2nd order differential equation

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
sara_87
Messages
748
Reaction score
0

Homework Statement



[tex]\epsilon[/tex]y'' - y' = 0
y(0)=0, y(1)= 1


The Attempt at a Solution



this s a perturbation problem, but I'm not sure how to go about doing it
if we let epsilon = 0 we'll go from 2nd order to first order so it's a boundary layer problem... i think
so if that's the case then i have to find the outer and inner layer
to find the outer layer i let epslon =0 and i get y' = 0

then to find the inner layer i let x=[tex]\epsilon[/tex]nX

then after a few steps i got to this:
1/[tex]\epsilon[/tex] d2y/dX2 - 1/[tex]\epsilon[/tex]dy/dX

and then solve it normally then find the general solution etc.

and also I'm letting y = y0+ epsilony1 + epsilony2...
and y' = ...
and y''= ...

and subbing them in and solving it too... i can do that, even though there's and epsilon?
and then I'm supposed to compare these together?
 
Physics news on Phys.org
i didn't mean for the epsilon to be floating like that
the epsilon is meant to be firmly on the ground... hope this helps
 
Last edited:
You have two sets of solutions, one outer solution and one inner solution.

To find the inner solution, set [tex]x=\delta (\varepsilon )X[/tex] and look for the dominant balance.