1.51 problem 58

Internal problem ID [14094]
Internal file name [OUTPUT/13775_Saturday_March_02_2024_02_49_26_PM_22347657/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 1. Introduction to Differential Equations. Exercises 1.1, page 10
Problem number: 58.
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]

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

1.51.1 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 {\ln \left (x \right )}{x} \end {align*}

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

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

1.51.2 Solving as quadrature ode

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

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

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {\ln \left (x \right )^{2}}{2} \] Verified OK.

1.51.3 Maple step by step solution

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

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)/x,y(1) = 0],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\ln \left (x \right )^{2}}{2} \]

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 13

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

\[ y(x)\to \frac {\log ^2(x)}{2} \]