Having examined advanced methods of solving differential equations, including partial differential equations, e.g.

Solving Partial Differential Equations: Three Easy Examples & One (Somewhat) Hard

it's useful to go back to basic DEs for the benefit of those who may either: a) want to review basic approaches or b) those who want to learn them for the first time. This is in preparation to showing specific applications to physcs. Let's begin with the simplest first order equation imaginable,
which is also variables separable:

dy = x dx

As with all differential equations, the solution is accomplished via the
process called *integration*. If we integrate both sides, we obtain:

Ã² dy = Ã² x dx

Yielding:

y = x^{ 2} / 2 + c

where c is some undefined (as yet) constant of integration. We call the above
the "general solution" to the differential equation. This general solution is, in fact, *a family of
parabolas:*

**General solution - family of parabola - for dy = x dx.**If we wanted to obtain the *particular* solution, we'd have to assign *boundary conditions* at the outset.
Usually these designate what values x, y are to have at a particular point, and
also often the first derivative (y' or dy/dx) at the same point.

__Kinematics-Dynamics:__

To fix ideas in terms of an actual application we consider an example from kinematics. This is the
basic differential equation for rate of displacement:

d
x/ dt = v or: Ã² dx = v Ã²dt

Integrating:

Ã²_{xo }^{x} dx = v Ã²_{0 }^{t} dt

Let position: x_{o} = 0 at time t = 0,

which is effectively the boundary condition. Now, v = at, so:

x - x_{o} = v Ã²_{0 }^{t} dt = a Ã²_{0 }^{t} t dt

**=** ½ a t^{
2} + C

But: C = x_{o } so the
general solution becomes

x = ½ a t^{ 2} + x_{o}

And the *particular *solution (since x_{o} = 0) is:

x = ½ a t^{ 2}

Another basic differential equation arises from Newton's 2nd law applied to the stretching force (F= - kx) on a spring, so that one may write:

F = ma = m(d^{2}x/dt ^{2})
= - kx

Or, isolating the 2nd derivative:

(d^{2}x/dt ^{2}) + (k/
m) x = 0

d^{2}x/dt ^{2} + w ^{2} x =
0

Where: w = Ã–(k/ m) is angular frequency

And k is the spring constant. This type of DE will have boundary conditions automatically set by the specific problem data or inputs, i.e. for mass, for k, for w, but sometime two variables may be given and the other(s) need to be obtained.

__Capacitor Discharge__:

Whenever a capacitor
C is discharged through a resistor R. If the p.d. across C is V, then we have: Q
= CV. The current discharged is then expressed:

I
= - dQ/ dt

(i.e. Q, the charge, is* decreasing* as the current I increases)

We
know that Q = CV but by Ohm’s law: V = IR

So:
Q = -CR (dQ/dt)

**On separating variables we can integrate:**

Ã²_{0 }^{t} dt/ CR = - Ã²_{Qo }^{Q} dQ/ Q

Where:

Ã²_{Qo
}^{Q} dQ/ Q = Ã²_{Q }^{Qo} dQ/ Q = ln (Q_{o}/Q)

So that:

t/ CR =
ln (Q_{o}/Q) or:

** **e** **^{-}^{(t/CR)} = Q/ Q_{o}

Finally:

Q = *Q*_{o} *e *^{-(t/CR)}

__Radioactive Decay__:

This bears an analogous form to capacitor discharge.

We define the activity of a radioactive source as:

A = dN/dt = - lN

Where *l** * is the decay constant. The negative sign
appended to the equation indicates that the amount N is decreasing with time t. Separate variables:

dN/N = - l dt

We can write the integral:

Ã²_{N }^{No} dN/ N = ln (N_{o}/N) = - l Ã²_{0 }^{t} dt

The radioactive decay, then, is based on
some original number of atoms N_{o }decaying with an activity l over time t:N = N_{o
} exp (-lt)

The half life is that time - call it t = T _{½} - for which half of the original atoms have disintegrated, i.e. N_{o } ® N_{o }** /**2

Therefore: N_{o } /2
= N_{o } exp (-l T _{½})

After
dividing N_{o} into both sides and taking natural
logarithms we get:

*l** *T_{½}** = **ln 2
*= *0.693

Or: T _{½} **
= **ln 2/ l** =
**0.693**/ **l

Using
this format, any sample or fossil with even a minuscule amount of radioactive
material can be dated. All
we need know is that over the period defined as T½ half
of the number of the remaining atoms decay and the activity A is in **Becquerels**
(Bq).

Thus, if **T**_{½
}= 15,000 yrs. for l = 200 Bq* **then if **l** = 50 Bq now the sample is 45,000 years old. *A
graphical depiction of generic radionuclide decay is shown below.

*Radioactive
Decay for a radionuclide with time T in millions of years (mY)*

Here the vertical axis shows a
relative scale for the amount or mass of
some, unnamed decaying isotope which
commences decay at some initial specified value, e.g. 1 gram then decreases to
half that original amount in one half life – and thereafter to half *that
amount* and so forth each additional half life.

__Projectile motion__: A series of separate differential equations can be set up using the diagram shown below, based on projectile motion - incorporating gravity but neglecting air drag effects.

**Diagram depicting projectile motion in 2 dimensions.**

Here we can write the DEs for time t=0, at position x=0, y = 0:v_{ x} =
dx/dt = v_{o} cos(q)

v_{y} = dy/dt = v_{o} sin(q)

The force components at time t are:

F_{ x} =
0 and: F_{ y} = - mg

With these incorporated the differential equations become:

i) m d^{2}x/dt ^{2 }= 0 and ii) m d^{2}y/ dt ^{2 }= - mg

Each equation above requires two integrations, yielding 4 constant of integration in all.

For eqn. (i) we get on integrating: dx/ dt = c1 Then:

Ã² dx = c1 Ã²dt

Yields: x = c1t + c2

Here for eqn. (ii) we have:

d^{2}y/ dt ^{2 }= - g So: dy/dt = - gt + c3

And: y = - ½ g t^{ 2} + c3t + c4

Based on the initial conditions given we have:

c1 = v_{o} cos(q), c2 = 0, c3 = v_{o} sin(q), c4 = 0

Then the position of the projectile at time t seconds after firing can be found from:

x = v_{o} cos(q) t and: y = - ½ g t^{ 2} + v_{o} sin(q) t

Clearly, the projectile attains maximum altitude when its y -component of velocity is 0, i.e.

dy/ dt = 0 = -gt + v_{o} sin(q)

This must occur at time:

t' = v_{o} sin(q)/ g

Then 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

__Suggested Problems:__

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

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

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

d^{2 }q/ dt ^{2 }= - q (g/ â„“)

Find the period and frequency

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.

5)A
radionuclide sample of N = 10^{15} atoms undergoes decay at the constant
average rate of dN/dt** **= 6.00
x 10^{11 }** **/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.

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

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