3.5 problem 5

Internal problem ID [12932]
Internal file name [OUTPUT/11585_Tuesday_November_07_2023_11_27_29_PM_60020797/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 1.4 page 61
Problem number: 5.
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]

\[ \boxed {w^{\prime }-\left (3-w\right ) \left (w+1\right )=0} \] With initial conditions \begin {align*} [w \left (0\right ) = 4] \end {align*}

3.5.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} w^{\prime } &= f(t,w)\\ &= -\left (w +1\right ) \left (w -3\right ) \end {align*}

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

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

3.5.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int -\frac {1}{\left (w +1\right ) \left (w -3\right )}d w &= \int {dt}\\ \frac {\ln \left (w +1\right )}{4}-\frac {\ln \left (w -3\right )}{4}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(w=4\) 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 (w +1\right )}{4}-\frac {\ln \left (w -3\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 (w+1\right )}{4}-\frac {\ln \left (w-3\right )}{4} &= t +\frac {\ln \left (5\right )}{4} \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (w+1\right )}{4}-\frac {\ln \left (w-3\right )}{4} = t +\frac {\ln \left (5\right )}{4} \] Verified OK.

3.5.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [w^{\prime }-\left (3-w\right ) \left (w+1\right )=0, w \left (0\right )=4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & w^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & w^{\prime }=\left (3-w\right ) \left (w+1\right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {w^{\prime }}{\left (3-w\right ) \left (w+1\right )}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {w^{\prime }}{\left (3-w\right ) \left (w+1\right )}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (w+1\right )}{4}-\frac {\ln \left (w-3\right )}{4}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} w \\ {} & {} & w=\frac {3 \,{\mathrm e}^{4 t +4 c_{1}}+1}{-1+{\mathrm e}^{4 t +4 c_{1}}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} w \left (0\right )=4 \\ {} & {} & 4=\frac {3 \,{\mathrm e}^{4 c_{1}}+1}{-1+{\mathrm e}^{4 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\ln \left (5\right )}{4} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\ln \left (5\right )}{4}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & w=\frac {15 \,{\mathrm e}^{4 t}+1}{-1+5 \,{\mathrm e}^{4 t}} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & w=\frac {15 \,{\mathrm e}^{4 t}+1}{-1+5 \,{\mathrm e}^{4 t}} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.11 (sec). Leaf size: 23

dsolve([diff(w(t),t)=(3-w(t))*(w(t)+1),w(0) = 4],w(t), singsol=all)
 

\[ w \left (t \right ) = \frac {15 \,{\mathrm e}^{4 t}+1}{-1+5 \,{\mathrm e}^{4 t}} \]

✓ Solution by Mathematica

Time used: 0.02 (sec). Leaf size: 26

DSolve[{w'[t]==(3-w[t])*(w[t]+1),{w[0]==4}},w[t],t,IncludeSingularSolutions -> True]
 

\[ w(t)\to \frac {15 e^{4 t}+1}{5 e^{4 t}-1} \]