Lesson 6-2: Dynamic Systems (Differential/Difference)

Solve linear ODEs/difference equations, compute steady states, and interpret behavior.

← Back to 12th Grade Math ← Back to Unit 6 Previous: Lesson 6-1 Next: Lesson 6-3 →

Learning Objectives

Solve first-order linear ODEs and difference equations.
Find steady states and determine stability by parameters.
Interpret solutions in applications (bank accounts, migration, growth).
Recognize logistic dynamics and long-run behavior.

Continuous Linear Dynamics

x'(t)=a x + b

If $a<0$ , the system converges to steady state $x^*=-b/a$ .
If $a>0$ , solutions grow unless $b$ counteracts.

x(t)=(x_0+ b/a) e^{a t} - b/a\quad (a\neq 0)

Discrete Linear Dynamics

x_{n+1}=r x_n + s

If $|r|<1$ , then $x_n\to s/(1-r)$ .
If $|r|>1$ , then $|x_n|$ diverges (except measure-zero initial conditions).

x_n=r^n (x_0 - s/(1-r)) + s/(1-r)\quad (r\neq 1)

Logistic Dynamics

P'(t)=r P (1-P/K)

→

P(t)=\frac{K}{1+A e^{-r t}}

Worked Examples

Example 1: Bank Account

Solve $x'(t)=r x + s,\ x(0)=x_0$ .

Solution:

Using integrating factor $e^{-r t}$ , we get $x(t)=(x_0+s/r)e^{r t}-s/r$ (for $r\neq 0$ ). Steady state $x^*=-s/r$ .

Example 2: Migration

Iterate $x_{n+1}=1.03 x_n - 200$ and observe convergence.

Solution:

Closed form $x_n=1.03^n (x_0 - s/(1-1.03)) + s/(1-1.03)$ with $s=200$ . Since $|1.03|>1$ , magnitude diverges unless $x_0=s/(1-1.03)$ .

Practice

Problem 1

Solve $x'(t)=-0.2 x + 3,\ x(0)=0$ and find the steady state and half-life.

Solution: $x(t)=15(1-e^{-0.2 t})$ , steady state $15$ ; half-life solves $x(t)=7.5$ → $t=\ln 2/0.2\approx 3.466$ .

Problem 2

Given $x_{n+1}=0.8 x_n + 10,\ x_0=0$ , compute $x_5$ and the limit.

Solution: Closed form $x_n=0.8^n (x_0-50)+50$ ⇒ $x_5=0.8^5(-50)+50\approx 32.768$ ; limit $50$ .

Problem 3

Continuous: solve $x'(t)=x-5,\ x(0)=1$ and determine $t$ with $x(t)=4$ .

Solution: $x(t)=(1-5)e^{t}+5=5-4 e^{t}$ ⇒ $4=5-4 e^{t}\Rightarrow e^{t}=1/4\Rightarrow t=-\ln 4$ .

Problem 4

Discrete: for $x_{n+1}=1.1 x_n - 5$ , find the closed form and discuss behavior.

Solution: $x_n=1.1^n (x_0-(-50)) - 50$ diverges in magnitude since $|1.1|>1$ .

Problem 5

Logistic: $P'(t)=0.6 P (1-P/100),\ P(0)=10$ . Estimate time to reach $P=80$ .

Solution: $P(t)=\frac{100}{1+A e^{-0.6 t}}$ , with $A=\frac{100-10}{10}=9$ . Solve $80=100/(1+9 e^{-0.6 t})$ → $e^{-0.6 t}=\tfrac{1}{36}$ → $t=\tfrac{\ln 36}{0.6}\approx 5.96$ .

Key Takeaways

Parameters govern stability and steady states.
Discrete steady state equals $s/(1-r)$ when $|r|<1$ .
Logistic growth saturates at carrying capacity $K$ .

Continue to Lesson 6-3.

← Back to Unit 6

Previous: 6-1 Next: 6-3