I hav a differential equations problem! The details are below!?

Question

A portion of a uniform chain of length 8 ft is loosely coiled around a peg at the edge of a high horizontal platform, and the remaining portion of the chain hangs at rest over the edge of the platform. Suppose the length of the overhanging chain is 3 ft, that the chain weighs 2 lb/ft, and that the positive direction is downward. Starting at t = 0 seconds, the weight of the overhanging portion causes the chain on the table to uncoil smoothly and to fall to the floor. If x9t0 denotes the length of the chain overhanging the table at time t > 0, then v= dx/dt is its velocity. When all resistive forces are ignored, it can be shown that a mathematical model relating v to x is given by xv(dv/dx + v ^2= 32x.
a) Rewrite this model in differential form. Proceed as in Problems 31- 36 and solve the DE for v in terms of x by finding an appropriate integrating factor. Find an explicit solution v(x).
b)Determine the velocity with which the chain leaves the platform. (and there is a platform

Answer 1

a) xv(dv/dx) + v^2 = 32 x
Let u = (xv^2)/2, then du/dx = v^2 / 2 + xv dv/dx.
So du/dx + u/x = 32x.
An integrating factor is e^(∫ 1/x dx) = e^(ln x) = x, so we get
x du/dx + u = 32x^2
=> d/dx (xu) = 32x^2
=> xu = (32/3) x^3 + C
=> u = (32/3) x^2 + C/x
=> (xv^2)/2 = (32/3) x^2 + C/x
=> v^2 = 64x / 3 + C/x^2
At time t=0, x = 3 and v = 0 so we get 0 = 64 + C/9, i.e. C = -9(64) = -576.
So v(x) = √ (64x/3 - 576/x^2)
= 8 √(x^3/3 - 9) / x.

b) The chain leaves the platform when x = 8, so v(x) = 8 √(8^3/3 - 9) / 8 = √(512/3 - 9) = √(485/3) = 12.7 ft/sec (1 d.p.)