shown above is a slope field for the differential equation $\\frac{dy}{dx}=y^{2}(4 - y^{2})$. if $y = g(x)$…

shown above is a slope field for the differential equation $\\frac{dy}{dx}=y^{2}(4 - y^{2})$. if $y = g(x)$ is the solution to the differential equation with the initial condition $g(-2)=-1$, then, $\\lim_{x\\to\\infty}g(x)$ is\n(a) $-\\infty$\n(b) -2\n(c) 0\n(d) 2\n(e) 3

shown above is a slope field for the differential equation $\\frac{dy}{dx}=y^{2}(4 - y^{2})$. if $y = g(x)$ is the solution to the differential equation with the initial condition $g(-2)=-1$, then, $\\lim_{x\\to\\infty}g(x)$ is\n(a) $-\\infty$\n(b) -2\n(c) 0\n(d) 2\n(e) 3

Answer

Explanation:

Step1: Find equilibrium solutions

Set (\frac{dy}{dx}=y^{2}(4 - y^{2}) = 0). Solve (y^{2}(4 - y^{2})=0), we get (y = 0,y=\pm2).

Step2: Analyze the behavior of the solution

We have the initial condition (y(-2)=-1). For the differential equation (\frac{dy}{dx}=y^{2}(4 - y^{2})), when (- 2<y<0), (\frac{dy}{dx}=y^{2}(4 - y^{2})>0) (since (y^{2}>0) and (4 - y^{2}>0) for (-2 < y<0)). When (y = - 2), (\frac{dy}{dx}=0). As (x\to\infty), the function (y = g(x)) approaches the equilibrium solution (y=-2) (because the slope (\frac{dy}{dx}) is positive for (-2 < y<0) and (y=-2) is a stable equilibrium (since for (y) slightly greater than (-2) (in the interval (-2<y<0)), (\frac{dy}{dx}>0) which means (y) is moving towards (y = - 2) as (x) increases)).

Answer:

B. -2