Two-Point Boundary Value Problem

[Roni1985]

Homework Statement

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

Homework Equations

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.

 

[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

 

[Roni1985]

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

Hello,


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

 

[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

 

[Roni1985]

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

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

thanks.

 

[lanedance]

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