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