# Understanding the integrating factor

I'm trying to understand solving ordinary linear differential equations by using integrating factor. I dont have my book yet so I'm reading from wikipedia https://en.wikipedia.org/wiki/Integrating_factor#Use_in_solving_first_order_linear_ordinary_differential_equations but I really don't get it.

I have my equation $y' + P(x)y = Q(x)$ , I multiply on both sides with my integrating factor $e^{\int{P(x)}dx}$. If I now integrate on both sides with respect to x and then I'm magically seems to get something simpler.

Can someone please help me understand whats happening in the transformation of the left hand side of the equation when it goes from a partial to a total derivative? Why does this work?

## 1 Answer

It works because exponential function has special property. By chain rule for function $f(x)$ you have:

$$(e^{f(x)})'=f'(x)e^{f(x)}$$

It's a very commonly used trick, so you should remember it. If you also use formula $(fg)'=fg'+f'g$ you get your "magic". For example:

$$y'(x)+2y(x)=0$$

Multiply both sides by $e^{2x}$ to get something like $(fg)'=fg'+f'g$:

$$y'(x)e^{2x}+2y(x)e^{2x}=y'(x)\cdot e^{2x}+y(x)\cdot (e^{2x})'=(y(x)e^{2x})'=0$$.

• Thank you very much I really appreciate it. I get it now =) – John Smith Sep 2 '14 at 4:41