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Orders and Degrees

  1. Jan 6, 2007 #1
    I was taking notes in class and the prof said that in the equation

    [tex]y^(double prime)-3y^(prime)+2y = 0[/tex] that 2 was the order.. is that due to the following?

    the first y = 0 when you take the derivative twice
    the second y = 0 when you derive it once
    and the first one = 2 when you derive it once

    Just a question.. shouldn't I derive the first one 3 time, and the second one 2 times, because you derive 2y once?
  2. jcsd
  3. Jan 6, 2007 #2


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    The order of a differential equation is defined as the highest derivative that the equation contains.

    Let's rewrite this as [tex]\frac{d^2y}{dx^2}-3\frac{dy}{dx}+2y=0[/tex]

    Now, since the highest derivative in this equation is [tex]\frac{d^2y}{dx^2}[/tex] the equation is a second order differential equation.

    I'm not really sure what you're doing here!
  4. Jan 6, 2007 #3
    the following is an attachment of my notes:

    I don't get why the order is 2 and not -3

    Attached Files:

  5. Jan 6, 2007 #4


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    Well, read my above post!

    Why do you think that the order is -3? The only way I can see you getting this is by looking at the coefficient in front of the y' term, and I'm not sure why you're doing that!
  6. Jan 6, 2007 #5
    because I don't understand
  7. Jan 6, 2007 #6
    think about it like the degree of a polynomial.
  8. Jan 6, 2007 #7
    you know what... I thought it was the number that comes out after you derive it a few times.. not how many times you derive it

    i get it now, thanks
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