1.6 problem 5.4 (i)

Internal problem ID [11972]
Internal file name [OUTPUT/10625_Saturday_September_02_2023_02_48_37_PM_85505114/index.tex]

Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section: Chapter 5, Trivial differential equations. Exercises page 33
Problem number: 5.4 (i).
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]

\[ \boxed {x^{\prime }=\sec \left (t \right )^{2}} \] With initial conditions \begin {align*} \left [x \left (\frac {\pi }{4}\right ) = 0\right ] \end {align*}

1.6.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime } + p(t)x &= q(t) \end {align*}

Where here \begin {align*} p(t) &=0\\ q(t) &=\sec \left (t \right )^{2} \end {align*}

Hence the ode is \begin {align*} x^{\prime } = \sec \left (t \right )^{2} \end {align*}

The domain of \(p(t)=0\) is \[ \{-\infty

1.6.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} x &= \int { \sec \left (t \right )^{2}\,\mathop {\mathrm {d}t}}\\ &= \tan \left (t \right )+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=\frac {\pi }{4}\) and \(x=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = 1+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -1 \end {align*}

Trying the constant \begin {align*} c_{1} = -1 \end {align*}

Substituting this in the general solution gives \begin {align*} x&=\tan \left (t \right )-1 \end {align*}

The constant \(c_{1} = -1\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= \tan \left (t \right )-1 \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ x = \tan \left (t \right )-1 \] Verified OK.

1.6.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }=\sec \left (t \right )^{2}, x \left (\frac {\pi }{4}\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int x^{\prime }d t =\int \sec \left (t \right )^{2}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & x=\tan \left (t \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\tan \left (t \right )+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (\frac {\pi }{4}\right )=0 \\ {} & {} & 0=1+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\tan \left (t \right )-1 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\tan \left (t \right )-1 \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 8

dsolve([diff(x(t),t)=sec(t)^2,x(1/4*Pi) = 0],x(t), singsol=all)
 

\[ x \left (t \right ) = \tan \left (t \right )-1 \]

✓ Solution by Mathematica

Time used: 0.016 (sec). Leaf size: 9

DSolve[{x'[t]==Sec[t]^2,{x[Pi/4]==0}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to \tan (t)-1 \]