2.11 Example 11 \(y^{\prime }=y\left ( 1-y\right ) \)

\begin{align*} y^{\prime } & =y\left ( 1-y\right ) \\ F & =p-y\left ( 1-y\right ) \\ & =0 \end{align*}

Since this is linear in \(p\), then there is no singular solution in the sense of envelope. We can not use the p-discriminant method. Mathematica gives singular solutions, but these are not the envelope. It calls them equilibrium solutions. Lets find them. 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. Mathematica 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 different values of \(c\).

Figure 8: Equilibrium solution envelope with general solution curves