- **Differential Equations**
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- - **2nd order with exponential and constant on right side**
(*http://www.physicsforums.com/showthread.php?t=239436*)

2nd order with exponential and constant on right sideHi everybody,
How do I solve this differential equation ??: y'' = a(Exp(-b*y)-1) ; where a, b are constants with the boundaries conditions : y'(x=0)=-K1 y'(x=L)=0 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 |

Welcome to PF!Hi ADGigus! Welcome to PF! :smile:
easy-peasy … ∫ y' (a Exp(-b y) - a) = ∫ay'Exp(-b y) - ∫ay' :smile: |

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. so 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... :-( |

Hi ADGigus! :smile:
Yes … I'm stuck too. :frown: Are you sure it's not y'' = a(Exp(-b*y-1)) ? :smile: |

it's okYes 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. |

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