ANSWERS: 1
  • Is the 'n' in brackets? y^(n) is different to y^n y^(n) means y differentiated n times. if y = cos kt then y^(1) = - k sin kt y^(2) = - k^2 cos kt = - k^2 y so if n was 2 and k was (5/2) then this would work. y^(3) = k^3 sin kt y^(4) = k^4 cos kt y^(5) = - k^5 sin kt y^(6) = - k^6 cos kt = - k^6 y so if n was 6 and k was cbrt(5/2) then this would work In general, the solution is k = (25/4)^/(1/n) where n = 4m+2

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