## Friday, June 25, 2010

### Differential Equations (Concluded)

We left off with two problems so will look at those now.

(1) Solve: x^2y dy – xy^2 dx – x^3y^2dx = 0

This is easy once one has access to a table of differentials or is able to work them out!

Factor to obtain:

xy(xdy – ydx) – x^2 y^2 dx = 0

Now, multiply by (x^-2y^-2):

(x dy – ydx)/ xy – x dx = 0

Then by the property of the differential d(ln y/x):

d(ln y/x) – xdx = 0

Integrating: ln(y/x) – x^2/2 = c

(2) Solve using any method for integrating factors:

x (dy/dx) - 3y = x^2

Put the equation into the form: dy/dx + Py = Q

Then: dy/dx – 3y/x = x

So: P = (-3/x) and Q = x

Therefore:

r = exp(INT Pdx) = exp (-3 ln x) = 1/ e ^3lnx = 1 /x^3

Therefore:

(1/x^3) y = INT (x /x^3) dx + C = -1/x + C

So: y = -x^2 + Cx^3

Now, we’re going to finish off this introductory view of differential equations by looking at a word problem:

A ship weighing 64,000 tons starts from rest under the impetus of a constant propeller thrust of 200,000 lbs. If the resistance of the water is 10,000 v lbs. find the velocity v in feet per second as a function of time, t.

The thrust = ma = 200,000 lbs. = kv at outset.

200,000 lbs. = 10,000 (v)

So v = 20 f/s initially.

The equation of motion needs to be as a function of time so:

mg – ma = kv or m(g – (dv/dt)) = kv

so: dv/ (g – av) = dt

-> -ln(g – av)/2a = t

the weight mg = 64,000 tons = (64,000 tons)(2 x 10^3 lb/ton) = 128 x 10^6 lbs.

so: m = W/g = (128 x 10^6 lb.)/ (32 f/s^2) = 4 x 10^6 sl

a = k/m = (10^4)/ (4 x 10^6 sl) = 2.5 x 10^-3

And: v = g/a[1 – exp(-2at)] = g/a[1 – exp(-0.0050t)]

Since g/a= 20 f/s:

v = 20[1 – exp(-0.0050t)]

This is the velocity in f/s as a function of time t

A problem to solve:

Find the x and y-coordinates of the points on a trajectory of a rocket shot at an angle of 80 degrees with an initial velocity v(o) = 100,000 f/s if the air resistance is 0.01 mv. Find the value of x and y after ten seconds.