1.11 Example 11

\[ y^{\prime }=y\left ( 1-y\right ) \] Since this is linear in \(y^{\prime }\) we can not use the p-discriminant method. By inspection we see this is separable. Hence the candidate singular solutions are obtain when \(y\left ( 1-y\right ) =0\). This is because this is what we have to divide both sides by to integrate. Therefore \begin {align*} y_{s} & =0\\ y_{s} & =1 \end {align*}

The general solution is found from \begin {align*} \int \frac {dy}{y\left ( 1-y\right ) } & =\int dx\\ -\ln \left ( y-1\right ) +\ln y & =x+c\\ \ln \frac {y}{y-1} & =x+c\\ \frac {y}{y-1} & =c_{1}e^{x}\\ y & =\left ( y-1\right ) c_{1}e^{x}\\ y-yc_{1}e^{x} & =-c_{1}e^{x}\\ y\left ( 1-c_{1}e^{x}\right ) & =-c_{1}e^{x}\\ y & =\frac {c_{1}e^{x}}{c_{1}e^{x}-1}\\ y & =\frac {c_{1}}{c_{1}-e^{-x}}\\ y & =\frac {1}{1-c_{2}e^{-x}} \end {align*}

Now we ask, can the singular solutions \(y_{s}=0,y_{s}=1\) be obtained from the above general solution for any value of \(c_{2}\)? We see when \(c_{2}=0\) then \(y=1\). Also when \(c_{2}=\infty \) then \(y=0\). So these are not really singular solutions. Some call these equilibrium solutions. But these should not be called singular solutions. Mathematica generates these when using the option IncludeSingularSolutions. But Maple does not give these when using the option singsol=all.

The following plot shows these equilibrium solutions with the general solution plotted using diļ¬€erent values of \(c\).