4.2 problem 1(b)

Internal problem ID [11368]
Internal file name [OUTPUT/10351_Wednesday_May_17_2023_07_49_46_PM_51759581/index.tex]

Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section: Chapter 1, First order differential equations. Section 1.3.1 Separable equations. Exercises page 26
Problem number: 1(b).
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]

\[ \boxed {x^{\prime }-{\mathrm e}^{-2 x}=0} \] With initial conditions \begin {align*} [x \left (0\right ) = 1] \end {align*}

4.2.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} x^{\prime } &= f(t,x)\\ &= {\mathrm e}^{-2 x} \end {align*}

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

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

4.2.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int {\mathrm e}^{2 x}d x &= \int {dt}\\ \frac {{\mathrm e}^{2 x}}{2}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(x=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {{\mathrm e}^{2}}{2} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {{\mathrm e}^{2}}{2} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {{\mathrm e}^{2}}{2} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {{\mathrm e}^{2 x}}{2} = t +\frac {{\mathrm e}^{2}}{2} \end {align*}

The constant \(c_{1} = \frac {{\mathrm e}^{2}}{2}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {{\mathrm e}^{2 x}}{2} &= t +\frac {{\mathrm e}^{2}}{2} \\ \end{align*}

Verification of solutions

\[ \frac {{\mathrm e}^{2 x}}{2} = t +\frac {{\mathrm e}^{2}}{2} \] Verified OK.

4.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }-{\mathrm e}^{-2 x}=0, x \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }={\mathrm e}^{-2 x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {x^{\prime }}{{\mathrm e}^{-2 x}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {x^{\prime }}{{\mathrm e}^{-2 x}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {1}{2 \,{\mathrm e}^{-2 x}}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=-\frac {\ln \left (\frac {1}{2 \left (t +c_{1} \right )}\right )}{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=1 \\ {} & {} & 1=-\frac {\ln \left (\frac {1}{2 c_{1}}\right )}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {{\mathrm e}^{2}}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {{\mathrm e}^{2}}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x=-\frac {\ln \left (\frac {1}{2 t +{\mathrm e}^{2}}\right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=-\frac {\ln \left (\frac {1}{2 t +{\mathrm e}^{2}}\right )}{2} \end {array} \]

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: 0.032 (sec). Leaf size: 13

dsolve([diff(x(t),t)=exp(-2*x(t)),x(0) = 1],x(t), singsol=all)
 

\[ x \left (t \right ) = \frac {\ln \left (2 t +{\mathrm e}^{2}\right )}{2} \]

✓ Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 17

DSolve[{x'[t]==Exp[-2*x[t]],{x[0]==1}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to \frac {1}{2} \log \left (2 t+e^2\right ) \]