= Solution
Take downward displacement as positive. At equilibrium, the lower spring carries $mg$ and extends by $mg/k$, while the upper spring supports both masses and extends by $2mg/k$. Write $x_1,x_2$ for the masses' downward displacements from these equilibrium positions. The extra upper-spring extension is $x_1$ and the extra lower-spring extension is $x_2-x_1$. Applying <Newton's second law> and cancelling the equilibrium forces gives
$$
\boxed{m\ddot x_1=-2kx_1+kx_2,\qquad m\ddot x_2=kx_1-kx_2.}
$$
The gravitational terms disappear because the constant equilibrium spring forces already balance the weights. Gravity sets the equilibrium extensions, rather than the coefficients of the linear oscillation equations for ideal springs.
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