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jeter31681
04-24-2007, 01:33 AM
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
04-24-2007, 02:41 PM
How many different boards did you put this on. I feel like I'm chasing you around! :p Well, since this may be people who haven't seen my other replies:
The general solution to y"+y= 0 is y(x)= Ccos(x)+ Dsin(x) whence y'(x)= -C sin(x)+ D cos(x). y(1)= Ccos(1)+ Dsin(1) and y(-1)= Ccos(-1)+ D sin(-1). y'(1)= -Csin(1)+ D cos(1) and y'(-1)= -Csin(-1)+ D cos(-1). Your conditions become y(1)+ y(-1)= Ccos(1)+ Dsin(1)+ Ccos(-1)+ Dsin(-1)= 0 and -Csin(1)+ D cos(1)- Csin(-1)+ Dcos(-1)= 2. Use the fact that cosine is an even function and sine is an odd function to reduce those two equations and solve for C and D.

Epsilon=One
04-24-2007, 02:51 PM
How many different boards did you put this on. I feel like I'm chasing you around! :p Well, since this may be people who haven't seen my other replies...Thanks for the "re-effort."

I'm one of those single focused "people."

jeter31681
04-24-2007, 02:55 PM
How many different boards did you put this on. I feel like I'm chasing you around! :p Well, since this may be people who haven't seen my other replies:
The general solution to y"+y= 0 is y(x)= Ccos(x)+ Dsin(x) whence y'(x)= -C sin(x)+ D cos(x). y(1)= Ccos(1)+ Dsin(1) and y(-1)= Ccos(-1)+ D sin(-1). y'(1)= -Csin(1)+ D cos(1) and y'(-1)= -Csin(-1)+ D cos(-1). Your conditions become y(1)+ y(-1)= Ccos(1)+ Dsin(1)+ Ccos(-1)+ Dsin(-1)= 0 and -Csin(1)+ D cos(1)- Csin(-1)+ Dcos(-1)= 2. Use the fact that cosine is an even function and sine is an odd function to reduce those two equations and solve for C and D.

two... :D