrt {10}\, \operatorname {arccoth}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )+\frac {i \pi \sqrt {10}}{2} = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -\sqrt {10}\, \operatorname {arccoth}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )+\frac {i \pi \sqrt {10}}{2} \end {align*}
Trying the constant \begin {align*} c_{1} = -\sqrt {10}\, \operatorname {arccoth}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )+\frac {i \pi \sqrt {10}}{2} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\sqrt {10}\, \operatorname {arctanh}\left (\tan \left (\frac {\theta }{2}\right ) \sqrt {10}\right ) = t -\sqrt {10}\, \operatorname {arccoth}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )+\frac {i \pi \sqrt {10}}{2} \end {align*}
The constant \(c_{1} = -\sqrt {10}\, \operatorname {arccoth}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )+\frac {i \pi \sqrt {10}}{2}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\sqrt {10}\, \operatorname {arctanh}\left (\tan \left (\frac {\theta }{2}\right ) \sqrt {10}\right ) &= t -\sqrt {10}\, \operatorname {arccoth}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )+\frac {i \pi \sqrt {10}}{2} \\ \end{align*}
Verification of solutions
\[ -\sqrt {10}\, \operatorname {arctanh}\left (\tan \left (\frac {\theta }{2}\right ) \sqrt {10}\right ) = t -\sqrt {10}\, \operatorname {arccoth}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )+\frac {i \pi \sqrt {10}}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\theta ^{\prime }+\frac {11 \cos \left (\theta \right )}{10}=\frac {9}{10}, \theta \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \theta ^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \theta ^{\prime }=\frac {9}{10}-\frac {11 \cos \left (\theta \right )}{10} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\theta ^{\prime }}{\frac {9}{10}-\frac {11 \cos \left (\theta \right )}{10}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {\theta ^{\prime }}{\frac {9}{10}-\frac {11 \cos \left (\theta \right )}{10}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\sqrt {10}\, \mathrm {arctanh}\left (\tan \left (\frac {\theta }{2}\right ) \sqrt {10}\right )=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \theta \\ {} & {} & \theta =-2 \arctan \left (\frac {\tanh \left (\frac {\left (t +c_{1} \right ) \sqrt {10}}{10}\right ) \sqrt {10}}{10}\right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} \theta \left (0\right )=1 \\ {} & {} & 1=-2 \arctan \left (\frac {\tanh \left (\frac {c_{1} \sqrt {10}}{10}\right ) \sqrt {10}}{10}\right ) \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\sqrt {10}\, \mathrm {arctanh}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\sqrt {10}\, \mathrm {arctanh}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & \theta =-2 \arctan \left (\frac {\tanh \left (\frac {\sqrt {10}\, t}{10}-\mathrm {arctanh}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )\right ) \sqrt {10}}{10}\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & \theta =-2 \arctan \left (\frac {\tanh \left (\frac {\sqrt {10}\, t}{10}-\mathrm {arctanh}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )\right ) \sqrt {10}}{10}\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`
✓ Solution by Maple
Time used: 0.344 (sec). Leaf size: 29
dsolve([diff(theta(t),t)=1-cos( theta(t)) + (1+cos(theta(t)))*(-1/10),theta(0) = 1],theta(t), singsol=all)
\[ \theta \left (t \right ) = -2 \arctan \left (\frac {\tanh \left (-\operatorname {arctanh}\left (\tan \left (\frac {1}{2}\right ) \sqrt {10}\right )+\frac {\sqrt {10}\, t}{10}\right ) \sqrt {10}}{10}\right ) \]
✓ Solution by Mathematica
Time used: 0.061 (sec). Leaf size: 36
DSolve[{theta'[t]==1-Cos[ theta[t]] + (1+Cos[theta[t]])*(-1/10),{theta[0]==1}},theta[t],t,IncludeSingularSolutions -> True]
\[ \theta (t)\to -2 \arctan \left (\frac {\tanh \left (\frac {t}{\sqrt {10}}-\text {arctanh}\left (\sqrt {10} \tan \left (\frac {1}{2}\right )\right )\right )}{\sqrt {10}}\right ) \]