# D'Alembert's Reduction of Order Method

1. Oct 13, 2014

### _N3WTON_

1. The problem statement, all variables and given/known data
Use the reduction of order metho to find a second linearly independent solution. What is the general solution of the differential equation?
$y'' - y = 0$
$y_1(x) = e^x$

2. Relevant equations
Reduction of order formula

3. The attempt at a solution
First, I set:
$y = ve^x$
$y' = ve^x + e^{x}v'$
$y'' = ve^x + 2e^{x}v' + e^{x}v''$
$(ve^x + 2e^{x}v' + e^{x}v'') - (ve^x) = 0$
$2e^{x}v' + e^{x}v'' = 0$
$e^{x}v'' + 2e^{x}v' = 0$
$v'' + e^{x}v' = 0$
$w = v'$
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.

2. Oct 13, 2014

### BvU

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

3. Oct 13, 2014

### vela

Staff Emeritus
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.

4. Oct 13, 2014

### _N3WTON_

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