3.4 problem 4

Internal problem ID [12931]
Internal file name [OUTPUT/11584_Tuesday_November_07_2023_11_27_22_PM_23130949/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: 4.
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]

\[ \boxed {y^{\prime }-\sin \left (y\right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

3.4.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= \sin \left (y \right ) \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=0\) is \[ \{-\infty

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[ \{-\infty

3.4.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{\sin \left (y \right )}d y &= \int {dt}\\ \ln \left (\tan \left (\frac {y}{2}\right )\right )&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \ln \left (\tan \left (\frac {1}{2}\right )\right ) = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \ln \left (\tan \left (\frac {1}{2}\right )\right ) \end {align*}

Trying the constant \begin {align*} c_{1} = \ln \left (\tan \left (\frac {1}{2}\right )\right ) \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \ln \left (\tan \left (\frac {y}{2}\right )\right ) = t +\ln \left (\tan \left (\frac {1}{2}\right )\right ) \end {align*}

The constant \(c_{1} = \ln \left (\tan \left (\frac {1}{2}\right )\right )\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \ln \left (\tan \left (\frac {y}{2}\right )\right ) &= t +\ln \left (\tan \left (\frac {1}{2}\right )\right ) \\ \end{align*}

Verification of solutions

\[ \ln \left (\tan \left (\frac {y}{2}\right )\right ) = t +\ln \left (\tan \left (\frac {1}{2}\right )\right ) \] Verified OK.

3.4.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\sin \left (y\right )=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sin \left (y\right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sin \left (y\right )}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{\sin \left (y\right )}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (-\cot \left (y\right )+\csc \left (y\right )\right )=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\arctan \left (\frac {2 \,{\mathrm e}^{t +c_{1}}}{\left ({\mathrm e}^{t +c_{1}}\right )^{2}+1}, -\frac {\left ({\mathrm e}^{t +c_{1}}\right )^{2}-1}{\left ({\mathrm e}^{t +c_{1}}\right )^{2}+1}\right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=\arctan \left (\frac {2 \,{\mathrm e}^{c_{1}}}{\left ({\mathrm e}^{c_{1}}\right )^{2}+1}, -\frac {\left ({\mathrm e}^{c_{1}}\right )^{2}-1}{\left ({\mathrm e}^{c_{1}}\right )^{2}+1}\right ) \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \end {array} \]

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: 1.594 (sec). Leaf size: 63

dsolve([diff(y(t),t)=sin(y(t)),y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \arctan \left (-\frac {2 \,{\mathrm e}^{t} \sin \left (1\right )}{\left (-1+\cos \left (1\right )\right ) {\mathrm e}^{2 t}-\cos \left (1\right )-1}, \frac {\left (1-\cos \left (1\right )\right ) {\mathrm e}^{2 t}-\cos \left (1\right )-1}{\left (-1+\cos \left (1\right )\right ) {\mathrm e}^{2 t}-\cos \left (1\right )-1}\right ) \]

✓ Solution by Mathematica

Time used: 0.008 (sec). Leaf size: 16

DSolve[{y'[t]==Sin[y[t]],{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \arccos (-\tanh (t-\text {arctanh}(\cos (1)))) \]