\(\Large{\blue{With\space variable \space masses}}\)

In the above pulley mass system mass $m$ decreases with a constant rate of $w$.

At $time=0s,m=M\Rightarrow m=M-wt$

If $x,y$ are the horizontal distances of $m,M$ from the pulley respectively and $T$ is the tension in the string $\Rightarrow y+x=constant$ $\Rightarrow\dot{y}=-\dot{x}$ $\Rightarrow \ddot{y}=-\ddot{x}$ $Applying \space newton's \space second \space law \space on \space each \space mass$ $M\ddot{y}=Mg-T..........[1]$ $\dfrac{dm\dot{x}}{dt}=mg-T$ $\Rightarrow \dfrac{dm\dot{y}}{dt}=T-mg..........[2]$ $[1]+[2]\Rightarrow\dfrac{dm\dot{y}}{dt}+M\ddot{y}=(M-m)g$ $\Rightarrow \dfrac{d[(M-wt)\dot{y}]}{dt}+M\ddot{y}=wtg$ $\Rightarrow \int_{t=0}^{t=t_1}{d[(M-wt)\dot{y}]}+\int_{t=0}^{t=t_1}M\ddot{y}{dt}=\int_{t=0}^{t=t_1}wtg{dt}$ $\Rightarrow M\dot{y}+(M-wt_1)\dot{y}=\dfrac{wt_1^2g}{2}$ $Since \space we \space can \space replace \space t_1 \space with\space t$ $\Rightarrow\dot{y}=\dfrac{wt^2g}{2(2M-wt)}$ $\Rightarrow \triangle y=\dfrac{wg}{2}\int_{0}^{t_1}\dfrac{t^2}{2M-wt}dt$ $let \space I=\int\dfrac{t^2}{\red{a-bt}}\red{dt}$ $=-\int\dfrac{t^2}{b}\red{d\ln|a-bt|}\space\space\space\blue{\because\dfrac{dt}{a-bt}=-\dfrac{-d\ln|a-bt|}{b}}$ $=-\dfrac{1}{b}\left({t^2\ln|a-2b|} -2\int t\ln|a-bt|dt\right)$ $= \dfrac{1}{b}\left(\red{2\int t\ln|a-bt|dt}-{t^2\ln|a-2b|} \right)$ $=\dfrac{1}{b}\left(\red{J}-{t^2\ln|a-2b|} \right)$ $\dfrac{J}{2}=\int t\ln|a-bt|dt$ $\blue{\because d[x\ln|a-bt|]-\dfrac{a}{b}d\ln|a-bt|-dt=\ln|a-bt|dt}$ $\therefore \dfrac{J}{2}=\int td[x\ln|a-bt|] -\dfrac{a}{b}\int td\ln|a-bt|-\int tdt$ $=\left(t^2\ln|a-bt|-\red{\int t\ln|a-bt|dt}\right)-\dfrac{a}{b}\left(t\ln|a-bt|-\int \ln|a-bt|dt\right)-\dfrac{t^2}{2}+\blue{C}$ $\Rightarrow\red{\dfrac{J}{2}}=t^2\ln|a-bt|-\red{\dfrac{J}{2}}-\dfrac{a}{b}\left(\cancel{t\ln|a-bt|}-\left(\cancel{t}-\dfrac{a}{b}\right)\ln|a-bt|+t\right)-\dfrac{t^2}{2}+\blue{C}$ $\Rightarrow J=t^2\ln|a-bt|-\dfrac{a^2}{b^2}\ln|a-bt|-\dfrac{at}{b}-\dfrac{t^2}{2}+\blue{C}$ $\Rightarrow J=\left(t^2-\dfrac{a^2}{b^2}\right)\ln|a-bt|-\left(\dfrac{at}{b}+\dfrac{t^2}{2}\right)+\blue{C}$ $\Rightarrow I=\dfrac{1}{b}\left(\red{\left(\cancel{t^2}-\dfrac{a^2}{b^2}\right)\ln|a-bt|-\left(\dfrac{at}{b}+\dfrac{t^2}{2}\right)+{C}}-\cancel{t^2\ln|a-2b|} \right)$ $=\dfrac{-1}{b}\left(\dfrac{ a^2}{b^2}\ln| a-bt|+\dfrac{ at}{b}+\dfrac{t^2}{2}+{C} \right)$ $=\dfrac{-1}{w}\left(\dfrac{4M^2}{w^2}\ln|2M-wt|+\dfrac{2Mt}{w}+\dfrac{t^2}{2}+{C} \right)$ $\Rightarrow \triangle y=\dfrac{wg}{2}\red{[I]_{0}^{t}}$ $=\dfrac{\cancel{w}g}{2\cancel{w}}\left(\dfrac{4M^2}{w^2}\ln\left(\left|\dfrac{2M}{2M-wt}\right|\right) -\dfrac{2Mt}{w}-\dfrac{t^2}{2}\right)$ $\Rightarrow \boxed{\triangle y={g}\left(\dfrac{2M^2}{w^2}\ln\left(\left|\dfrac{2M}{2M-wt}\right|\right) -\dfrac{Mt}{w}-\dfrac{t^2}{4}\right)}$