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[SOLVED] s1.2.4 Find the equilibrium solution y_e

karush

Well-known member
Jan 31, 2012
2,886
$\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

Administrator
Staff member
Feb 24, 2012
13,775
$\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
Jan 31, 2012
2,886
$\displaystyle ay-b=0\implies y_e=\frac{b}{a}$

uhmm how did $y$ become $y_e$
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\displaystyle ay-b=0\implies y_e=\frac{b}{a}$

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.