2.5 problem 5

2.5.1 Solving as quadrature ode
2.5.2 Maple step by step solution

Internal problem ID [12903]
Internal file name [OUTPUT/11556_Tuesday_November_07_2023_11_27_01_PM_72649872/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 1.3 page 47
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]

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

2.5.1 Solving as quadrature ode

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

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

Summary

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

Figure 67: Slope field plot

Verification of solutions

\[ y = \frac {{\mathrm e}^{2 t} c_{1}^{2}}{-1+{\mathrm e}^{2 t} c_{1}^{2}} \] Verified OK.

2.5.2 Maple step by step solution

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

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: 0.0 (sec). Leaf size: 14

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

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

Solution by Mathematica

Time used: 0.404 (sec). Leaf size: 33

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

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