# 1st order linear differential equations - integrating factors

1. Sep 27, 2009

### austeve

Hi guys, this is my first post but have read the forums for a long time - a quick search didnt bring up anything that could help me. So i was wondering if someone could please explain something to me.

I have a differential equation
(excuse the pictures, I dont know how to use the equation writer in this?)
and basically I have to get it into a seperable form and simplify like I have done in the picture.

In the picture I explain what I want to know, basically
how come the dy/dx e^(8x) + 8ye^(8x) can be written as d[e^(8x)]/dx ... where do the other terms go? I have been doing a few of these problems by following examples, and its not explained and I have been taking it for granted that this happens ... but I want to know why?

2. Sep 27, 2009

### lurflurf

That is due to the product rule for differentiation
u'v+uv'=(uv)'
y' (e^(8x)) + y(8e^(8x))=y' (e^(8x)) + y(e^(8x))'=[y(e^(8x)) ]'

3. Sep 27, 2009

### CompuChip

It's simply the product rule for differentiation:

$$\frac{d[f(x) g(x)]}{dx} = \frac{df(x)}{dx} g(x) + f(x) \frac{dg(x)}{dx}$$

Here follows a somewhat technical explanation of why you do this, hopefully you can follow.

So you start out with a differential equation of the form
dy/dx + p(x) y = f(x). ​
If you have some function I(x) which is nowhere zero, you can multiply the whole equation by I(x) to get
dy/dx I(x) + p(x) I(x) y = f(x) I(x). ​
The trick is now to choose I(x) such that the left hand side is precisely a derivative, so you can directly integrate both sides. Well, by the product rule
d[y I(x)]/dx = dy/dx I(x) + y dI/dx. ​
So you are happy when you can get I(x) to satisfy dI(x)/dx = I(x) p(x). This equation can be solved; for example, using separation:
(dI(x)/dx) / I(x) = p(x),​
d(ln I(x))/dx = p(x)​
because I(x) was nowhere zero. Integrate both sides:
ln I(x) = ∫ p(x) dx​
and solve for I(x):
I(x) = exp[ ∫ p(x) dx ] ​

Thus, you have obtained the equation
d[y I(x)]/dx = I(x) p(x)​
and you can now solve it by integrating:
y I(x) = ∫ I(x) p(x) dx,​
provided you can do the integral on the right.

Hope that was not too advanced for you? :)

4. Sep 27, 2009

### austeve

=D
The light bulb is starting to brighten now! The above steps were not clear in my examples, but by seeing exactly where
I(x) = exp[ ∫ p(x) dx ] ​

came from made it alot easier to understand. Thankyou very much for the help!