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Disagreement with my teacher about linear DE's

  1. Jan 25, 2013 #1
    I think I may have just phrased what I meant poorly but..

    If I take any linear DE, consider the independent variable (t) to be some constant, and consider y and all its derivatives to be just standard variables, and I graph this function, it always is a linear object (line, plane, etc) with dimension depending on how many y/derivatives of y are in the differential equation.

    I don't see how this could be false. This is a direct consequence of y and all derivatives having a linear relation, which is the definition of a linear DE.
     
  2. jcsd
  3. Jan 25, 2013 #2
    t can't be considered a constant, considering it to be a constant would cause all meaning to be lost from the derivatives.

    Though, if I'm getting you correctly, you're replacing t with c (where c is an arbitrary constant) and f, f', f'', etc. with with independent variables and seeing if the resulting equation is linear wrt f, f', f'', and the other derivatives? I'm pretty sure it will be, but I don't really see why there's any reason to do this, especially since you just get meaningless jargon (again, due to the fact that the derivatives now mean nothing.) No offence, though, interesting idea!
     
  4. Jan 25, 2013 #3
    Yes, this is what I mean.

    I know that t can't "actually" be considered constant, but for the sake of attributing meaning to why the term "linear" is used I thought of it as being "ignored" moreso than constant.

    He stated that in math we use words that don't always fit (linear differential equation) but with this process it fits perfectly, although I'm not very good at expressing my ideas.

    Like..
    cos(yy') = 2t is nonlinear, and the graph of

    Y = arccos(C)/X = K/X

    Is not a line , where X is Y', C is a constant for 2t, and K is arccos(C) which is some other constant.

    Meanwhile

    2y' + ty = 0

    Is linear and

    Y = -2X/C

    Is a line with the same changes as before.
     
  5. Jan 25, 2013 #4

    lurflurf

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    2y' + ty = 0
    is linear in the in the sense that if
    y1 and y2 are solutions
    c1y1+c2y2 is a solutions
     
  6. Jan 25, 2013 #5
    This is true but I don't see what this has to do with the validity of what I am describing.
     
  7. Jan 25, 2013 #6

    lurflurf

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    I do not know what you are describing. Linear DE's are linear, that is why they are so named. Perhaps you are talking about the fact that what some call affine others call linear.

    If y1 and y2 are solutions
    then c1y1+c2y2 is a solutions
    is what linear means

    (if Ly=f instead of Ly=0
    If y1 and y2 are solutions
    then (c1y1+c2y2)/(c1+c2) is a solutions
    is what linear means
     
    Last edited: Jan 25, 2013
  8. Jan 25, 2013 #7
    What I am describing is just an "idea" that arose when we were discussing the origins of the term linearity.

    I understand that linear DEs are linear regardless of this idea (for the exact reasons you said) I am just providing an origin for what led me to think about it. I am not asking what makes a linear DE linear, I'm asking if the idea I've outlined in my first post is a true one.
     
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