Internal problem ID [5725]
Internal file name [OUTPUT/4973_Sunday_June_05_2022_03_15_38_PM_88735436/index.tex]
Book: Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold
Scientific. Singapore. 1995
Section: Chapter 1. First order differential equations. Section 1.1 Separable equations problems.
page 7
Problem number: 12.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {\left (1+z^{\prime }\right ) {\mathrm e}^{-z}=1} \]
In canonical form the ODE is \begin {align*} z' &= F(t,z)\\ &= f( t) g(z)\\ &= -1+{\mathrm e}^{z} \end {align*}
Where \(f(t)=1\) and \(g(z)=-1+{\mathrm e}^{z}\). Integrating both sides gives \begin{align*} \frac {1}{-1+{\mathrm e}^{z}} \,dz &= 1 \,d t \\ \int { \frac {1}{-1+{\mathrm e}^{z}} \,dz} &= \int {1 \,d t} \\ -z +\ln \left (-1+{\mathrm e}^{z}\right )&=t +c_{1} \\ \end{align*} Which results in \begin{align*} z &= \ln \left (-\frac {1}{{\mathrm e}^{t +c_{1}}-1}\right ) \\ \end{align*} Since \(c_{1}\) is constant, then exponential powers of this constant are constants also, and these can be simplified to just \(c_{1}\) in the above solution. Which simplifies to \[ z = \ln \left (-\frac {1}{{\mathrm e}^{t +c_{1}}-1}\right ) \] gives \[ z = \ln \left (-\frac {1}{{\mathrm e}^{t} c_{1} -1}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} z &= \ln \left (-\frac {1}{{\mathrm e}^{t} c_{1} -1}\right ) \\ \end{align*}
Verification of solutions
\[ z = \ln \left (-\frac {1}{{\mathrm e}^{t} c_{1} -1}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (1+z^{\prime }\right ) {\mathrm e}^{-z}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & z^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & z^{\prime }=-\frac {{\mathrm e}^{-z}-1}{{\mathrm e}^{-z}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {z^{\prime } {\mathrm e}^{-z}}{{\mathrm e}^{-z}-1}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {z^{\prime } {\mathrm e}^{-z}}{{\mathrm e}^{-z}-1}d t =\int \left (-1\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left ({\mathrm e}^{-z}-1\right )=-t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} z \\ {} & {} & z=-\ln \left ({\mathrm e}^{t -c_{1}}+1\right ) \end {array} \]
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: 15
dsolve((1+diff(z(t),t))*exp(-z(t))=1,z(t), singsol=all)
\[ z \left (t \right ) = \ln \left (-\frac {1}{c_{1} {\mathrm e}^{t}-1}\right ) \]
✓ Solution by Mathematica
Time used: 0.722 (sec). Leaf size: 28
DSolve[(1+z'[t])*Exp[-z[t]]==1,z[t],t,IncludeSingularSolutions -> True]
\begin{align*} z(t)\to \log \left (\frac {1}{2} \left (1-\tanh \left (\frac {t+c_1}{2}\right )\right )\right ) \\ z(t)\to 0 \\ \end{align*}