Excellent questions both. First, the derivative question: Suppose
we're solving *y*'(*x*) - tan(*x*)*y*(*x*) =
sin(*x*). The integrating factor is the exponential of an
antiderivative of the coefficient of the *y* term, so let's find
that first.
*Int*[ -tan(*x*) ]*dx* =
*Int*[ -sin(*x*)/cos(*x*) ]*dx* =
ln(cos(*x*))
(by substituting *w*=cos(*x*)), so the integrating factor is
*e*^{ln(cos(x))} =
cos(*x*). Multiplying both sides of the equation by this will,
we know, turn the left-hand side into
(cos(*x*) *y*(*x*))'
because that's how the integrating factor was chosen. (On the
in-class worksheet we showed that this was the case for the parachuter
example.) Thus
(cos(*x*) *y*(*x*))' = cos(*x*) sin(*x*),
and the rest of the problem is just integration.
*f*(*x*) = *Int*[ 4sin(*x*) ]*dx*
is a random antiderivative of 4sin(*x*) -- it could be
*f*(*x*) = -4 cos(*x*) + 4,
*f*(*x*) = -4 cos(*x*) + 9,
*f*(*x*) = -4 cos(*x*) - 27, etc. Suppose I told you
that we wanted the antiderivative to have the property that
*f*(0) = 12. None of the functions we've written down work for
this -- for the first, *f*(0) = 0, and so on. So you'd go
back to the general antiderivative, *f*(*x*) = -4
cos(*x*) + *C*, and plug in *x*=0:
*f*(0) = -4 cos(0) + *C* = 12, or
-4 + *C* = 12,
so *C*=16. The function we want is *f*(*x*) = -4
cos(*x*) + 16. Now suppose that the function we started with was
*f*(*x*) = *Int*[ 4sin(*x*^{2}) ]*dx*.
There isn't any way of antidifferentiating the right-hand side by
hand, so we're sort of stuck. Undeterred, I still want
*f*(0)=12. How can we do this? Well, if we put limits on the
integral (with one limit being *x*) we still have an
antiderivative: for example, one such would be
*f*(*x*) = *Int*_{0}^{x}[ 4sin(*t*^{2}) ]*dt*.
(We changed the dummy variable in the integrand to make it different
from the limit.) For this function,
*f*(*0*) = *Int*_{0}^{0}[ 4sin(*t*^{2}) ]*dt* = 0,
which isn't quite 12. In fact, it's just like the first example we
did above with the antiderivative of 4 sin(*x*). To get the
antiderivative with *f*(0) = 12, we just add the appropriate
constant: take
*f*(*x*) = *Int*_{0}^{x}[ 4sin(*t*^{2}) ]*dt* + *C*.
Then
*f*(0) = *Int*_{0}^{0}[ 4sin(*t*^{2}) ]*dt* + *C* = 12, so

0 +*C* = 12,
and *C* = 12. The antiderivative we want is
*f*(*x*) = *Int*_{0}^{x}[ 4sin(*t*^{2}) ]*dt* + 12.
By writing the antiderivative with limits in this manner we can
exactly say what the function solving an IVP is. Pretty cunning, huh?!

Easy for me to say, of course... What about that second question? Let's think about antiderivatives for a moment. The function

0 +

Gavin's DiffEq Clarification 000126

Last Modified: Wed Jan 26 21:16:34 CST 2000

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