Gavin's DiffEq Class Clarification: Jan 26

Question: What was that about the exact derivative when using the IFA? And where Did that specific antiderivative come from?
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 eln(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.

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

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(x2) ]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) = Int0x[ 4sin(t2) ]dt.
(We changed the dummy variable in the integrand to make it different from the limit.) For this function,
f(0) = Int00[ 4sin(t2) ]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) = Int0x[ 4sin(t2) ]dt + C.
Then
f(0) = Int00[ 4sin(t2) ]dt + C = 12, so
0 + C = 12,
and C = 12. The antiderivative we want is
f(x) = Int0x[ 4sin(t2) ]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?!

Gavin's DiffEq Clarification 000126