The ratio of the deflections of the free end of a cantilever due to an isolated load at $${\frac{1}{3}^{{\text{rd}}}}$$ and $${\frac{2}{3}^{{\text{rd}}}}$$ of the span, is

For deflection at any point a from the fixed end is given as: δ= (Pa^2(3l-a))/(6EI). Now you can put the value a*(1/3) and a*(2/3) in the formula and work out for the ratio.

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deflection at free end in a cantilever due to concentreted load=WL^3/3EI

then deflection at free end due to pt. load at any pt X (measured at a point X from the fixed end) is given by as below=(WX^3/3EI)+(WX^2(L-X)/2EI)

Now putting the value L/3 and 2L/3 in the above equation we can get the answer

using differtial equation

EI(d2y/dx2) = M = wx , x = distance from free end of cantilliver
EIy= 1/6wx3-1/2wL2x+1/3wL3
put values of x = 2/3L and 1/3L

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