Excellent, if vague, question! Let's take this in several sections:
What are differential equations and where do they come from?
Let's just leave it where we had it in class: they're just equations
with derivatives in them, for example,
y'(x) = 1 - y(x).
Where do they come from? They arise whenever it makes sense to
describe how something changes. As an example, the interest we earn on,
and therefore the rate of change for, an investment is proportional to
the amount of money invested. Say we have an investment of $100
invested at a 5% continuously compounded rate: then the rate at which
we are getting money right now is (0.05)(100) = $5/year. Of
course, because it's accumulating continuously, in a second we'll have
a (very little) bit more than $100 and the rate of accumulation will
be higher. But in any event the rate is just the derivative, so we
could say that
P'(t) = .05 P(t),
where at this instant
P =100.
What am I supposed to do to solve a differential equation?
Ideally, we want the function that is in the equation -- so for
the first example above, we'd want to find the function
y(x) and for the second, P(t).
Unfortunately, we don't (yet) know how to find these. However, we can
check to see if something is a solution. Let's do another algebraic
example. Suppose I said "can you test whether y = 4 is a
solution to
y4 -
8 y3 +
14 y2 +
8 y = 15?"
You'd say "sure!", because you know you can plug in the
y and
see if it works. Here, we get
(4)4 -
(8) (4)3 +
(14) (4)2 +
(8)(4) ?=? 15
256 - 8(64) + (14)(16) + (8)(4) ?=? 15
256 - 512 + 224 + 32 ?=? 15
0 ?=? 15
But 15 isn't equal to 0. So
y = 4 isn't a solution. We
can do the same thing for differential equations: is
y =
1 - 12
e-x a solution to
y' = 1 -
y? Plug it in:
(1 - 12 e-x)' ?=? 1 -
(1 - 12 e-x)
12 e-x ?=? 1 - 1 +
12e-x
12e-x ?=?
12e-x
Here, 12
e-x is in fact equal
to 12
e-x, so it is a solution.
Why can't I just integrate both sides of the differential equation
to solve it?
You can't do this if the dependent variable appears on the
right-hand side of the differential equation: consider
y' = 1 - y again. What happens if we
integrate? We get (with "int" indicating an integral sign)
int( y'(x) dx ) =
int( (1 - y(x)) dx )
y(x) = x - int( y(x) dx)
But we don't know what
y(
x) is, so we can't do the
second integral on the right-hand side.
What does a differential equation tell me?
It does the same thing (sort of) as a derivative -- except you
might have to use values of both the dependent (function) and the
independent variable to find the slope. (In the case of the equation
y' = 1 - y you only need to know the dependent
variable, y.)
How does that line-thing come in?
If we can't solve the differential equation (find y(x)
or P(t), etc.), we might be able to find approximate
values for it. We can do this the same way we would when using a
tangent line to approximate successive values of a function: if I tell
you that f(0) = 1 and f'(0) = 2, then you
could write a tangent line at that point (the slope is m =
2 and the y-intercept is 1):
y = 2 x + 1.
Then, if we asked "what's
f(.1)?", it would be reasonable to
guess that it's about
y = 2 (.1) + 1 = 1.2
because the tangent line is close to the function. This is exactly
what we did in class -- the change in the function is the
2 (.1), which is just the slope times the change in
x, and we add that to the current function value (1). So we
can guess that
f(.1) is about 1.2. How is this related to the
differential equation? If the differential equation for
f(
x) involves
f(
x), once we've got an
estimate for
f(.1) we can estimate
f'(.1) and therefore
repeat the process with another tangent line...