1.25 problem 25

Internal problem ID [1894]
Internal file name [OUTPUT/1895_Sunday_June_05_2022_02_38_00_AM_29794131/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: 25.
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]

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

1.25.1 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)\\ &= {\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

1.25.2 Solving as quadrature ode

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

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

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

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

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

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

Summary

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

Verification of solutions

\[ -{\mathrm e}^{-y} = x -1 \] Verified OK.

1.25.3 Maple step by step solution

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

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: 0.031 (sec). Leaf size: 12

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

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

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 13

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

\[ y(x)\to -\log (1-x) \]