2nd order Differential Eq. - Reduction of Order

  • Thread starter leonida
  • Start date
  • #1
10
0
I have a problem with differential equations - 2nd order - reduction of order

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

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

when pluged into original equation i have

[tex](x-1)e^xV''-xe^xV'=0[/tex] with substitution [tex] V'=u[/tex]

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 ex . book gives me solution as x, which, upon check is the right solution.. help how to get there is appreciated !
 

Answers and Replies

  • #2
106
2
Don't forget the product rule when differentiating.
 
  • #3
10
0
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
[tex]y=Ve^x[/tex]
[tex]y'=V'e^x + Ve^x[/tex]
[tex]y''=V''e^x + 2V'e^x+Ve^x[/tex]
pluging back in original equations gives me
[tex](x-1)V''e^x+2V'e^x+V'xe^x=0[/tex]
setting all the members with V=0 and factoring out ex, since it cannot be zero i am left with
[tex](x-1)V''+V'(x+2)=0[/tex]
substituting V'=u
[tex]u'=-u (x+2)/(x-1)[/tex]
[tex]du/u=-(x+2)dx/(x-1)[/tex]
this gives me wild answer, and i need to be at y2=x


help please...
 

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