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Reduction of Order

  1. Aug 19, 2010 #1
    1. The problem statement, all variables and given/known data

    solve y"-4y'+4y=0 y1=e^(2x) using reduction of order

    3. The attempt at a solution
    y2=uy=ue^2x
    y2'=u'e^2x+2ue^2x
    y2"=u"e^2x+4u'e^2x+4ue^2x

    I then substitute that into the original equation to get

    u"e^2x+4u'e^2x+4ue^2x-4u'e^2x-8ue^2x+ue^2x=0

    simplify to get
    u"e^2x=0

    from here I do not know what to do...I do know the answer is suppose to be xe^2x, but I don't know how that is done.
     
  2. jcsd
  3. Aug 19, 2010 #2

    rock.freak667

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    Homework Helper

    From u"e2x=0, you can divide by e2x and solve u''=0.
     
  4. Aug 19, 2010 #3
    ahh...so then u"=0 makes u'=c and then later u=xc1+c2 and
    y2=uy1
    y2=xc1*e^2x+c2*e^2x

    but what then? how do I solve for c1 and c2?
     
  5. Aug 19, 2010 #4

    Mark44

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    Staff: Mentor

    You need initial conditions in order to solve for the constants c1 and c2.
     
  6. Aug 19, 2010 #5
    however, in my solutions manual it says the solution comes out to be xe^2x, and I have no idea how that came to be. except for the use of this equation
    y2=y1S e^(-SP(x)dx)/y1^2 dx
     
  7. Aug 19, 2010 #6

    Mark44

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    Staff: Mentor

    The general solution of your diff. equation is y = c1e^(2x) + c2xe^(2), for any values of c1 and c2. The simplest pair of linearly independent solutions is the pair with c1 = c2 = 1, so maybe they just arbitrarily chose that one.
     
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