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2nd order differential equation using reduction of order

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Use the method of reduction of order to find a second solution of the given differential equation

    t2y'' - 4ty' + 6y, t>0; y1(t) = t2


    3. The attempt at a solution

    Here's what I have so far:
    y = vt2
    y' = 2tv + t2v'
    y'' = 2v + 4tv' + t2v''

    so

    t2 (2v + 4tv' + t2v'') - 4t (2tv + t2v') + 6(vt2) = 0

    t4v'' = 0

    I'm stuck here. Am I supposed to divide by y, which would make it t2v'' = 0
    wouldn't that make v'' = 0 anyways? What do I do now? I know i'm supposed to integrate it some how but I am just utterly lost from here...

    PLEASE HELP!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 9, 2008 #2

    Hootenanny

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    You're absolutely spot on here. Let's look at your final line

    t4v''=0

    Now the boundary conditions state that t is positive, hence this means that t4 must be non-zero. Furthermore, this implies that

    v''=0

    Now you have a second order differential equation

    [tex]\frac{d^2 v}{dt^2}=0[/tex]

    Simply integrate twice with respect to t. Do you follow?
     
  4. Oct 9, 2008 #3
    So the 2nd general solution should be y2 = t6 / 18 ?
     
  5. Oct 9, 2008 #4

    Hootenanny

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    How did you get that?
     
  6. Oct 9, 2008 #5
    integrate twice with respect to t

    integral of t4v'' = t5 / 3
    integral of t5 / 3 = t6 / 18?
     
  7. Oct 9, 2008 #6

    Hootenanny

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    That's wrong. As I said in my previous post, you need only solve the ODE

    v''=0

    And then multiply the solution by y1 to obtain the general solution.
     
  8. Oct 9, 2008 #7
    How do i solve the ODE?
     
  9. Oct 9, 2008 #8

    Hootenanny

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    As I said before integrate twice with respect to t.
     
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