A slope field for $\dfrac{dy}{dx} = y(y-2)$ is shown. The equilibrium $y = 0$ is stable, while $y = 2$ is unstable. A solution starts at $y(0) = 3$. What happens to $y$ as $x o \infty$?

Question

A slope field for $\dfrac{dy}{dx} = y(y-2)$ is shown. The equilibrium $y = 0$ is stable, while $y = 2$ is unstable. A solution starts at $y(0) = 3$. What happens to $y$ as $x 	o \infty$?

Accepted Answer

A. $y$ increases without bound.

Answer

$y$ increases without bound.

Answer

$y$ approaches $2$.

Answer

$y$ approaches $0$.

Answer

$y$ decreases without bound.

A slope field for $\dfrac{dy}{dx} = y(y-2)$ is shown. The equilibrium $y = 0$ is stable, while $y = 2$ is unstable. A solution starts at $y(0) = 3$ . What happens to $y$ as $x \to \infty$ ?

Explanation

Core Concept

Correct Answer

More Unit 7: Differential Equations practice questions

A slope field for dydx=y(y−2)\dfrac{dy}{dx} = y(y-2)dxdy​=y(y−2) is shown. The equilibrium y=0y = 0y=0 is stable, while y=2y = 2y=2 is unstable. A solution starts at y(0)=3y(0) = 3y(0)=3. What happens to yyy as x→∞x \to \inftyx→∞?

Explanation

Core Concept

Correct Answer

More Unit 7: Differential Equations practice questions

A slope field for $\dfrac{dy}{dx} = y(y-2)$ is shown. The equilibrium $y = 0$ is stable, while $y = 2$ is unstable. A solution starts at $y(0) = 3$ . What happens to $y$ as $x \to \infty$ ?