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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.
 one year ago
 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
 one year ago

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zonazooBest ResponseYou've already chosen the best response.0
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?
 one year ago

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

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

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

zonazooBest ResponseYou've already chosen the best response.0
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?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

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

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

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

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago
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