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

ADGigus Jun9-08 07:22 AM

2nd order with exponential and constant on right side
 
Hi 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

tiny-tim Jun9-08 08:21 AM

Welcome to PF!
 
Hi ADGigus! Welcome to PF! :smile:

easy-peasy

∫ y' (a Exp(-b y) - a)

= ∫ay'Exp(-b y) - ∫ay' :smile:

ADGigus Jun9-08 08:48 AM

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... :-(

tiny-tim Jun9-08 09:05 AM

Hi ADGigus! :smile:

Yes I'm stuck too. :frown:

Are you sure it's not y'' = a(Exp(-b*y-1)) ? :smile:

ADGigus Jun9-08 09:14 AM

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.


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