1. **State the problem:** We want to find values of $k$ such that the function $y = 5 \cos(kt)$ satisfies the differential equation $$16 \frac{d^2 y}{dt^2} = -4y.$$\n\n2. **Write the given function and its derivatives:**\nThe function is $$y = 5 \cos(kt).$$\nFirst derivative: $$\frac{dy}{dt} = -5k \sin(kt).$$\nSecond derivative: $$\frac{d^2 y}{dt^2} = -5k^2 \cos(kt).$$\n\n3. **Substitute into the differential equation:**\nThe equation is $$16 \frac{d^2 y}{dt^2} = -4y.$$\nSubstitute $$\frac{d^2 y}{dt^2} = -5k^2 \cos(kt)$$ and $$y = 5 \cos(kt):$$\n$$16 (-5k^2 \cos(kt)) = -4 (5 \cos(kt)).$$\n\n4. **Simplify both sides:**\n$$-80 k^2 \cos(kt) = -20 \cos(kt).$$\n\n5. **Divide both sides by $-20 \cos(kt)$ (assuming $\cos(kt) \neq 0$):**\n$$\cancel{-80 k^2} \cancel{\cos(kt)} \div \cancel{-20} \cancel{\cos(kt)} = \cancel{-20} \cancel{\cos(kt)} \div \cancel{-20} \cancel{\cos(kt)}$$\nwhich simplifies to $$4 k^2 = 1.$$\n\n6. **Solve for $k^2$:**\n$$k^2 = \frac{1}{4}.$$\n\n7. **Find $k$ values:**\n$$k = \pm \frac{1}{2}.$$\n\n**Final answer:** The values of $k$ are $\frac{1}{2}; -\frac{1}{2}$.