1.28 problem 3(d)

Internal problem ID [6132]
Internal file name [OUTPUT/5380_Sunday_June_05_2022_03_35_47_PM_40340673/index.tex]

Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.2 THE NATURE OF SOLUTIONS. Page 9
Problem number: 3(d).
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]

\[ \boxed {\left (x^{2}-1\right ) y^{\prime }=1} \] With initial conditions \begin {align*} [y \left (2\right ) = 0] \end {align*}

1.28.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(x) &=0\\ q(x) &=\frac {1}{x^{2}-1} \end {align*}

Hence the ode is \begin {align*} y^{\prime } = \frac {1}{x^{2}-1} \end {align*}

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

1.28.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { \frac {1}{x^{2}-1}\,\mathop {\mathrm {d}x}}\\ &= -\operatorname {arctanh}\left (x \right )+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=2\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = -\operatorname {arccoth}\left (2\right )+\frac {i \pi }{2}+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \operatorname {arccoth}\left (2\right )-\frac {i \pi }{2} \end {align*}

Trying the constant \begin {align*} c_{1} = \operatorname {arccoth}\left (2\right )-\frac {i \pi }{2} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-\operatorname {arctanh}\left (x \right )+\operatorname {arccoth}\left (2\right )-\frac {i \pi }{2} \end {align*}

The constant \(c_{1} = \operatorname {arccoth}\left (2\right )-\frac {i \pi }{2}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\operatorname {arctanh}\left (x \right )+\operatorname {arccoth}\left (2\right )-\frac {i \pi }{2} \\ \end{align*}

Verification of solutions

\[ y = -\operatorname {arctanh}\left (x \right )+\operatorname {arccoth}\left (2\right )-\frac {i \pi }{2} \] Verified OK.

1.28.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (x^{2}-1\right ) y^{\prime }=1, y \left (2\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{x^{2}-1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \frac {1}{x^{2}-1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-\mathrm {arctanh}\left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\mathrm {arctanh}\left (x \right )+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (2\right )=0 \\ {} & {} & 0=-\mathrm {arctanh}\left (\frac {1}{2}\right )+\frac {\mathrm {I} \pi }{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\mathrm {arctanh}\left (\frac {1}{2}\right )-\frac {\mathrm {I} \pi }{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\mathrm {arctanh}\left (\frac {1}{2}\right )-\frac {\mathrm {I} \pi }{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\mathrm {arctanh}\left (x \right )+\mathrm {arctanh}\left (\frac {1}{2}\right )-\frac {\mathrm {I} \pi }{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\mathrm {arctanh}\left (x \right )+\mathrm {arctanh}\left (\frac {1}{2}\right )-\frac {\mathrm {I} \pi }{2} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.063 (sec). Leaf size: 15

dsolve([(x^2-1)*diff(y(x),x)=1,y(2) = 0],y(x), singsol=all)
 

\[ y \left (x \right ) = -\operatorname {arctanh}\left (x \right )+\operatorname {arctanh}\left (\frac {1}{2}\right )-\frac {i \pi }{2} \]

✓ Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 27

DSolve[{(x^2-1)*y'[x]==1,{y[2]==0}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{2} (\log (3-3 x)-\log (x+1)-i \pi ) \]