1. ## Integration confusion

I'm having a look at a "proof" or motivating illustration of the Integrating Factor method for solving a first-order linear ODE, but can't quite understand the precise way in which one goes from step (1) below to step (2) here. I understand that one is to integrate both sides, but I think I'm a bit unclear as to how the integration is being done from $t_0$ to $t$ when the integrand itself is to involve t, and how the term [TEX]I(t_0)w(t_0)[TEX] arrives in the LHS. And presumably, on the RHS we are free to just change to a dummy variable in the integrand while we integrate over t?

(1) $\frac{d}{dt}I(t)w(t) = I(t)r(t)$

Integrating both sides gives

(2) $I(t)w(t) - I(t_0)w(t_0) = \int_{t_0}^t{I(\sigma)r(\sigma)d \sigma}$

2. ## Re: Integration confusion

In general

$f(t)-f(t_0)=\displaystyle{\int_{t_0}^t} \left(\dfrac{d}{d\sigma} f(\sigma)\right) ~d\sigma$

This is just the fundamental theorem of calculus.

Just apply this directly to your formula.

writing the first line as

$\dfrac{d}{d\sigma}I(\sigma)w(\sigma)=I(\sigma) r(\sigma)$

we substitute into the integral formula above to get

$I(t)w(t)-I(t_0)w(t_0)=\displaystyle{\int_{t_0}^t} \left(\dfrac{d}{d\sigma}I(\sigma)w(\sigma)\right) ~d\sigma$

$I(t)w(t)-I(t_0)w(t_0)=\displaystyle{\int_{t_0}^t}I(\sigma) r(\sigma) ~d\sigma$

3. ## Re: Integration confusion

Originally Posted by icarus92
I'm having a look at a "proof" or motivating illustration of the Integrating Factor method for solving a first-order linear ODE, but can't quite understand the precise way in which one goes from step (1) below to step (2) here. I understand that one is to integrate both sides, but I think I'm a bit unclear as to how the integration is being done from $t_0$ to $t$ when the integrand itself is to involve t, and how the term [TEX]I(t_0)w(t_0)[TEX] arrives in the LHS. And presumably, on the RHS we are free to just change to a dummy variable in the integrand while we integrate over t?
We can change to a dummy variable on both sides!

(1) $\frac{d}{dt}I(t)w(t) = I(t)r(t)$
$\int_{t_0}^t \frac{d}{ds}I(s)w(s) ds= \int_{t_0}^t I(\sigma)r(\sigma) d\sigma$

Integrating both sides gives

(2) $I(t)w(t) - I(t_0)w(t_0) = \int_{t_0}^t{I(\sigma)r(\sigma)d \sigma}$