Question

У"-у,_6y 8. y'(0)-0 -cost y(0)-0 9. y"-y,-6y-3 sin2t y'(0) y(0)-1 0

Answer 1

y^11 - y^1 - 6y = cos t y(0) = 0, y^1 (0) = 0----(1) rightarrow auxillary equation of given circle is m^2 - m - 6 = 0 (m-3) (m + 2) = 0 m-3 = 0 or m = -2 rightarrow m = 3,-2 therefore the complimentary solution is given by y(t) = c_1 e^-2t + c_2 e^3t now we find particular solution consider y = A cos t + B isn t ----(*) is solution of given circle y^1 = -A sin t + B cos t y^11 = -A cos t - B sin t therefore equation (1) become (-A cos t - B sin t) - (- A sin t + B cos t) - 6 (A cos t + B sin t) = cos t rightarrow (-A - B - 6A) cos t + (- B + A - 6B) sin t = cos t compairing coefficient on both sides -7A - B = 1----(3) A - 7B = 0----(4) (3) + 7 times (4) -7A - B = 1 7A - 49B = 0 -50B = 1 rightarrow B = -1/50 therefore A = 7B A = -7/50 from equation (*) y_p (t) = -7/50 cos t - 1/ 50 sin t