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

shown above is a slope field for the differential equation \\( \\frac{d y}{d x}=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{d y}{d x}=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 sign of (\frac{dy}{dx})

For (y\in(-2,0)\cup(0,2)), (\frac{dy}{dx}=y^{2}(4 - y^{2})>0). For (y>2) or (y < - 2), (\frac{dy}{dx}=y^{2}(4 - y^{2})<0).

Step3: Use the initial condition

Given (y = g(x)) with (g(-2)=-1). Since (-2<-1<0) and (\frac{dy}{dx}>0) for (y\in(-2,0)), the function (y = g(x)) is increasing. As (x\to\infty), the function (y = g(x)) approaches the upper - bound of the interval where (\frac{dy}{dx}>0) among the equilibrium solutions.

Answer:

B. - 2