# Calc 2 differential equations

1. Apr 13, 2013

### nick.martinez

im in calculus 2 right now and we are doing differential equaitons right now. im confused as to why when i find the integrating factor I(x)=e^(∫p(x) and when i multiply both sides i get
e^∫(p(x))[(dy/dx)+p(x)*y]=d(e^(p(x)dx)*y) how are they equal. i will give an example.

(dy/dx)+y=x*e^(x) the integrating factor is e^(∫dx)=e^(x) then i multiply both sides by
e^(x) which gives

e^(x)*[(dy/dx)+y]=e^(x)*[x*e^(x)]

which is equal to:

d[e^(x)*y]/dx=e^(x)*x; basically my question is how do you get here? how does e^(x)*dy/dx just disappear after i multiply the integrating factor to both sides?

2. Apr 14, 2013

### Simon Bridge

Lets see... your example is: $\renewcommand{\dyx}[1]{\frac{d #1}{dx}}$

$$\dyx{y}+y=xe^{x}$$ - integrating factor is: $e^{\int dx}=e^x$
multiplying both sides by the integrating factor gives you:

$$e^x\left [ \dyx{y}+y \right ] = e^x\left [ xe^x \right ]$$ ... to from there to the end result - you must first multiply out the brackets.

But you can see that the two expressions are the same by doing the differentiation on the LHS:

$$\dyx{} [e^x y]$$

This is how you choose the integrating factor - you have to notice that this relation will work.
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx