Two-Point Boundary Value Problem

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Jamin2112
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Homework Statement



y'' + ßy = 0, y'(0)=0, y'(L)=0

Homework Equations



Meh

The Attempt at a Solution



I so already did the ß>1 and ß<1; I'm stuck on the ß=0. It seems easy enough. y'' = 0 -----> y' = A -----> 0=A, 0=A (from the two initial conditions) ------> No non-trivial solution.

But...

The answer in the back of the book says ß0=0, y0(x)=1; ...


??
 
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Jamin2112 said:

Homework Statement



y'' + ßy = 0, y'(0)=0, y'(L)=0

Homework Equations



Meh

The Attempt at a Solution



I so already did the ß>1 and ß<1; I'm stuck on the ß=0. It seems easy enough. y'' = 0 -----> y' = A -----> 0=A, 0=A (from the two initial conditions) ------> No non-trivial solution.

Starting with y'' = 0 you need to integrate twice to get y, getting two constants.
 
LCKurtz said:
Starting with y'' = 0 you need to integrate twice to get y, getting two constants.

The boundaries it gives me are in terms of y' :wink:


Of course I know
y'' = 0 -----> y' = B ----> y = Ax + b
 
Jamin2112 said:
The boundaries it gives me are in terms of y' :wink:


Of course I know
y'' = 0 -----> y' = B ----> y = Ax + b

But the point is: does your boundary condition force only the trivial solution or can you get a non-trivial solution in this case?