Applications Of Differential Equations To Real World Problems

In this post we look at examples of how differential equations can be applied to solve some real world problems.

1)  A 100 gallon tank is full of pure water. Let pure water run into the tank at the rate of 2 gals/ min. and a brine solution containing 1/2 lb. of salt run in at the rate of 2 gals/min. The mixture flows out of the tank through an outlet tube at the rate of 4 gals/min. Assuming perfect mixing, what is the amount of salt in the tank after t minutes?

Solution:

Let s be the amount of salt in the tank in pounds at time t. Then:

s/ 100 = concentration of salt (i.e. as a proportion of total gallons of pure water in tank initially)

Then: ds/ dt = net rate of change =  (rate of gain in lbs/min -   rate of loss in lbs/min)

We can further write:

ds/dt = 1 -  4s/ 100 = 1 - s/25

Writing the basic differential equation to solve: ds/ (25 - s) = 

dt/ 25

This requires integrating both sides:

ò ^s _o  ds/ (25 – s) =   ò ^t _o  dt/25

Note the integration is taken from 0 to s on the left side and from 0 to t on the right. This leads to:

ln(25 – s)  - ln (25) = - t/25

And finally:

s =  25 (1 – e ^–t/25)

Let's take a time t = 25 minutes, what do we get?

 s =  25 (1 – e ^{–25 /25})=  25 (1 – e ^-1) = 25 (1 – 0.3678) = 25(0.6322) = 15.8 lbs.

 2) A block of mass m = 2.0 kg rests on a smooth horizontal surface attached to a spring. The spring has the property that it is stretched 0.05 m by a force of 10 N. If the block is displaced 0.05 m from the equilibrium position and released, find: the frequency and period of the motion.

Solution: We will apply Newton’s 2^nd law of motion, so:

F = ma = -kx where k is the spring constant.

Since F = 10N when x = 0.05 m, then:

k = F/x   =  (10 N)/ (0.05m) = 200 Nm^-1

Now rewrite the force balance equation ma = -kx:

m(d²x/dt ²) = - kx  or:     m(d²x/dt ²)  +  kx  = 0

Divide through by the mass, m:

(d²x/dt ²)  +  (k/ m)  x  = 0

Which is the basic equation for a simple harmonic oscillator, i,.e.:

(d²x/dt ²)  +  w ²  x  = 0

Where: w =  Ö(k/ m)  =  Ö(200 Nm^-1 / 2 kg) =

Ö (100 Nm^-1 /kg)  = 10 s^-1   or 10 radians/ sec

The frequency, f is related to  angular frequency w :

 w = 2 pf  so:

f = w / 2 p  =  (10 s^-1  ) / 2 p   = 5/p  s^-1 

The period T = 2 p/w = 2 p/(10 s^-1 ) = p/5 s = 0.63 s

3) A sled and its occupants weigh 1,000 lbs. It is coasting down a  5^o 45’ incline. The force of friction is 40.2 lbs. and the air resistance (drag) is at any given time 2 times the velocity in feet per second. Find an expression for the velocity after t seconds from rest and the specific velocity after 10 secs.

Recall: Weight = mass x g

And take g (acceleration of gravity) to be 32.17 f/s/s

The force exerted by the pull of gravity is vertically downward (see if you can sketch a free body diagram with forces acting)  and we may write:

F = mg sin q  = 1000 sin 5^o 45’  

= 100.2  

The total force acting is therefore:

 F = 100.2 - 40.2 - 2v = 60 - 2v  


WHY is this? We have two negative contributions (40.2 and 2v) on the LHS because the force of friction and drag both act opposite to the direction of motion. '2v' because the drag is stated as 'twice the velocity'.   The starting DE becomes:  

(1000)/ g (dv/dt) =       60- 2v

Re-arranging to separate variables:

dv / (30 - v)  =  (32.17) dt/ 500  = 0.06434 dt

For which the solution is obtained by integration, i.e.

