Answer
Using newton law
"f=ma=m\\frac{d^2x}{dt^2}=k\/x^3"
Particle goes 0 to x0 in time 0 to the
Solving above equation
"\\frac{dx}{dt}=-k\/2mx^2+c"
At t=0 V=0 so C=0
Again integrate
"\\frac{dx}{dt}=-k\/2mx^2"
"\\int_0^{x_0}2x^2 dx=-\\int _0^tkdt\/m"
"\\frac{2x_0^3}{3}=-kt\/m"
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