2nd order Differential Eq. - Reduction of Order

[leonida]

I have a problem with differential equations - 2nd order - reduction of order

my problem is as follows:
$$(x − 1)y" − xy' + y = 0 , x > 1 ; y_1(x) = e^x$$

solving this type of diff. eq. says to use $$y=y_1(x)V(x)$$ which gives me $$y=Ve^x$$ differentiating y gives me
$$y'=V'e^x$$ &
$$y''=V''e^x$$

when pluged into original equation i have

$$(x-1)e^xV''-xe^xV'=0$$ with substitution $$V'=u$$

from this point on i am not sure whether i should omit (x-1) since x>1 and cannot be zero, or should i include it. But no matter which road i take, i get a solution that includes some combination of e^x . book gives me solution as x, which, upon check is the right solution.. help how to get there is appreciated !

 

[206PiruBlood]

Don't forget the product rule when differentiating.

 

[leonida]

206,
thanks for the help. i always forget to apply this rule.. but still this does not lead me to the answer...

so.. i Have
$$y=Ve^x$$
$$y'=V'e^x + Ve^x$$
$$y''=V''e^x + 2V'e^x+Ve^x$$
pluging back in original equations gives me
$$(x-1)V''e^x+2V'e^x+V'xe^x=0$$
setting all the members with V=0 and factoring out e^x, since it cannot be zero i am left with
$$(x-1)V''+V'(x+2)=0$$
substituting V'=u
$$u'=-u (x+2)/(x-1)$$
$$du/u=-(x+2)dx/(x-1)$$
this gives me wild answer, and i need to be at y₂=xhelp please...