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We have the following ODE:

\begin{equation} \begin{cases} \frac{dy}{dx}=y(1-y)\\ y(0)=1/2 \end{cases} \end{equation}

We can use the following equality: $$\frac{1}{y(1-y)}=1/2\bigg(\frac{1}{y}+\frac{1}{1-y}\bigg)$$

Then we do the following:

$$\frac{dx}{dy}=\frac{1}{2y}+\frac{1}{2-2y}$$ $$dx=1/2\frac{1}{y}dy+\frac{1}{2-2y}dy$$ $$1/2\ln y-1/2\ln (1-y)=x+C$$ $$C+x=\ln\frac{\sqrt{y}}{\sqrt{1-y}}$$ $$Ce^x=\frac{\sqrt{y}}{\sqrt{1-y}}$$ $$Ce^{2x}=\frac{y}{1-y}$$

But I am struggling to separate this. And then I find it even harder to answer the following:

"Note that y=1 is composed of critical points, determine with the use of the ODE if these CPs are stable".

This is difficult to know, because the system has no eigenmatrix, and without an eigenmatrix, the are no eigenvalues, so I cannot tell if the "system" is stable. Besides, it is not a "system" either.

Any ideas?

Thanks!

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  • $\begingroup$ @moo simplification spotted. thanks $\endgroup$ Commented Oct 18, 2022 at 11:46
  • $\begingroup$ my fault....... $\endgroup$ Commented Oct 18, 2022 at 11:49
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    $\begingroup$ You can't do math without morning's coffee @moo $\endgroup$ Commented Oct 18, 2022 at 11:53
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    $\begingroup$ and not to mention MILK to the coffee @moo $\endgroup$ Commented Oct 18, 2022 at 11:54
  • $\begingroup$ The logistic equation is also Bernoulli, $(y^{-1})'=1-y^{-1}$ is then linear in $u=y^{-1}$. $\endgroup$ Commented Oct 18, 2022 at 12:18

3 Answers 3

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From the last line:

$$Ce^{2x}-Ce^{2x}y=y$$

$$y(1+Ce^{2x})=Ce^{2x}$$

$$y=\frac{Ce^{2x}}{1+Ce^{2x}}$$

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  • $\begingroup$ You may want to mention that $C = 1$(can be found by $y(0)=\frac{1}{2}$) $\endgroup$
    – Hersh
    Commented Oct 18, 2022 at 12:20
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  • The partial fraction for $1/y(1-y)$ is just $$ \frac{1}{y(1-y)}=\frac{A}{y}+\frac{B}{1-y}=\frac{1}{y}-\frac{1}{1-y} $$

  • The IVP has a ODE that is separable, i.e., \begin{align*} \frac{1}{\tau(1-\tau)}d\tau&=ds,\\ \int_{1/2}^{y}\frac{1}{\tau(1-\tau)}d\tau=\int_{0}^{x}ds,\\ y(x)=\frac{e^{x}}{e^{x}+1}, \end{align*} where the final step is just partial fraction to $1/\tau(1-\tau)$.

  • Since $y'=y(1-y)$, so for $y=1$ we have $y'=0$. Hence indeed that is a critical point.

  • The ODE $y'=y(1-y)$ can be written as $y'=y-y^{2}$ so the ODE is non linear.

  • The autonomous system has equilibrium point for $y(1-y)=0$, i.e., $y=0$ or $y=1$.

  • Roughly speaking, an equilibrium point $y_0$ is stable if solutions of the IVP starting near $y_0$ do not wander too far away from $y_0$ in future time.

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$\displaystyle{\frac{1}{y(1-y)}=\frac{1-y+y}{y(1-y)}=\frac{1-y}{y(1-y)}+\frac{y}{y(1-y)}=\frac{1}{y}+\frac{1}{1-y}}$

Solving further the DE, we get $\displaystyle{\frac{y}{1-y}=e^x}$

Edit: Regarding the system

$$\text{Let}\ F=\frac{\mathrm dy}{\mathrm dx}=y(1-y)$$ So, we see $y=0\ \text{and}\ y=1$ are CPs. For $y>1,\ F<0\ \text{and for}\ 0<y<1, F>0$ suggesting that the system has a sink at $y=1$.

Likewise, we see that system has a source at $y=0$.

Thus, $y=1$ is stable equilibrium point(or CP as you say) and $y=0$ is unstable equilibrium point.

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