Internal problem ID [12869]
Internal file name [OUTPUT/11522_Monday_November_06_2023_01_31_17_PM_23065635/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Exercises section 1.2. page
33
Problem number: 9.
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]
\[ \boxed {y^{\prime }-{\mathrm e}^{-y}=0} \]
Integrating both sides gives \begin {align*} \int {\mathrm e}^{y}d y &= t +c_{1}\\ {\mathrm e}^{y}&=t +c_{1} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\ln \left (t +c_{1} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (t +c_{1} \right ) \\ \end{align*}
Verification of solutions
\[ y = \ln \left (t +c_{1} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-{\mathrm e}^{-y}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{-y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\mathrm e}^{-y}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{{\mathrm e}^{-y}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {1}{{\mathrm e}^{-y}}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\ln \left (t +c_{1} \right ) \end {array} \]
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: 0.016 (sec). Leaf size: 8
dsolve(diff(y(t),t)=exp(-y(t)),y(t), singsol=all)
\[ y \left (t \right ) = \ln \left (t +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.369 (sec). Leaf size: 10
DSolve[y'[t]==Exp[-y[t]],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \log (t+c_1) \]