#$&* course Mth 279 5/10 Query 20 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Using variation of parameters, solve the equation y '' + 36 y = csc^3 ( 6 t ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First solving for the homogeneous solution: P(a) = a^2 + 36 = 0 y_c(t) = Asin(6t) + Bcos(6t) Then assuming y_p(t) = sin(6t)u1(t) + cos(6t)u2(t) and sin(6t)u1'(t) + cos(6t)u2'(t) = 0 [sin(6t) cos(6t)] [u1'(t)] = [ 0 ] [6cos(6t) -6sin(6t)] [u2'(t)] [csc^3(6t) ] W(t) = -6sin^2(6t) - 6cos^2(6t) = -6(sin^2(6t) + cos^2(6t)) = -6 u1'(t) = -y2(t)g(t)/W(t) = -cos(6t)csc^3(6t)/-6 = -cos(6t)/(-6sin^3(6t)) = cot(6t)csc^2(6t)/6 u2'(t) = y1(t)g(t)/W(t) = sin(6t)csc^3(6t)/-6 = sin(6t)/(-6sin^3(6t)) = -csc^2(6t)/6 [u1'(t)] = [cot(6t)csc^2(6t)/6] [u2'(t)] [-csc^2(6t)/6] u1(t) = (1/6)Int(cot(6t)csc^2(6t))dt = (-1/72)cot^2(6t) + c u2(t) = (-1/6)Int(csc^2(6t))dt = (1/36)cot(6t) + c2 ----------------------------------------------------------------------- y_c(t) = Asin(6t) + Bcos(6t) y_p(t) = (-1/72)cot^2(6t) sin(t) + (1/36)cot(6t) cos(t) y_p(t) = (-1/72)(cos^2(6t)/sin^2(6t)) sin(t) + (1/36)(cos(6t)/sin(6t)) cos(t) y_p(t) = (-1/72)(cos^2(6t)/sin(6t)) + (1/36)(cos^2(6t)/sin(6t)) y_p(t) = (-1/72)(cos(6t)cot(6t)) + (1/36)(cos(6t)cot(6t)) y_p(t) = (1/72)(cos(6t)cot(6t)) y(t) = Asin(6t) + Bcos(6t) + (1/72)cos(6t)cot(6t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:"