5.1 problem 1 and 13 (i)

5.1.1 Existence and uniqueness analysis
5.1.2 Solving as quadrature ode
5.1.3 Maple step by step solution

Internal problem ID [12950]
Internal file name [OUTPUT/11603_Tuesday_November_07_2023_11_51_54_PM_78927745/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 1.6 page 89
Problem number: 1 and 13 (i).
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]

\[ \boxed {y^{\prime }-3 y \left (y-2\right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

5.1.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= 3 y \left (y -2\right ) \end {align*}

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

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

5.1.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{3 y \left (y -2\right )}d y &= \int {dt}\\ \frac {\ln \left (y -2\right )}{6}-\frac {\ln \left (y \right )}{6}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {i \pi }{6} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {i \pi }{6} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {i \pi }{6} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y -2\right )}{6}-\frac {\ln \left (y \right )}{6} = t +\frac {i \pi }{6} \end {align*}

The constant \(c_{1} = \frac {i \pi }{6}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (y-2\right )}{6}-\frac {\ln \left (y\right )}{6} &= t +\frac {i \pi }{6} \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (y-2\right )}{6}-\frac {\ln \left (y\right )}{6} = t +\frac {i \pi }{6} \] Verified OK.

5.1.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-3 y \left (y-2\right )=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=3 y \left (y-2\right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y \left (y-2\right )}=3 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y \left (y-2\right )}d t =\int 3d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y-2\right )}{2}-\frac {\ln \left (y\right )}{2}=3 t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {2}{{\mathrm e}^{6 t +2 c_{1}}-1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {2}{{\mathrm e}^{2 c_{1}}-1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\mathrm {I}}{2} \pi \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\mathrm {I}}{2} \pi \hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {2}{{\mathrm e}^{6 t}+1} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {2}{{\mathrm e}^{6 t}+1} \end {array} \]

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: 14

dsolve([diff(y(t),t)=3*y(t)*(y(t)-2),y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {2}{1+{\mathrm e}^{6 t}} \]

Solution by Mathematica

Time used: 0.012 (sec). Leaf size: 16

DSolve[{y'[t]==3*y[t]*(y[t]-2),{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {2}{e^{6 t}+1} \]