Find the standard form of the equation of the hyperbola satisfying the given conditions.
Endpoints of transverse axis: $(0,-16),(0,16)$; asymptote: $y=2x$

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 transverse axis is along the y-axis, since the endpoints of the transverse axis are (0,-16) and (0,16). Therefore, the length of the transverse axis is $2a=16-(-16)=32$, so $a=16$.

Step 3 ：The equation of the asymptote is $y=2x$, which can be rewritten as $\frac{y}{x}=2$. This is the slope of the asymptote, and for a hyperbola with its transverse axis along the y-axis, the slope of the asymptotes is $\pm \frac{a}{b}$. Therefore, $\frac{a}{b}=2$, and since we know that $a=16$, we can solve for $b$.

Step 4 ：Let's calculate $b$ and then write the equation of the hyperbola in standard form. $a=16$, $slop{e}_{asymptote}=2$, $b=8.0$

Step 5 ：Now that we have the values of $a$ and $b$, we can write the equation of the hyperbola in standard form. The equation of a hyperbola with its transverse axis along the y-axis is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$.

Step 6 ：Let's substitute $a=16$ and $b=8$ into this equation.

Step 7 ：Final Answer: The equation of the hyperbola is $\frac{y^2}{{16}^2}-\frac{x^2}{8^2}=1$, or simplified, $\frac{y^2}{256}-\frac{x^2}{64}=1$. So, the final answer is $\boxed{{\displaystyle \frac{y^2}{256}-\frac{x^2}{64}=1}}$.