1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

2nd order Differential Eq. - Reduction of Order

  1. Oct 6, 2012 #1
    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 !
     
  2. jcsd
  3. Oct 6, 2012 #2
    Don't forget the product rule when differentiating.
     
  4. Oct 7, 2012 #3
    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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 2nd order Differential Eq. - Reduction of Order
  1. 2nd order diff eq. (Replies: 1)

Loading...