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y=c1sin x + c2cos x is a solution to y''+y=0. a. determine whether there are one or more members of this family that satisfy the conditions y(0)=0 and y(pi)=0. b. Show that the zero function, y(triple bar)0, is the only member of the family that satisfies the conditions y(0)=0 and y(pi/2)=0.

Mathematics
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For a... are you just supposed to substitute the 0 for x, and then the pi for x... b/c then you get 1 and -1, not zero... but what does that tell me?
and for b, Im unsure of what it is really asking me?
y(0)=c2, not 1 you are trying to find conditions for c1 and c2 that satisfy the conditions

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y(0)=c2 y(pi)=-c2 what requirement does this put on c2 ? what about on c1 ?
well if y(0)=c2, then c2 must be equal to 0 right? and I guess it really doesnt matter what c1 is does it?
exactly, so the answer for the first part is that c2=0 and c1 can be any real number. for the next part, what system of equations do you get by plugging in x=0 and x=pi/2 ?
well when x =0 then all you have is y=c2 and if you plug in pi/2 you get y=c1
yes, and we are told that y(0)=0 and y(pi/2)=0, so what are c1 and c2 ?
well then both have to be 0?
yep, so then we just have that y(x)=0 for all x, also known as the zero function http://mathworld.wolfram.com/ZeroFunction.html
okay, thank you. makes sense. but what is the difference between y=0, and y(triplebar)0?
I have never heard of the triple bar notation. I think they are just using it to emphasize that y(x)=c1y2(x)+c2y2(x)=(0)y1+(0)y1=0 as opposed to just having either the c1y1 or c2y2 term =0
alright.. well thank you again. much appreciated.
it's like a trivial solution vector (c1=0, c2=0, c3=0, ..., cn=0)\[f(x)=\vec 0\]I usually see the triple bar thing used for definitions, but here I suppose it means this. That's all I know :p
you're welcome!

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