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

The equation is $\square$.

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. \( a = 16 \), \( slope_{asymptote} = 2 \), \( b = 8.0 \)

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 \( \boxed{\frac{y^2}{256} - \frac{x^2}{64} = 1} \).