Question Part Points Submissions Used 1 0 /1 2/5 Total 0 /1 In this problem, x =

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Question Part Points Submissions Used 1 0/1 2/5 Total 0/1 In this problem, x = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions-- x(pi/6)=1/2, x'(pi/6)= 0 Question Part Points Submissions Used 1 0/1 2/5 Total 0/1 In this problem, x = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions-- x(pi/6)=1/2, x'(pi/6)= 0 Question Part Points Submissions Used 1 0/1 2/5 Total 0/1 Question Part Points Submissions Used 1 0/1 2/5 Total 0/1 Question Part Points Submissions Used 1 0/1 2/5 Total 0/1 Question Part Points Submissions Used 1 0/1 2/5 Total 0/1 Question Part Points Submissions Used 1 0/1 2/5 1 0/1 2/5 Total 0/1 In this problem, x = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions-- x(pi/6)=1/2, x'(pi/6)= 0 In this problem, x = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions-- x = c1 cos t + c2 sin t x'' + x = 0. x(pi/6)=1/2, x'(pi/6)= 0 Question Part Points Submissions Used

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sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
sin(x+PI/6) = sin(x) sin(PI/6)+ cos(x) cos(PI/6)
sin(PI/6)=1/2
cos(PI/6)=%u221A3/2

sin(x+PI/6) = sin(x) (1/2) + cos(x) (%u221A3/2)

sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
sin(x-PI/6) = sin(x) sin(PI/6)- cos(x) cos(PI/6)

sin (x+(pi/6)) - sin (x-(pi/6)) =sin(x) (1/2) + cos(x) (%u221A3/2) -sin(x) (1/2)+ cos(x) (%u221A3/2)
2 cos(x) (%u221A3/2) = 1/2
cos(x) %u221A3 = 1/2
cos(x) = %u221A3/2

x=PI/3 & x = 5PI/3

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