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\[d^2x/dt^2=g-kx/m\] To do with Hooke's law with gravity added.
you just wanna solve the DE ?
Yes
\[x''+\frac {k}mx=g\]the complimentary soution is\[y_c=c_1\cos(\omega t)+c_2\sin(\omega t)\]where \(\omega=\sqrt{\frac km}\) now we just need the particular solution...
we can also turn the complimentary into\[y_c=R\cos(\omega t-\delta)\]where\[R=\sqrt{c_1^2+c_2^2}\]and\[\delta=\tan^{-1}(\frac{c_1}{c_2})\]...
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typo*\[\delta=\tan^{-1}(\frac {c_2}{c_1})\]
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