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A problem about reduction of the order of a linear ODE

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that if y1 is a solution to the ODE y'''+a2y''+a1y'+a0y=0 then the substitution y=uy1 reduces the order of the equation to a 2nd order linear ODE.

    3. The attempt at a solution
    well, I calculated first, second and third derivatives of y and plugged them in the equation and after cancellation and some tedious algebraic operations I obtained this new equation:
    (3y1''+2a2y1'+ a1y1)u' + (3y1' + a2y1)u'' + u'''y1=0

    Now I'm stuck and don't know what I should do next at this point. We still have the third derivative of u in the equation and I don't know how to cancel that.
     
    Last edited: Nov 4, 2011
  2. jcsd
  3. Nov 4, 2011 #2

    LCKurtz

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    The first term should be y'''.

    Presumably you mean y = uy1.
    I didn't work it all out but, assuming you have checked your work, you can just let v = u' and you will have a second order equation in v.
     
  4. Nov 4, 2011 #3
    Yes.

    Yes.

    That's a good idea. but do you think the professor wanted us to do that at this step? I mean does he accept this as the solution to his problem?
     
    Last edited: Nov 4, 2011
  5. Nov 4, 2011 #4

    LCKurtz

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    Of course, you need an "= 0" on the right side to make it an equation. But yes, or course, you have reduced the problem of solving a 3rd order equation to one of solving a 2nd order equation.
     
  6. Nov 4, 2011 #5
    Fine. the next question asks me that If we have two linearly independent solutions y1,y2 for this ODE we can find the 3rd solution. Can I use the idea of order reduction from this problem to solve that? If yes, how?
    Thanks for helping.
     
  7. Nov 4, 2011 #6

    LCKurtz

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    I haven't ever tried that since it never comes up. My guess is that uy2 would lead to a second solution to your u equation allowing you to reduce it to a first order. So I expect the answer is yes. I guess all you can do is try it.
     
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