ò ^v _o  dv/ (30 – v) = ò ^t _o  (0.06434) dt

Or:

ln (30 - v)  - ln 30 =  -0.06434 t

Taking natural logs of each side:

(30 - v)/ 30 =  e ^–0.06434t

v   =  30 (1 -  e ^–0.06434t  )


After 10 seconds:  v = 30 (1 -   e ^–0.6434 ) = 30 (1 -0.5255) = 30 (0.4745)

v(10) = 14.24 feet/sec

4)  

 A rocket is launched at an angle of 80 degrees with an initial velocity of 100,000 f/s if the air resistance is 0.01mv. Find the value of x and y after 10 seconds.

Data:    q = 80^o,   k = 0.01mv   v = 10⁵ fs^-1 and t = 10s

The differential equation can be written:

m(d²x/dt²) =  - k (dx/dt)/ v = - (0.01mv)x’/ v  = - 0.01mx’

Similarly:

my” = -mg – (0.01mv)y’/v  =   -mg – 0.01my’

Yielding the simultaneous pair:

x” + 0.01x’ = 0

y” + 0.01 y’ = -g

We know:

x’_o = v_o cos (q) = v_o cos (80) = (10⁵) cos 80

y’_o = v_o sin (q) = v_o sin (80) = (10⁵) sin 80

General Solutions:

x = c1 + c2 (e ^-0.01t)

y = c3 + c4((e ^-0.01t)  - 100 gt

x’ = -0.01c2 (e ^-0.01t)

y’ = -0.01c4((e ^-0.01t)  - 100 g

At t = 0, x = 0 so:

0 =  c1 + c2 (e ^-0.01t)

But:

x’_o  = v_o cos (80) = (10⁵) cos 80 =  -0. 01c2 (e ^-0.01t)

Then: c2 = - (10⁷) cos 80

So: 0 = c1 + ( - (10⁷) cos 80) or c1 =  (10⁷) cos 80

At t = 0, y = 0:

Then: 0 = c3 + c4 (e ^-0.01t)

And: y’_o  = v_o sin (80) = (10⁵) sin 80 = -0.01c4(e ^-0.01t)  - 100 g

So: c4 = -(10⁷) sin 80 – 10000g

\ c3 = - c4 (e ^-0.01t) = 10⁷ sin 80 + 10000g

To obtain the x and y-coordinates:

First, the x-coordinate:

x = c1 + c2 (e ^-0.01t) = 10⁷ cos 80 – (10⁷ cos 80)( e ^-0.01t)

x = 10⁷ cos 80 [1 - ( e ^-0.01t)] =  10⁷ (0.17365) [1 - ( e ^-0.01t)]

or:   x = 1736500 [1 - ( e ^-0.01t)]

The y-coordinate:

y = c3 + c4 ( e ^-0.01t)

Or:  y = 10⁷ sin 80 + 10000g – (10⁷ sin 80 + 10000g(e ^-0.01t))

y = 10⁷ sin 80 + 10000g(1 - e ^-0.01t) – 100 gt

y = [10⁷ (0.9848) + 320,000](1 - e ^-0.01t) – 100 gt

y = 1016800(1 - e ^-0.01t) – 100 gt

After 10 seconds:

x = 1736500 [1 - ( e ^-0.01t)] = x = 1736500 [1 - ( e ^-0.01(10))]

x = 165, 200 ft. (» 31. 3 miles)

y = 1016800(1 - e ^-0.01(10)) – 100(32)(10)

y =  930,580 ft.  ( » 177 miles)

Suggested Problems:

1) A 5 lb. weight hangs on a spring whose spring constant is 10. The weight is pulled 6 inches further down and released. Find the equation of motion, its period and frequency. If g = 32 ft/sec/sec give the units for the spring constant.

2) A circuit consists of an inductance L and resistance R  connected in series with effective voltage E.  Find a solution for the current i, 

a) Using the integrating factor exp (ò a dt), if   a  =   R/L

b) More specifically,  if E  =  E_o sin  w t