Solutions To Physics-Based Differential Equation Problems

 1) Find the general solution to the equation:  dy/ dx = -x/y

Solution:

Rewrite:  y dy =  - x dx

Integrate both sides:

ò y dy =  - ò x dx

Then: ½ y ²  = ½ x ² +  c

 2) Find the general solution to the equation: dy/dx = e ^{x - y}

Solution:

Rewrite:  e ^y dy   =  e ^x  dx    

Integrate both sides:

ò e ^y dy   =  ò e ^x  dx    =>

e ^y    =   e ^x  + c

3) The linearized equation for a simple pendulum with small oscillations is given by:

d² q/ dt ²  =  -  q (g/ ℓ)  

Find the period and frequency

Solution:

By comparison with simple harmonic oscillator, e.g. in spring example, we have:

w  =  Ö(g/ ℓ)

Therefore:

T  = 1/f   =  2p  Ö ℓ/g  

f   =  1/ 2p  Ö g/ ℓ    

4) A capacitor of C = 500 mF is initially charged to a potential difference of 10.0 V and connected to a resistor R = 100 kW and allowed too discharge. Find:

a)     The initial charge on the capacitor

b)    The time constant for discharge through the resistor R.

c)    The time for the initial charge to decay to one half its initial value.

d)   The energy stored in the capacitor by this time.

Solutions:

 a))  CV  =  (500 mF) (10.0 V)  =  (5 x 10^—4 F) (10.0 V) = 5 x 10^—3 C

b)    t  =  CR =   (5 x 10^—4 F)  (10 ⁵  W.)  =  5 x 10 ¹ s

c)    Q =  Q_o  e ^-(t/CR)     

Or:   Q/  Q_o   =   ½  = e ^-(t/CR)     

    2  =   e ^(t/CR)

             t  = CR ln 2   =     (50 s)  (0.693)  =   34.7  s

d)      Q =   ½ Q_o   = 2.5 x 10^—3 C

W  =   Q ² / 2C  =   (2.5 x 10^—3 C) ²  /  2 (5 x 10^—4 F) = 6.25 x 10^—3

5)A radionuclide sample of N = 10¹⁵ atoms undergoes decay at the constant average rate of dN/dt =  6.00 x  10¹¹  /s.  From this information, find:

a) The Activity A

b) The decay constant l

c) The half- life  (T _½) of the sample in minutes.

Solutions:

a) Given:    dN/dt = - lN

A = - lN   =   - ( 6.00 x 10^-4 s^-1 )( 10 ¹⁵ )=   -  6.00 x 10¹¹/s  i.e.

6.00 x 10¹¹    decays per second

b)  So that: 

l   =    1/N | dN/dt |   =  10 ^-15 (6.00 x 10¹¹/s)

l   =  6.00 x 10^-4 s^-1

c)  Half life: T_½  =  ln 2/l    = 0.693/ l 

=  0.693/ (6.00 x  10^-4 s^-1)  =  1160 s  (or 19.3 minutes) 

6) Using the example for the projectile motion, along with the accompanying diagram, find:

i) The maximum height attained when the launch angle q = 30 degrees.

ii) Write an equation for the slope (dy/dx)  of the path of the projectile at any point.

iii) Write the condition to obtain the range  R (maximum horizontal distance) of the projectile and find it.

iv) From (iv) what can be inferred about the angle F ?

Solutions:

i) The maximum altitude is:

y _max =  - ½ g t' ²   + v_o sin(q) t' =

- ½ g [v_o sin(q)/  g] ²  +  v_o sin(q) [v_o sin(q)/  g]

=  v_o sin(q) ² /  2 g    

q = 30 degrees:

=  v_o sin(30) ² /  2 g   

But sin (30) = 1/2 so:

y _max =  v_o (1/2) ² /  2 g   

=   v_o (1/4) /  2 g   =  v_o /  8 g

ii) Slope at any point: 

dy/dx =  (dy/dt)/ (dx/dt) =

 - ½ g t ² + v_o sin(q)/v_o cos(q)

iii) Range (R) occurs when y = 0  

 This occurs at double the time for max. altitude since trajectory is symmetric.

So: 2t'  =  2 ( v_o sin(q) / g ) 

Then x = R  =  v_o cos(q) 2 ( v_o sin(q) / g ) 

=  ( v_o ² / g) sin(2q)

iv)  The slope dy/dx is (expanded from (ii)):

dy/dx = -2 v_o sin(q) + v_o sin(q)/ v_o cos(q) 

= -  [sin(q)/ cos(q)] =   -  tan (q)

This shows the angle F =   p   -  q

i.e. projectile returns to Earth at same angle it left.