# Just need help simplifying

• March 9th 2010, 07:37 PM
steph3824
Just need help simplifying
This is for a reduction of order problem. I won't post the whole problem because I only need help simplifying this..

$(x-1)(e^{x}v''+e^{x}2v'+ve^{x})-x(e^{x}v'+ve^{x})+ve^{x}=0$

So what the heck does this simplify out to because I don't think I'm doing it right
• March 9th 2010, 07:44 PM
pickslides
Hi Steph, expand it all out and group together in terms of $v, v'$ and $v''$
• March 9th 2010, 07:52 PM
steph3824
That's what I'm doing but I don't think I am doing it correctly
• March 10th 2010, 07:27 AM
steph3824
I can get it down to a point where I get $xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$

After this I don't know how to simplify it anymore..help please?
• March 10th 2010, 12:19 PM
pickslides
Quote:

Originally Posted by steph3824
I can get it down to a point where I get $xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$

This is not correct as your $v$ term is no longer present.

Have another go at expanding there should be 9 terms. You can then group by $v, v'$ and $v"$

$xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$
$xv''e^{x}-v''e^{x}+2xv'e^{x}-2v'e^{x}-xv'e^{x}=0$
$(xe^{x}-e^{x})v"+(2xe^{x}-2e^{x}-xe^{x})v'=0$
$(xe^{x}-e^{x})v"+(xe^{x}-2e^{x})v'=0$