1.27 problem 3(c)

Internal problem ID [6131]
Internal file name [OUTPUT/5379_Sunday_June_05_2022_03_35_46_PM_85888940/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(c).
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]

\[ \boxed {y^{\prime }=\ln \left (x \right )} \] With initial conditions \begin {align*} [y \left ({\mathrm e}\right ) = 0] \end {align*}

1.27.1 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) &=\ln \left (x \right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime } = \ln \left (x \right ) \end {align*}

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

1.27.2 Solving as quadrature ode

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

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

The solutions are \begin {align*} c_{1} = 0 \end {align*}

Trying the constant \begin {align*} c_{1} = 0 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-x +\ln \left (x \right ) x \end {align*}

The constant \(c_{1} = 0\) gives valid solution.

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -x +\ln \left (x \right ) x \] Verified OK.

1.27.3 Maple step by step solution

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

Maple trace

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

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 10

dsolve([diff(y(x),x)=ln(x),y(exp(1)) = 0],y(x), singsol=all)
 

\[ y \left (x \right ) = x \left (\ln \left (x \right )-1\right ) \]

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 11

DSolve[{y'[x]==Log[x],{y[Exp[1]]==0}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to x (\log (x)-1) \]