4.19 problem 9

Internal problem ID [11385]
Internal file name [OUTPUT/10368_Wednesday_May_17_2023_07_50_07_PM_22720490/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: 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]

\[ \boxed {x^{\prime }-x \left (4+x\right )=0} \] With initial conditions \begin {align*} [x \left (0\right ) = 1] \end {align*}

4.19.1 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)\\ &= x \left (x +4\right ) \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

4.19.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{x \left (x +4\right )}d x &= \int {dt}\\ \frac {\ln \left (x \right )}{4}-\frac {\ln \left (x +4\right )}{4}&= 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 {\ln \left (5\right )}{4} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -\frac {\ln \left (5\right )}{4} \end {align*}

Trying the constant \begin {align*} c_{1} = -\frac {\ln \left (5\right )}{4} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (x \right )}{4}-\frac {\ln \left (x +4\right )}{4} = t -\frac {\ln \left (5\right )}{4} \end {align*}

The constant \(c_{1} = -\frac {\ln \left (5\right )}{4}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (x\right )}{4}-\frac {\ln \left (4+x\right )}{4} &= t -\frac {\ln \left (5\right )}{4} \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (x\right )}{4}-\frac {\ln \left (4+x\right )}{4} = t -\frac {\ln \left (5\right )}{4} \] Verified OK.

4.19.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }-x \left (4+x\right )=0, x \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=x \left (4+x\right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {x^{\prime }}{x \left (4+x\right )}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {x^{\prime }}{x \left (4+x\right )}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (x\right )}{4}-\frac {\ln \left (4+x\right )}{4}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=-\frac {4 \,{\mathrm e}^{4 t +4 c_{1}}}{{\mathrm e}^{4 t +4 c_{1}}-1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=1 \\ {} & {} & 1=-\frac {4 \,{\mathrm e}^{4 c_{1}}}{{\mathrm e}^{4 c_{1}}-1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\frac {\ln \left (5\right )}{4} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\frac {\ln \left (5\right )}{4}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x=-\frac {4 \,{\mathrm e}^{4 t}}{-5+{\mathrm e}^{4 t}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=-\frac {4 \,{\mathrm e}^{4 t}}{-5+{\mathrm e}^{4 t}} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 16

dsolve([diff(x(t),t)=x(t)*(4+x(t)),x(0) = 1],x(t), singsol=all)
 

\[ x \left (t \right ) = \frac {4}{-1+5 \,{\mathrm e}^{-4 t}} \]

✓ Solution by Mathematica

Time used: 0.014 (sec). Leaf size: 21

DSolve[{x'[t]==x[t]*(4+x[t]),{x[0]==1}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to -\frac {4 e^{4 t}}{e^{4 t}-5} \]