4.8 problem 10

4.8.1 Solving as quadrature ode
4.8.2 Maple step by step solution

Internal problem ID [539]
Internal file name [OUTPUT/539_Sunday_June_05_2022_01_43_54_AM_75461840/index.tex]

Book: Elementary differential equations and boundary value problems, 10th ed., Boyce and DiPrima
Section: Section 2.5. Page 88
Problem number: 10.
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]

\[ \boxed {y^{\prime }-y \left (1-y^{2}\right )=0} \]

4.8.1 Solving as quadrature ode

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

Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {\sqrt {\left ({\mathrm e}^{2 c_{1} +2 t}-1\right ) {\mathrm e}^{2 c_{1} +2 t}}}{{\mathrm e}^{2 c_{1} +2 t}-1}\\ &=\frac {\sqrt {\left (c_{1}^{2} {\mathrm e}^{2 t}-1\right ) c_{1}^{2} {\mathrm e}^{2 t}}}{c_{1}^{2} {\mathrm e}^{2 t}-1}\\ y_2&=-\frac {\sqrt {\left ({\mathrm e}^{2 c_{1} +2 t}-1\right ) {\mathrm e}^{2 c_{1} +2 t}}}{{\mathrm e}^{2 c_{1} +2 t}-1}\\ &=-\frac {\sqrt {\left (c_{1}^{2} {\mathrm e}^{2 t}-1\right ) c_{1}^{2} {\mathrm e}^{2 t}}}{c_{1}^{2} {\mathrm e}^{2 t}-1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {\left (c_{1}^{2} {\mathrm e}^{2 t}-1\right ) c_{1}^{2} {\mathrm e}^{2 t}}}{c_{1}^{2} {\mathrm e}^{2 t}-1} \\ \tag{2} y &= -\frac {\sqrt {\left (c_{1}^{2} {\mathrm e}^{2 t}-1\right ) c_{1}^{2} {\mathrm e}^{2 t}}}{c_{1}^{2} {\mathrm e}^{2 t}-1} \\ \end{align*}

Figure 149: Slope field plot

Verification of solutions

\[ y = \frac {\sqrt {\left (c_{1}^{2} {\mathrm e}^{2 t}-1\right ) c_{1}^{2} {\mathrm e}^{2 t}}}{c_{1}^{2} {\mathrm e}^{2 t}-1} \] Verified OK.

\[ y = -\frac {\sqrt {\left (c_{1}^{2} {\mathrm e}^{2 t}-1\right ) c_{1}^{2} {\mathrm e}^{2 t}}}{c_{1}^{2} {\mathrm e}^{2 t}-1} \] Verified OK.

4.8.2 Maple step by step solution

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

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: 0.0 (sec). Leaf size: 29

dsolve(diff(y(t),t) = y(t)*(1-y(t)^2),y(t), singsol=all)
 

\begin{align*} y \left (t \right ) &= \frac {1}{\sqrt {c_{1} {\mathrm e}^{-2 t}+1}} \\ y \left (t \right ) &= -\frac {1}{\sqrt {c_{1} {\mathrm e}^{-2 t}+1}} \\ \end{align*}

Solution by Mathematica

Time used: 0.676 (sec). Leaf size: 100

DSolve[y'[t]== y[t]*(1-y[t]^2),y[t],t,IncludeSingularSolutions -> True]
 

\begin{align*} y(t)\to -\frac {e^t}{\sqrt {e^{2 t}+e^{2 c_1}}} \\ y(t)\to \frac {e^t}{\sqrt {e^{2 t}+e^{2 c_1}}} \\ y(t)\to -1 \\ y(t)\to 0 \\ y(t)\to 1 \\ y(t)\to -\frac {e^t}{\sqrt {e^{2 t}}} \\ y(t)\to \frac {e^t}{\sqrt {e^{2 t}}} \\ \end{align*}