Anonymous

TomV's answer is good, but just to clarify a key point... For *small* deflections (y<<L) we can assume the x-coordinate of the free end is approximately x=L. This isn't accurate enough for large deflections. The question was unclear.

Anonymous

You have d²y/dx² = k(L-x)^2 Integrate once. you get dy/dx = -1/3 k (l-x)³ + c Integrate again y = y0 +1/12 k(1-x)⁴ + c x +d This is the general solution. Here c and d are constants.