Find the standard form of the equation of the hyperbola satisfying the given conditions. Foci at $(0,-4)$ and $(0,4)$; vertices at $(0,2)$ and $(0,-2)$

The equation is $◻$.

Step 1 ：The standard form of the equation of a hyperbola with its center at the origin (0,0) is given by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ if the transverse axis is along the x-axis, and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ if the transverse axis is along the y-axis.

Step 2 ：In this case, the foci and vertices are along the y-axis, so the transverse axis is along the y-axis. The distance from the center to a focus is $c$, and the distance from the center to a vertex is $a$.

Step 3 ：Given that the foci are at (0,-4) and (0,4), the distance from the center to a focus is $c=4$.

Step 4 ：Given that the vertices are at (0,2) and (0,-2), the distance from the center to a vertex is $a=2$.

Step 5 ：We can find $b$ using the relationship $c^2=a^2+b^2$.

Step 6 ：Let's calculate $b^2$ using this relationship. $a=2,c=4$, so $b^2=c^2-a^2=12$.

Step 7 ：The value of $b^2$ is 12. Now we can write the standard form of the equation of the hyperbola. Since the transverse axis is along the y-axis, the equation is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$.

Step 8 ：Final Answer: The equation of the hyperbola is $\boxed{{\displaystyle \frac{y^2}{4}-\frac{x^2}{12}=1}}$.