# Two-Point Boundary Value Problem

1. May 8, 2010

### Roni1985

1. The problem statement, all variables and given/known data

y''+$$\lambda$$y=0
y'(0)=0
y'(pi)=0

2. Relevant equations

3. The attempt at a solution

What's puzzling me is the case when we check if the eigenvalue is zero.
y''=0
y'=C1
y=C1x+C2

Now when I check the first boundary value I get C1=0
now How do I check the second one ? with the pi...
It doesn't make sense plugging into the first derivative again because I have no x value (only a constant).
The answers show this:

lambda=0 is an eigenvalue and the general solution is y0(x)=1

How did they get this ?

Thanks.

2. May 8, 2010

### lanedance

y'=C1
y'(0)=0, y'(pi)=0 --> C1 = 0
so
y=C2, constant, which up to a scalar multiple is the same as 1

3. May 8, 2010

### Roni1985

Hello,

I didn't really understand your last statement.
the constant is the scalar multiple of 1 ?
I think I'm missing something...

4. May 8, 2010

### lanedance

whenever you find it solution to a DE it is only determined up to a scalar multiple

in this case any constant satisfies the DE & bc's

5. May 8, 2010

### Roni1985

Oh, right....
but why did they choose
Yo(x)=1
an arbitrary choice ?

thanks.

6. May 8, 2010

### lanedance

based on what you've said, yeah i think so

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