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Homework Help: 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...
     
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