x\'\'+x=0 is x=c 1 cos(t)+c 2 sin(t). Find c 1 and c 2 so that x(/4)=2 and x\'(/

Question

x''+x=0 is x=c1cos(t)+c2sin(t). Find c1 and c2 so that x(/4)=2 and x'(/4)=22.

Explanation / Answer

x=c1cos(t)+c2sin(t) given x(pi/4)=2^(1/2) substitute t = pi/4 in x we get x(pi/4) = C1cos(pi/4) + C2sin(pi/4) we know cos(pi/4) = sin(pi/4) = 1/2^(1/2) = 2^(-1/2) and x(pi/4)=2^(1/2) so 2^(1/2) = C1 2^(-1/2) + C2 2^(-1/2) 2^(1/2) * 2^(1/2) = C1 2^(-1/2) *2^(1/2) + C2 2^(-1/2) * 2^(1/2) (by multiplying with 2^(1/2) on both sides so 2 = C1 + C2 ------>>>>>(1) and x' = [C1cos(t)]+[C2sin(t)] = -C1 sin(t) + C2 cos(t) (since derivative of sin(t) is cos(t) and that of cos(t) is -sin(t)] given x'(pi/4) = 2^(3/2) so substituting this in above x' we get 2^(3/2) = -C1 cos(pi/4) + C2 sin(pi/4) = [-C1 + C2]2^(-1/2) again multiplying with 2^(1/2) on both sides we get 2^2 = -C1 + C2 = 4 ------>>>>>(2) by adding equations (1) and (2) we get 2 * C2 = 6 ---->>>> C2 = 3 and C1 = 1 - 3 = -2 so C1 = -2 and C2 = 3

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