1.29 problem 3(e)

Internal problem ID [6133]
Internal file name [OUTPUT/5381_Sunday_June_05_2022_03_35_49_PM_75045527/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(e).
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]

\[ \boxed {x \left (x^{2}-4\right ) y^{\prime }=1} \] With initial conditions \begin {align*} [y \left (1\right ) = 0] \end {align*}

1.29.1 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 \left (x^{2}-4\right )} \end {align*}

Hence the ode is \begin {align*} y^{\prime } = \frac {1}{x \left (x^{2}-4\right )} \end {align*}

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

1.29.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { \frac {1}{x \left (x^{2}-4\right )}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x^{2}-4\right )}{8}+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = \frac {\ln \left (3\right )}{8}+\frac {i \pi }{8}+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -\frac {\ln \left (3\right )}{8}-\frac {i \pi }{8} \end {align*}

Trying the constant \begin {align*} c_{1} = -\frac {\ln \left (3\right )}{8}-\frac {i \pi }{8} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x^{2}-4\right )}{8}-\frac {\ln \left (3\right )}{8}-\frac {i \pi }{8} \end {align*}

The constant \(c_{1} = -\frac {\ln \left (3\right )}{8}-\frac {i \pi }{8}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x^{2}-4\right )}{8}-\frac {\ln \left (3\right )}{8}-\frac {i \pi }{8} \\ \end{align*}

Verification of solutions

\[ y = -\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x^{2}-4\right )}{8}-\frac {\ln \left (3\right )}{8}-\frac {i \pi }{8} \] Verified OK.

1.29.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x \left (x^{2}-4\right ) y^{\prime }=1, y \left (1\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1}{x \left (x^{2}-4\right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \frac {1}{x \left (x^{2}-4\right )}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (x -2\right )}{8}+\frac {\ln \left (x +2\right )}{8}-\frac {\ln \left (x \right )}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\ln \left (x -2\right )}{8}+\frac {\ln \left (x +2\right )}{8}-\frac {\ln \left (x \right )}{4}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=\frac {\ln \left (3\right )}{8}+\frac {\mathrm {I} \pi }{8}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\frac {\ln \left (3\right )}{8}-\frac {\mathrm {I} \pi }{8} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\frac {\ln \left (3\right )}{8}-\frac {\mathrm {I} \pi }{8}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (x -2\right )}{8}+\frac {\ln \left (x +2\right )}{8}-\frac {\ln \left (x \right )}{4}-\frac {\ln \left (3\right )}{8}-\frac {\mathrm {I} \pi }{8} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (x -2\right )}{8}+\frac {\ln \left (x +2\right )}{8}-\frac {\ln \left (x \right )}{4}-\frac {\ln \left (3\right )}{8}-\frac {\mathrm {I} \pi }{8} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.031 (sec). Leaf size: 29

dsolve([x*(x^2-4)*diff(y(x),x)=1,y(1) = 0],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\ln \left (x +2\right )}{8}+\frac {\ln \left (-2+x \right )}{8}-\frac {\ln \left (x \right )}{4}-\frac {\ln \left (3\right )}{8}-\frac {i \pi }{8} \]

✓ Solution by Mathematica

Time used: 0.007 (sec). Leaf size: 26

DSolve[{x*(x^2-4)*y'[x]==1,{y[1]==0}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{8} \left (\log \left (\frac {1}{3} \left (4-x^2\right )\right )-2 \log (x)\right ) \]