6.3 problem 3

Internal problem ID [12672]
Internal file name [OUTPUT/11325_Friday_November_03_2023_06_30_25_AM_58471505/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: 3.
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 ) = 2] \end {align*}

6.3.1 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

6.3.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=2\) 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 \(c_{1}\) found above in the general solution gives \begin {align*} -\ln \left (y -1\right ) = x \end {align*}

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

Summary

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

Verification of solutions

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

6.3.3 Maple step by step solution

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

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: 10

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

\[ y \left (x \right ) = {\mathrm e}^{-x}+1 \]

✓ Solution by Mathematica

Time used: 0.04 (sec). Leaf size: 12

DSolve[{y'[x]==-y[x]+1,{y[0]==2}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-x}+1 \]