B.1.2.3 Find the equilibrium solution and....

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In summary, we have a differential equation $\dfrac{dy}{dt}=ay-b$ and we want to find the equilibrium solution $y_e$. By rewriting the equation as $y'=ay-b=0$, we can solve for the equilibrium solution $y_e=\dfrac{b}{a}$. If $Y(t)=y-y_e$, then $Y(t)$ represents the deviation from the equilibrium solution. The differential equation satisfied by $Y(t)$ is $Y'=aY+b$. To integrate this equation, we can let $u=ay+b$ and $du=ady$. Then, the equation becomes $\dfrac{dy}{u}=\dfrac{dx}{a}$. After integrating, we get
  • #1
karush
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$\tiny{b.1.2.3}$
Screenshot 2021-05-11 6.36.22 PM.png

Consider the differential equation
$\displaystyle \dfrac{dy}{dt}=ay-b$
Find the equilibrium solution $y_e$ rewrite as
$y'=ay-b=0$
then
$ay-b=0\implies y_e=\dfrac{b}{a}$
(b) Let $Y(t)=y-y_e$;
thus $Y(t)$ is the deviation from the equilibrium solution.
the differential equation satisfied by $Y(t)$.

so far but ?here is the book answer

Screenshot 2021-05-11 6.30.12 PM.png
 
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  • #2
karush said:
$\tiny{b.1.2.3}$
View attachment 11141
Consider the differential equation
$\displaystyle \dfrac{dy}{dt}=ay-b$
Find the equilibrium solution $y_e$ rewrite as
$y'=ay-b=0$
then
$ay-b=0\implies y_e=\dfrac{b}{a}$
(b) Let $Y(t)=y-y_e$;
thus $Y(t)$ is the deviation from the equilibrium solution.
the differential equation satisfied by $Y(t)$.

so far but ?
If Y= y-y_e the y= Y+ y_e.
Y'= y'= ay- b= a(Y+ y_e)= a(Y+ b/a)= aY+ b.
Y also satisfies the equation Y'= aY+ b.

With y'= dy/dx= ay+ b, dy/(ay+b)= dx.
To integrate the left, let u= ay+ b so that du= ady. dy= (1/a)du and the equation becomes
(1/a) du/u= dx
(1/a) ln(u)= x+ C
ln(u)= ax+ aC
u= e^{ax+ aC}= e^{aC}e^{ax}= C'e^{ax} where C'= e^{aC}
ay+ b= C'e^{ax}
ay= C'e^{ax}- b
y= (C'/a)e^{ax}- b/a= C''e^{ax}- b/a where C''= C'/a.

If a increases, the equilbrium, b/a, decreases but e^{ax} increases faster so y increases faster.
If a decreases. the equilibrium, b/a, increases but e^{ax} decreases slow so y increases slower.

here is the book answer

View attachment 11140
 
  • #3
mahalo helps a lot
 

FAQ: B.1.2.3 Find the equilibrium solution and....

1. What is an equilibrium solution?

An equilibrium solution is a point where the value of a function does not change over time. In other words, it is a stable point where the forces acting on a system are balanced and there is no net change.

2. How do you find the equilibrium solution?

The equilibrium solution can be found by setting the derivative of the function equal to zero and solving for the variable. This will give the value of the variable at the equilibrium point.

3. Why is finding the equilibrium solution important?

Finding the equilibrium solution is important because it helps us understand the behavior of a system over time. It allows us to identify the stable points where the system will eventually settle, and also helps us make predictions about the future behavior of the system.

4. Can there be more than one equilibrium solution?

Yes, there can be multiple equilibrium solutions for a given function. This can happen when the function has multiple critical points, where the derivative is equal to zero. Each critical point represents a potential equilibrium solution.

5. How does the equilibrium solution affect the stability of a system?

The equilibrium solution is a stable point where the forces acting on a system are balanced. This means that if the system is disturbed from this point, it will eventually return to it. Therefore, the equilibrium solution plays a crucial role in determining the stability of a system.

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