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Differential equation---a conceptual problem

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data
    I am reading a note on differential equation.There is a point that I don't understand,hopefully someone can explain
    (Please see the attched)
    2. Relevant equations


    3. The attempt at a solution
    The notes wrote " a1t + b1 x + c1 = a1T + b1X
    a2t + b2t + c2 = a2T + b2X

    the following sentence dX =dx + 0 and dT = dt+0
    What does it mean?
    I guess it was something like dX/dT = d(x+ħ)/d(t+τ)?But then I have no idea.This is the first time I have met this kind of case...
    Can someone explain,thx so much 10848927_1222013937814418_6698737448882156997_o.jpg
     
  2. jcsd
  3. Oct 13, 2015 #2

    Ray Vickson

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    It says to choose ##\tau## and ##\eta## such that ##\bar{x} = x + \eta## and ##\bar{t} = t + \tau## obey the two equations you wrote above. That tells you how to find ##\tau## and ##\eta##, and since the ##a_i## and ##b_i## are constants, so are ##\tau## and ##\eta##. Thus, ##d \bar{x} = dx + d \eta = dx##, because ##d \eta = 0## (##\eta## is a constant).
     
  4. Oct 13, 2015 #3
    But why does dn = 0? I have only learnt dy/dx,I dont know what does it mean when they are separated(and so I post this thread :/)...
     
  5. Oct 13, 2015 #4

    Ray Vickson

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    You don't know why the derivative of a constant is zero? Go back to your elementary calculus notes!
     
  6. Oct 13, 2015 #5
    Yes,I understand dy/dx = 0 if y is a constant.But the note said dy = 0,so the bottom dx is vanished,and that's make me confused
     
  7. Oct 13, 2015 #6

    Ray Vickson

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    I don't see any ##y## or ##dy## anywhere in the note.
     
  8. Oct 13, 2015 #7
    Sorry,I mean d(something)
    From the notes:
    dX = dx + 0,dT = dt + 0

    The problem is that I don't know what's going on because usually when you differentiate a function,say y = x^2,you write:
    dy/dx = d(x^2)/dx = 2x ,the Leibniz's notation is written as dy/dx.In my notes,just like dX = dx + 0,and the "denominator" in the standard dy/dx notation vanishes,so that's why I don't understand the statement above.
     
  9. Oct 13, 2015 #8

    phyzguy

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    Writing it this way is a shorthand called implicit differentiation. You have:
    [tex] \bar x = x + \eta; \frac{d \bar x}{d x} = 1 + 0[/tex]
    Now multiply through by dx and write:
    [tex] d \bar x = dx + 0 [/tex]

    After a bit of practice, you just skip the intermediate step and write immediately:
    [tex] d \bar x = dx + 0 [/tex]
     
  10. Oct 13, 2015 #9
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