Solving "y''(x) +y(x)=0" w/ BC's of y(1)+y(-1)=0 & y'(1)+y'(-1)=2

[jt316]

how would i find the exact solution to:

y''(x) +y(x)=0

with BC's of:

y(1)+y(-1)=0

y'(1)+y'(-1)=2

a little confussing because the BC's I am used to dealing with are like:
y(0)=0
y(1)=0

 

[HallsofIvy]

You really have to think a little. You know, I hope, that the general solution to y"+ y= 0 is y(x)= Ccos(x)+ Dsin(x) so that y'(x)= -Csin(x)+ Dcos(x). Now put those functions into your boundary conditions:
y(1)+ y(-1)= Ccos(1)+ Dsin(1)+ Ccos(-1)+ Dsin(-1)= 0
y'(1)+ y'(-1)= -Csin(1)+ Dcos(1)- Csin(-1)+ Dcos(-1)= 2.
Using the fact that cosine is an even function and sine is an odd function will simplify these a lot.

Didn't we just do this in another thread?

 

[jt316]

You really have to think a little. You know, I hope, that the general solution to y"+ y= 0 is y(x)= Ccos(x)+ Dsin(x) so that y'(x)= -Csin(x)+ Dcos(x). Now put those functions into your boundary conditions:
y(1)+ y(-1)= Ccos(1)+ Dsin(1)+ Ccos(-1)+ Dsin(-1)= 0
y'(1)+ y'(-1)= -Csin(1)+ Dcos(1)- Csin(-1)+ Dcos(-1)= 2.
Using the fact that cosine is an even function and sine is an odd function will simplify these a lot.

Didn't we just do this in another thread?

I see where I was messing up. I found y(1),y(-1),y'(1) & y'(-1) but I was trying to plug the boundary conditions into these equations instead of pluging these into my boundary conditions. Thanks for the insight.