# Thread: Just need help simplifying

1. ## 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..

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

So what the heck does this simplify out to because I don't think I'm doing it right

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

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

4. I can get it down to a point where I get $\displaystyle xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$

After this I don't know how to simplify it anymore..help please?

5. Originally Posted by steph3824
I can get it down to a point where I get $\displaystyle xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$
This is not correct as your $\displaystyle v$ term is no longer present.

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

This is how you do it with your previous answer

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

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

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

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