1.26 problem 26

Internal problem ID [1895]
Internal file name [OUTPUT/1896_Sunday_June_05_2022_02_38_03_AM_24856868/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 5, page 21
Problem number: 26.
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]

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

1.26.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= -\left (-1+{\mathrm e}^{y}\right ) {\mathrm e}^{-y} \end {align*}

The \(y\) domain of \(f(x,y)\) when \(x=0\) is \[ \{-\infty

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is \[ \{-\infty

1.26.2 Solving as quadrature ode

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

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

The solutions are \begin {align*} c_{1} = -\ln \left (-1+{\mathrm e}\right ) \end {align*}

Trying the constant \begin {align*} c_{1} = -\ln \left (-1+{\mathrm e}\right ) \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\ln \left (-1+{\mathrm e}^{y}\right ) = x -\ln \left (-1+{\mathrm e}\right ) \end {align*}

The constant \(c_{1} = -\ln \left (-1+{\mathrm e}\right )\) gives valid solution.

Summary

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

Verification of solutions

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

1.26.3 Maple step by step solution

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

Maple trace

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

✓ Solution by Maple

Time used: 0.359 (sec). Leaf size: 32

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

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

✓ Solution by Mathematica

Time used: 0.007 (sec). Leaf size: 18

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

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