4.4 problem 8

Internal problem ID [12944]
Internal file name [OUTPUT/11597_Tuesday_November_07_2023_11_34_05_PM_68301495/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 1.5 page 71
Problem number: 8.
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]

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

4.4.1 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)\\ &= y \left (y -1\right ) \left (y -3\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

4.4.2 Solving as quadrature ode

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

The solutions are \begin {align*} c_{1} = -\frac {\ln \left (2\right )}{6} \end {align*}

Trying the constant \begin {align*} c_{1} = -\frac {\ln \left (2\right )}{6} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y \right )}{3}-\frac {\ln \left (y -1\right )}{2}+\frac {\ln \left (y -3\right )}{6} = t -\frac {\ln \left (2\right )}{6} \end {align*}

The constant \(c_{1} = -\frac {\ln \left (2\right )}{6}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (y\right )}{3}-\frac {\ln \left (y-1\right )}{2}+\frac {\ln \left (y-3\right )}{6} &= t -\frac {\ln \left (2\right )}{6} \\ \end{align*}

Verification of solutions

\[ \frac {\ln \left (y\right )}{3}-\frac {\ln \left (y-1\right )}{2}+\frac {\ln \left (y-3\right )}{6} = t -\frac {\ln \left (2\right )}{6} \] Verified OK.

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: 1.703 (sec). Leaf size: 133

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

\[ y \left (t \right ) = \frac {\left (2 \,{\mathrm e}^{6 t}-4\right ) \left (1-{\mathrm e}^{6 t}+\sqrt {{\mathrm e}^{6 t} \left ({\mathrm e}^{6 t}-2\right )}\right )^{\frac {2}{3}}+\left (\left (i \sqrt {3}-1\right ) \left (1-{\mathrm e}^{6 t}+\sqrt {{\mathrm e}^{6 t} \left ({\mathrm e}^{6 t}-2\right )}\right )^{\frac {1}{3}}-i \sqrt {3}-1\right ) \left ({\mathrm e}^{6 t}-\sqrt {{\mathrm e}^{6 t} \left ({\mathrm e}^{6 t}-2\right )}-2\right )}{\left (1-{\mathrm e}^{6 t}+\sqrt {{\mathrm e}^{6 t} \left ({\mathrm e}^{6 t}-2\right )}\right )^{\frac {2}{3}} \left (2 \,{\mathrm e}^{6 t}-4\right )} \]

✓ Solution by Mathematica

Time used: 0.068 (sec). Leaf size: 104

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

\[ y(t)\to \frac {\sqrt [3]{2 \sqrt {e^{6 t} \left (e^{6 t}-2\right )^3}+8 e^{6 t}-2 e^{12 t}-8}}{e^{6 t}-2}-\frac {2^{2/3}}{\sqrt [3]{\sqrt {e^{6 t} \left (e^{6 t}-2\right )^3}+4 e^{6 t}-e^{12 t}-4}}+1 \]