Quentin Johnson

2022-01-21

Find the critical points and sketch the phase portrait of the given autonomous first order differential equation. Classify the each critical point as asmyptotically stable, unstable, or semi-stable.

${y}^{\prime}={y}^{2}(4-{y}^{2})$

eninsala06

Beginner2022-01-21Added 37 answers

Please take into consideration the autonomous first order differential equation provided.

${y}^{\prime}={y}^{2}(4-{y}^{2})$

The autonomous first order differential equation offered has the following form:

$\frac{dy}{dx}=f\left(y\right)={y}^{2}(4-{y}^{2})$

We can get the critical points by solving f(y) = 0

${y}^{2}(4-{y}^{2})=0$

$y=0{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}y=\pm 2$

Therefore, the critical points are y = -2, 0 and 2.

Now, as shown below, determine the asymptotically stable, unstable, or semi-stable state

${y}^{2}(4-{y}^{2})>0\text{}\text{}when\text{}\text{}4-{y}^{2}0$

Therefore,

${y}^{2}<4$

$-2<y<2$

Thus, the f(y) is increasing in (-2,2)

Now,

${y}^{2}(4-{y}^{2})<0\text{}\text{}when\text{}\text{}4-{y}^{2}0$

Therefore,

${y}^{2}>4$

$y<-2\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}y2$

$(-\mathrm{\infty},-2)\cup (2,\mathrm{\infty})\text{}the\text{}f\left(y\right)\text{}\text{is decreasing}$

Now classify them as shown below:

at $y=0,\text{}\text{semi-stable}\text{}\{\begin{array}{c}\text{above increasing}\\ \text{below increasing}\end{array}$

at $y=2,\text{}\text{asymptoti stable}\text{}\{\begin{array}{c}\text{above increasing}\\ \text{below increasing}\end{array}$

at $y=-2,\text{}\text{ustable}\text{}\{\begin{array}{c}\text{above increasing}\\ \text{below increasing}\end{array}$

William Appel

Beginner2022-01-22Added 44 answers

You helped me at the most important moment, thanks

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