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2nd order with exponential and constant on right side

  1. Jun 9, 2008 #1
    Hi everybody,

    How do I solve this differential equation ??:

    y'' = a(Exp(-b*y)-1) ;

    where a, b are constants

    with the boundaries conditions :


    without the constant term I can do

    y''*y' = y' a Exp(-b y)

    then integrate it

    [tex]\ {1/2} (y')^2= {a/b} ~Exp(-b y)[/tex]

    and so on and finaly find something in hyperbolic function Tanh()....

    but With the constant term " -a " on the right side, I don't know how to start.

    Thank very much for your help
  2. jcsd
  3. Jun 9, 2008 #2


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    Welcome to PF!

    Hi ADGigus! Welcome to PF! :smile:

    easy-peasy …

    ∫ y' (a Exp(-b y) - a)

    = ∫ay'Exp(-b y) - ∫ay' :smile:
  4. Jun 9, 2008 #3
    ok but my problem is in the followings...

    Thank you for your very quick answer and the welcome.

    Yes I agree til there it's easy but it's the following that gives me troubles.

    starting from :

    [tex] y'' = a Exp(-b y) -a [/tex]

    [tex] y ' y'' = y ' a Exp(-b y) -a y ' [/tex]

    after the integration on y on both sides, I have :

    [tex] 1/2~(y ')^2 = -a/b~ Exp(-b y) -a y + K [/tex]

    [tex]y' =\sqrt{-2a (1/b ~ Exp(-b y)-y) + K}[/tex]

    and then I'm really stuck... :-(
    Last edited: Jun 9, 2008
  5. Jun 9, 2008 #4


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    Hi ADGigus! :smile:

    Yes … I'm stuck too. :frown:

    Are you sure it's not y'' = a(Exp(-b*y-1)) ? :smile:
  6. Jun 9, 2008 #5
    it's ok

    Yes I'm sure it's correct,

    it's coming from the expression of a Schottky diode (current density vs Voltage)

    so it's :

    J = a(Exp(-bV) -1)

    because you want 0 current at bias of 0 Volts....

    I already try many way to solve this equations, til now I'm fully stuck

    Even numerically I don't find any correct way to solve it.

    May be there is some polynomial approximation to use, but I don't really find anything fine.

    if we take the developpement of exponential at the first order it's easy (1+V), but it's insufficient, I lose the diode physic and it does like it's something linear, so ohmic. And It's not enough for my calculation.
    Last edited: Jun 9, 2008
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