6.2 problem 2

6.2.1 Existence and uniqueness analysis
6.2.2 Solving as quadrature ode
6.2.3 Maple step by step solution

Internal problem ID [12671]
Internal file name [OUTPUT/11324_Friday_November_03_2023_06_30_25_AM_44070852/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 2. The Initial Value Problem. Exercises 2.3.2, page 63
Problem number: 2.
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]

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

6.2.1 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) &=1\\ q(x) &=1 \end {align*}

Hence the ode is \begin {align*} y^{\prime }+y = 1 \end {align*}

The domain of \(p(x)=1\) is \[ \{-\infty

6.2.2 Solving as quadrature ode

Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 1\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=1 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 1 \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 1 \] Verified OK.

6.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+y=1, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-y+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-y+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-y+1}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (-y+1\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-{\mathrm e}^{-x -c_{1}}+1 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-{\mathrm e}^{-c_{1}}+1 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 5

dsolve([diff(y(x),x)=-y(x)+1,y(0) = 1],y(x), singsol=all)
 

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

Solution by Mathematica

Time used: 0.002 (sec). Leaf size: 6

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

\[ y(x)\to 1 \]