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 one year ago
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.
 one year ago
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.

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zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0For 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?

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0and for b, Im unsure of what it is really asking me?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1y(0)=c2, not 1 you are trying to find conditions for c1 and c2 that satisfy the conditions

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1y(0)=c2 y(pi)=c2 what requirement does this put on c2 ? what about on c1 ?

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0well if y(0)=c2, then c2 must be equal to 0 right? and I guess it really doesnt matter what c1 is does it?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1exactly, 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 ?

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0well when x =0 then all you have is y=c2 and if you plug in pi/2 you get y=c1

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1yes, and we are told that y(0)=0 and y(pi/2)=0, so what are c1 and c2 ?

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0well then both have to be 0?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1yep, so then we just have that y(x)=0 for all x, also known as the zero function http://mathworld.wolfram.com/ZeroFunction.html

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0okay, thank you. makes sense. but what is the difference between y=0, and y(triplebar)0?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1I 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

zonazoo
 one year ago
Best ResponseYou've already chosen the best response.0alright.. well thank you again. much appreciated.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1it'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
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