D'Alembert's Reduction of Order Method

[_N3WTON_]

Homework Statement

Use the reduction of order metho to find a second linearly independent solution. What is the general solution of the differential equation?
y'' - y = 0
y_1(x) = e^x

Homework Equations

Reduction of order formula

The Attempt at a Solution

First, I set:
y = ve^x
y' = ve^x + e^{x}v'
y'' = ve^x + 2e^{x}v' + e^{x}v''
(ve^x + 2e^{x}v' + e^{x}v'') - (ve^x) = 0
2e^{x}v' + e^{x}v'' = 0
e^{x}v'' + 2e^{x}v' = 0
v'' + e^{x}v' = 0
Then I made a substitution:
w = v'
So the equation becomes:
w' + e^{x}w = 0
At this point, I tried to find an integration factor. However, the integrating factor I obtained is a bit unusual, which leads me to believe that I have made a mistake somewhere. This is the integrating factor I obtained:
p(x) = e^x
u(x) = e^{\int e^x} = e^{e^x}
At this point, due to the odd integrating factor, I am not sure what I have done wrong or how to continue the problem.

 

[BvU]

$e^{x}v'' + 2e^{x}v' = 0$ does not lead to $v'' + e^{x}v' = 0$ !

 

[vela]

I hope this mistake was a total brainfart. If not, you really need to get a handle on basic algebra. It'll make learning this stuff a lot easier when you're not wasting time tracking down avoidable errors like this one.

 

[_N3WTON_]

I hope this mistake was a total brainfart. If not, you really need to get a handle on basic algebra. It'll make learning this stuff a lot easier when you're not wasting time tracking down avoidable errors like this one.

It totally was, I promise my algebra is not that bad :p ...anyway thanks you both for the help