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📚 Quick Study Guide
- 🔍 Equilibrium Points: Equilibrium points, also known as critical points or stationary points, are the values of $y$ for which $\frac{dy}{dt} = 0$. These points represent constant solutions to the differential equation.
- 📈 Finding Equilibrium Points: To find equilibrium points, set the differential equation equal to zero and solve for $y$. For example, given $\frac{dy}{dt} = f(y)$, solve $f(y) = 0$.
- 📊 Classifying Equilibrium Points: Equilibrium points can be classified as stable, unstable, or semi-stable. This classification depends on the behavior of solutions near the equilibrium point.
- ➡️ Stability Analysis:
- Stable: Solutions near the equilibrium point converge towards it as $t$ increases.
- Unstable: Solutions near the equilibrium point move away from it as $t$ increases.
- Semi-stable: Solutions approach the equilibrium point from one side but move away from it from the other side.
- 🧪 Using the Sign of $\frac{dy}{dt}$:
- If $\frac{dy}{dt} > 0$ for $y < y_e$ and $\frac{dy}{dt} < 0$ for $y > y_e$, then $y_e$ is stable.
- If $\frac{dy}{dt} < 0$ for $y < y_e$ and $\frac{dy}{dt} > 0$ for $y > y_e$, then $y_e$ is unstable.
- If $\frac{dy}{dt}$ has the same sign on both sides of $y_e$, then $y_e$ is semi-stable.
- 📝 Linearization (Advanced): For more complex systems, linearization around the equilibrium point can be used to determine stability. Calculate the Jacobian matrix and analyze its eigenvalues.
Practice Quiz
-
Consider the differential equation $\frac{dy}{dt} = y(y-2)$. Which of the following are the equilibrium points?
- $y = 0$ only
- $y = 2$ only
- $y = 0$ and $y = 2$
- $y = -2$ and $y = 2$
-
For the differential equation $\frac{dy}{dt} = y - y^2$, classify the stability of the equilibrium point $y = 0$.
- Stable
- Unstable
- Semi-stable
- Cannot be determined
-
Given $\frac{dy}{dt} = (y-1)(y-3)$, classify the stability of $y=1$.
- Stable
- Unstable
- Semi-stable
- Asymptotically stable
-
Which of the following differential equations has equilibrium points at $y = -1$ and $y = 1$?
- $\frac{dy}{dt} = y^2 - 1$
- $\frac{dy}{dt} = y^2 + 1$
- $\frac{dy}{dt} = y - 1$
- $\frac{dy}{dt} = -y - 1$
-
Consider $\frac{dy}{dt} = -(y-2)^2$. Classify the equilibrium point at $y = 2$.
- Stable
- Unstable
- Semi-stable
- Asymptotically stable
-
For the equation $\frac{dy}{dt} = y^3 - 4y$, find all equilibrium points.
- $y = 0$ only
- $y = 2$ and $y = -2$ only
- $y = 0$, $y = 2$, and $y = -2$
- No equilibrium points
-
If $\frac{dy}{dt} = f(y)$ and $f'(y_e) < 0$ at an equilibrium point $y_e$, what can be said about the stability of $y_e$?
- $y_e$ is stable
- $y_e$ is unstable
- $y_e$ is semi-stable
- No conclusion can be made
Click to see Answers
- C
- B
- B
- A
- C
- C
- A
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