Some Basic Applications Of Differential Equations

(1)  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

(2)  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.

(3 )Stretched spring: A 1 kg mass which naturally stretches a spring 5 cm is pulled down 5 cm farther and released. Find the equation of motion, its period and frequency.

Solution:  Note that: 5 cm = 0.05 m,

To get k, the spring constant, note: F = ma = -kx

A 1 kg mass exerts a downward force F = 10 N on the spring  (assuming g = 10 ms^-2). Since:

F = ma = mg = (1 kg) (10 ms^-2) = 10 kgms^-2  =  10 N

So: k = 10N/ 0.05m =  200 N/m

i.e. if 10N stretches the spring 0.05m then 200N stretches it 1 m

The equation of motion is then:  mx” +  kx  = mg

Or, since m = 1 kg and mg = 10 N:    x” +  200 x  = 10

 

Then the general solution is:

x = c1 sin (10)t + c2 cos (10) t + 0.05

At t = 0, x = 10 cm (5cm + 5cm) = 0.1m and:

 

0.10= 0 + c2 cos (0) + 0.05 = c2 + 0.05

Or: c2 = 0.10 – 0.05 = 0.05

Further: x’ = 0 at t = 0 so that (remember your 1^st derivatives of sine, cos!)

 

0 = 10 c1 cos 10(0) – 10 c2sin (10) (0) = 10 c1 cos (10)

or: c1 = 0

Therefore: x = 0.05 cos (10)t + 0.05

The period T = 2 p/w =     2 p/ (10) = p/ 5  s

The frequency f = 1/T = 1/ (p/ 5) =  5/ p   

 (4) Consider the L-C-R series circuit shown.

 Here C is a capacitor (which can store charge Q), R is a resistor, and L is an inductance. The effect of inductance is to oppose any change in current and contributes to a voltage drop L (dI/dt). The capacitor meanwhile exhibits a voltage drop of V = C/Q. By Kirchoff’s law (the total p.d. is the sum of the drops around the loop) it must be true that:

 

L dI/dt + RI  + Q/C  - E = 0

Where E is the electromotive force. But note that the current I is the rate of change of charge Q, so that: I = dQ/dt

Then: dI/dt = d²Q/dt²

 

So we need to rewrite the differential equation as:

 

L (d²Q/dt² )  + R (dQ/dt) + Q/C = E

 

We know how to form auxiliary equations so we write this one:

LCm² + (RC) m + 1 = 0

Let a = (LC), b = (RC) and c = 1 then use the quadratic formula:

 

{-b + Ö (b² – 4ac)}/ 2a

 

To obtain: m = {- RC +  Ö (R² C²  – 4LC)}/ 2LC

The form of the complementary function will then depend on the form of the discriminant, (R² C²  – 4LC).  The particular integral depends on E, i.e. if E = const. than Q = CE is a particular solution.

Problems for the Math Maven:

 

1) A  5 lb. weight hangs vertically on a spring whose spring constant k = 10. The weight is pulled 6 inches farther down and released. Find the equation of motion, its period and the frequency. What would the units be for k?

 

2)  For the circuit shown, consider the special case of NO capacitor. Then the differential equation of interest becomes:  L (dI/dt) + RI = E

Find the solution if we let E = E_o sin wt.

3)     A ball weighing 0.75  lb. is thrown vertically upward from a point 6 ft above the surface of the Earth, with an initial velocity of 20 ft/ sec. As it rises it is acted upon by air resistance which is numerically equal to v /64 in pounds.  Where v is the velocity in ft. per second.  How high will the ball rise?