2nd order differential equation

  • Thread starter sara_87
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  • #1
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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 thats 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 theres and epsilon?
and then i'm supposed to compare these together?
 

Answers and Replies

  • #2
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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:
  • #3
hunt_mat
Homework Helper
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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.
 

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