5.36 problem 37 (v)

5.36.1 Solving as quadrature ode
5.36.2 Maple step by step solution

Internal problem ID [12985]
Internal file name [OUTPUT/11638_Tuesday_November_07_2023_11_53_52_PM_77068426/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 1.6 page 89
Problem number: 37 (v).
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]

\[ \boxed {y^{\prime }-\cos \left (\frac {\pi y}{2}\right )=0} \]

5.36.1 Solving as quadrature ode

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

Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {2 \arctan \left (\frac {{\mathrm e}^{\pi c_{1} +\pi t}-1}{{\mathrm e}^{\pi c_{1} +\pi t}+1}, \frac {2 \,{\mathrm e}^{\frac {1}{2} \pi c_{1} +\frac {1}{2} \pi t}}{{\mathrm e}^{\pi c_{1} +\pi t}+1}\right )}{\pi }\\ &=\frac {2 \arctan \left (\frac {c_{1}^{2} {\mathrm e}^{\pi t}-1}{c_{1}^{2} {\mathrm e}^{\pi t}+1}, \frac {2 c_{1} {\mathrm e}^{\frac {\pi t}{2}}}{c_{1}^{2} {\mathrm e}^{\pi t}+1}\right )}{\pi } \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 \arctan \left (\frac {c_{1}^{2} {\mathrm e}^{\pi t}-1}{c_{1}^{2} {\mathrm e}^{\pi t}+1}, \frac {2 c_{1} {\mathrm e}^{\frac {\pi t}{2}}}{c_{1}^{2} {\mathrm e}^{\pi t}+1}\right )}{\pi } \\ \end{align*}

Figure 106: Slope field plot

Verification of solutions

\[ y = \frac {2 \arctan \left (\frac {c_{1}^{2} {\mathrm e}^{\pi t}-1}{c_{1}^{2} {\mathrm e}^{\pi t}+1}, \frac {2 c_{1} {\mathrm e}^{\frac {\pi t}{2}}}{c_{1}^{2} {\mathrm e}^{\pi t}+1}\right )}{\pi } \] Verified OK.

5.36.2 Maple step by step solution

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

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: 0.031 (sec). Leaf size: 48

dsolve(diff(y(t),t)=cos(Pi/2*y(t)),y(t), singsol=all)
 

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

Solution by Mathematica

Time used: 0.846 (sec). Leaf size: 31

DSolve[y'[t]==Cos[Pi/2*y[t]],y[t],t,IncludeSingularSolutions -> True]
 

\begin{align*} y(t)\to \frac {2 \arcsin \left (\coth \left (\frac {1}{2} \pi (t+c_1)\right )\right )}{\pi } \\ y(t)\to -1 \\ y(t)\to 1 \\ \end{align*}