show above is a slope field for the differential equation dy/dx = y²(4 - y²). if y = g(x) is the solution to…

Explanation:

Step1: Find equilibrium solutions

Set $\frac{dy}{dx}=y^{2}(4 - y^{2})=0$. Then $y^{2}(2 - y)(2 + y)=0$. The equilibrium - solutions are $y = 0$, $y = 2$, and $y=-2$.

Step2: Analyze the sign of $\frac{dy}{dx}$ in intervals

Consider the intervals:

• When $y\lt - 2$, $\frac{dy}{dx}=y^{2}(4 - y^{2})\lt0$ (since $y^{2}\gt0$ and $4 - y^{2}\lt0$).
• When $-2\lt y\lt0$, $\frac{dy}{dx}=y^{2}(4 - y^{2})\gt0$ (since $y^{2}\gt0$ and $4 - y^{2}\gt0$).
• When $0\lt y\lt2$, $\frac{dy}{dx}=y^{2}(4 - y^{2})\gt0$.
• When $y\gt2$, $\frac{dy}{dx}=y^{2}(4 - y^{2})\lt0$.

Step3: Use the initial condition

We have the initial condition $g(-2)=-1$. Since $\frac{dy}{dx}\gt0$ for $-2\lt y\lt0$, the function $y = g(x)$ is increasing when $y$ is in the interval $(-2,0)$. As $x\to\infty$, the function $y = g(x)$ will approach the nearest equilibrium solution in the direction of its growth. The nearest equilibrium solution for an increasing function starting from $y=-1$ is $y = 0$.

Answer:

d. 0