Just need help simplifying

**steph3824** · March 9th 2010, 07:37 PM

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

**pickslides** · March 9th 2010, 07:44 PM

Hi Steph, expand it all out and group together in terms of $v, v'$ and $v''$

**steph3824** · March 9th 2010, 07:52 PM

That's what I'm doing but I don't think I am doing it correctly

**steph3824** · March 10th 2010, 07:27 AM

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?

**pickslides** · March 10th 2010, 12:19 PM

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 $v"$

This is how you do it with your previous answer

$xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$

$xv''e^{x}-v''e^{x}+2xv'e^{x}-2v'e^{x}-xv'e^{x}=0$

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

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

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