1.50 problem 57

Internal problem ID [14093]
Internal file name [OUTPUT/13774_Saturday_March_02_2024_02_49_26_PM_40001740/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: 57.
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]

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

1.50.1 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 {\cos \left (\frac {1}{x}\right )}{x^{2}} \end {align*}

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

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

1.50.2 Solving as quadrature ode

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

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

The solutions are \begin {align*} c_{1} = 2 \end {align*}

Trying the constant \begin {align*} c_{1} = 2 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-\sin \left (\frac {1}{x}\right )+2 \end {align*}

The constant \(c_{1} = 2\) gives valid solution.

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\sin \left (\frac {1}{x}\right )+2 \] Verified OK.

1.50.3 Maple step by step solution

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

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: 12

dsolve([diff(y(x),x)=cos(1/x)/x^2,y(2/Pi) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = -\sin \left (\frac {1}{x}\right )+2 \]

✓ Solution by Mathematica

Time used: 0.008 (sec). Leaf size: 13

DSolve[{y'[x]==Cos[1/x]/x^2,{y[2/Pi]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to 2-\sin \left (\frac {1}{x}\right ) \]