# [SOLVED]s1.2.4 Find the equilibrium solution y_e

#### karush

##### Well-known member
$\textsf{Consider the differential equation$\displaystyle \frac{dy}{dt}=ay-b$}$

(a) Find the equilibrium solution $y_e$
rewrite as
$y'-ay=b$
$\displaystyle -\exp\int a \, da=e^{a^{2}/2}$
$\color{red}{y_e=b/a}$

(b) Let $Y(t)=y-y_e$; thus $Y(t)$ is the deviation from the equilibrium solution.
Find the differential equation satisfied by $Y(t)$.
????
$\color{red}{Y' = aY}$
ok stopped in my tracks.. red is book answer

#### MarkFL

Staff member
$\textsf{Consider the differential equation$\displaystyle \frac{dy}{dt}=ay-b$}$

(a) Find the equilibrium solution $y_e$
rewrite as
$y'-ay=b$
$\displaystyle -\exp\int a \, da=e^{a^{2}/2}$
$\color{red}{y_e=b/a}$

(b) Let $Y(t)=y-y_e$; thus $Y(t)$ is the deviation from the equilibrium solution.
Find the differential equation satisfied by $Y(t)$.
????
$\color{red}{Y' = aY}$
ok stopped in my tracks.. red is book answer
Okay, we are given:

$$\displaystyle \d{y}{t}=ay-b$$

Any equilibrium solutions are found from:

$$\displaystyle \d{y}{t}=0$$

$$\displaystyle ay-b=0\implies y_e=\frac{b}{a}\quad\checkmark$$

Next, we are given:

$$\displaystyle Y(t)=y-y_e$$

This implies:

$$\displaystyle \d{Y}{t}=\d{y}{t}\implies \d{Y}{t}=ay-b=a(Y+y_e)-b=aY+a\frac{b}{a}-b=aY$$

Make sense?

#### karush

##### Well-known member
 $\displaystyle ay-b=0\implies y_e=\frac{b}{a}$ uhmm how did $y$ become $y_e$

Last edited:

#### MarkFL

 $\displaystyle ay-b=0\implies y_e=\frac{b}{a}$ uhmm how did $y$ become $y_e$
The solution we are finding in this case is $$y_e$$, since we have set the derivative to zero.