Math Revisited: Worked Problems for Differential Equations

In examining differential equations it is worth looking at applications in terms of word problems. Some examples are presented below:

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:

(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- Find the energy eigenvalues for an infinite square well of width a.

Solution:

We sketch the system as shown below:





















One notes the box limits are from 0 to a and this is the x -direction. Inside the box the potential V = 0 and outside, V = ¥, hence the term, "infinite square well".  Therefore, outside the interval (0,a) one has the quantum wave form:

y*y = 0

That is the product of the wave function by its complex conjugate = 0. The Schrodinger wave equation is written (for one dimension): H^ y = E y

where H^ denotes the Hamiltonian operator:

H^ =  [-iħ d/dx] ²/2m

Then the differential 1-D Schrodinger equation to solve becomes:

dy²/dx²  + K² y = 0,   where K =  Ö [2mE] / ħ

And this second order differential equation has the solution:

y = C sin Kx + D cos Kx   (C, D constants)

We set boundary conditions thus:

At x = 0+, y (0+) = D (since sin(0) = 0)

At x = 0, y (0) = 0
As a cautionary note, one must postulate that the wave function is continuous (since we want y*y to represent something physically observable).

Then:

y (0+) = y (0-) therefore D = 0

and:

y = C sin Kx

Now, at x = a+, y (a+) = 0

At x = a-, y (a-) = C sin K(a)

For continuity: y (a+) = y (a-)

Therefore: C sin K(a-) = 0    but C can't = 0, or no particle!

Therefore: sin K(a-) = 0

Þ  sin (Ö[2mE] / ħ) a = 0, and  ((Ö[2mE] / ħ) a = n π

Therefore:

 a = nπ/K and n = + 1, + 2, ..etc.

Thus:   2mE = n² π² ħ²

E = { n² π² ħ² } / 2ma²

Which yields the energy eigenvalues:

E = { n² π² ħ² }/ 2ma² (E_n = 1, 2, 3, .....n)

Thus, E_n  is restricted to the discrete values:

π² ħ²/ 2ma²,     4 π² ħ²/ 2ma² ,    9 π² ħ²/ 2ma²   etc.