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Reduction of order

  1. Oct 15, 2006 #1
    Use reduction of order to find a second solution of the given differential equation
    t^2y"-4ty'+6y=0 t>0 y_1(t)=t^2

    first I put in in the standard form
    y"-4/ty'+6/t^2y =0
    then y= v(t)t^2
    y'=v'(t)t^2+v(t)2t
    y"=y"(t)t^2+v'(t)2t+v'(t)2t+v(t)2 = t^2v"(t)+4tv'(t)+2v(t)
    putting these into the original equation gives me
    t^2v"(t)+4tv'(t)+2v(t)-4tv'(t)-8v(t)+6t^2v(t)
    simplifying gives me
    t^2v"(t)+6t^2v(t)-6v(t)
    I don't know what to do from here. can someone please help?
     
  2. jcsd
  3. Oct 16, 2006 #2

    HallsofIvy

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    The is wrong. You have added 6y but it should be [itex]\frac{6}{t^2}y[/itex], just 6v, not 6t^3 v.
    You should have
    [tex]t^2v"+ 4tv'+ 2v- 4tv'- 8v+ 6v= t^2v"= 0[/itex]
    That should be easy!

     
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