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- Thread starter neptune12XII
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fzero

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As you mention, the antiderivative of a function ##f(x)## is the function

$$ F(x) = \int^x_{x_0} f(x') dx'.$$

Here ##x'## is what's called a "dummy" variable, since once we integrate over it, the expression no longer depends on it. This formula is justified in standard calculus texts, so I won't say more about it.

Now if we consider differential equations, we can quickly realize that the simplest types of differential equation to solve are ones of the form

$$ \frac{df(x)}{dx} = g(x), ~~~~~(1)$$

i.e., the expression containing the derivatives is actually a total derivative. Then we can see that the solution to the equation is just

$$ f(x) = \int_{x_0}^x g(x') dx',~~~~(2)$$

where ##x_0## is a constant that supplies the constant of integration for the general solution. Equations of the form (1) are sometimes called "integrable," since they have a direct solution (even if we can't simplify the actual integral on the right-hand side of (2).)

Now an equation of the form

$$ T'(t) + k T(t) = 0 ,~~~~(3)$$

for constant ##t## is also integrable, because

$$ -k = \frac{T'(t)}{T(t)} = \frac{d}{dt} \left( \ln T(t) \right).$$

We will find that

$$ T(t) = A e^{-k t},$$

for a constant ##A##.

Now that we have these pieces in place, we can actually explain why integrating factors exist.

Consider our equation

$$ \frac{df(x)}{dx} + p(x) f(x) = g(x).$$

The left-hand side is not a total derivative, but we can ask if some multiple of it is. Namely, can we find a function ##m(x)## such that

$$ \frac{d}{dx} \Bigl( m(x) f(x) \Bigr) = m(x) \frac{df(x)}{dx} + m(x) p(x) f(x) = m(x) g(x).~~~~~(4)$$

If we expand the derivative on the left, we find that such an ##m(x)## must satisfy the differential equation

$$ \frac{d m(x) }{dx} + p(x) m(x) = 0.$$

This equation of the same form as the time equation (3), except now instead of the constant ##k##, we have the function ##p(x)##. We can still use the logarithm to solve for

$$ m(x) = \exp \left( \int^x p(x') dx' \right).$$

Once ##m(x)## is determined, we can also integrate (4) to find

$$ f(x) = \frac{1}{m(x)} \left( \int^x m(x') g(x') dx' + C \right).$$

To get to your question, we can try to apply this to equation (3), where we didn't really need an integration factor. We simply have ##p(t) =k##, so that ##m(t) = e^{kt}##. Since ##g(t)=0##, we simply have ##f(x) = C e^{-kt}##. So the integrating factor method still works, it is just overkill. Also, we really needed to know how to solve equations of the form (3) in order to derive the method in the first place.

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