Write down a differential equation of the form dy/dt = ay + b whose solutions have the required behavior as t → ∞. all other solutions diverge from y = 2.

We want [tex]y=2[/tex] to be an unstable equilibrium solution - in other words, we want all solutions with initial condition [tex]y(t_0)=c[/tex] for [tex]c[/tex] near 2 to diverge from [tex]y=2[/tex] - so [tex]-dfrac ba=2[/tex].

Divergence from [tex]y=2[/tex] would require that for [tex]y>2[/tex], any solution is increasing, and for [tex]y<2[/tex], and solution is decreasing. In order for that to happen, we need

[tex]\dfrac{\mathrm dy}{\mathrm dt}=ay+b>0\implies y>-\dfrac ba[/tex]

if [tex]y>2[/tex], and

[tex]\dfrac{\mathrm dy}{\mathrm dt}=ay+b<0\implies y<-\dfrac ba[/tex]

if [tex]y<2[/tex].

Now we just pick [tex]a,b[/tex] to satisfy these conditions. An easy choice is [tex]a=1[/tex], which forces [tex]b=-2[/tex], so that one possible ODE would be

[tex]\dfrac{\mathrm dy}{\mathrm dt}=y-2[/tex]

I've attached a plot of the slope field to demonstrate the behavior of the solutions.

okpalawalter8 okpalawalter8 · Answer 2 · 2021-09-09T18:20:59+02:00

We have that the differential equation of the form dy/dt = ay + b whose solutions have the required behavior as t → ∞. all other solutions diverge from y = 2 is

[tex]\frac{dy}{dt}=y-2[/tex]

From the Question we are told that

[tex]dy/dt = ay + b[/tex]

[tex]t \rightarrow \infty[/tex]

[tex]y = 2.[/tex]

Generally,

Where

[tex]y = 2.[/tex]

[tex]dy/dt = ay + b[/tex]

[tex]dy/dt = a2 + b[/tex]

[tex]dy/dt = 0[/tex]

Therefore

[tex]a2+ b=0[/tex]

[tex]b=-2a[/tex]

Substituting

[tex]dy/dt = a(y-2)\\\\\\\frac{dy}{y-6}=adt[/tex]

[tex]y=2+e^{at}[/tex]

[tex]t \rightarrow \infty[/tex] Then all other solutions diverge from

[tex]y=2\\\\a=1[/tex]

Therefore

[tex]y(t)=e^{1.\infy+6}\\\\y(t)=\infty +2\\\\y(t) to \infty\\\\\frac{dy}{dt}=1.(y-2)[/tex]

[tex]\frac{dy}{dt}=y-2[/tex]

For more information on this visit

https://brainly.com/question/22271063?referrer=searchResults