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ODE in terms of y and x

  1. Mar 31, 2014 #1
    Which the difference between diff equations of kind: [tex]\frac{dy}{dx} = \exp(x)[/tex] [tex]\frac{dy}{dx} = 1/x[/tex]
    and diff equations of kind:
    [tex]\frac{dy}{dx} = y[/tex] [tex]\frac{dy}{dx} = \frac{1}{\exp(y)}[/tex] ?
     
  2. jcsd
  3. Mar 31, 2014 #2

    Simon Bridge

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    Nothing.
    Well... apart from the obvious.

    You can call the variables anything you like.
    Try solving them and see what you get.
     
  4. Apr 1, 2014 #3
    A general linear diff equation of 1st order have a form: [tex]f(x)y'(x)+g(x)y(x)=h(x)[/tex] The 3 first equations fits into this definition, but the last no. However, the last equation is linear too. This is a contradiction?
     
  5. Apr 1, 2014 #4
    Nothing,
    As already mentioned both are the same.
    What do you mean by that?
     
  6. Apr 1, 2014 #5
    Note that when you want solve an equation you must specify the independent variable , first tell us what variable your are assuming as independent one,
     
    Last edited: Apr 1, 2014
  7. Apr 1, 2014 #6

    Mark44

    Staff: Mentor

    No, the last equation is not linear. In a linear differential equation y, y', y'', ... occur to the first power only, but can be multiplied by functions of the independent variable. With the right side being e-y, the equation is not linear.
     
  8. Apr 1, 2014 #7

    Mark44

    Staff: Mentor

    I'm assuming that the independent variable is x, and y is the dependent variable. That seems fairly clear from what Jhenrique wrote.
     
  9. Apr 1, 2014 #8
    But the antiderivative's solution of ##\frac{dy}{dx} = \frac{1}{x}## is log(x) and the antiderivative's solution of ##\frac{dy}{dx} = \frac{1}{\exp(y)}## is log(x) too. So why the first is linear and the second no?
     
  10. Apr 1, 2014 #9

    Mark44

    Staff: Mentor

    Two things:
    1. The solution of a differential equation has nothing to do with whether the equation is linear or not.
    2. The two equations do NOT have the same solution.

    ## \frac{dy}{dx} = e^{-y}##
    ##\Rightarrow e^y dy = dx##
    ##\Rightarrow \int e^y dy = \int dx##
    ##\Rightarrow e^y = x + C##
    ##\Rightarrow y = ln(x + C)##
    This solution is different from the one for dy/dx = 1/x.
     
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