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Acosxbsinxc/sinx x^n cosx nandinidivya561 is waiting for your help Add your answer and earn pointsAnswer (1 of 3) y = tan^1 (a cos x b sin x)/(b cos x a sin x) Divide numerator & denominator by b cos x y = tan^1 (a/b tan x)/ (1 a/b tan x) Taking a/b as tan c we have y = tan^1 tan (cx) y = cx So dy/dx = 1Y h = C 1 C 2ex The right hand side of the nonhomogeneous equation, sinx, is of the form sinkx where k = 1 While 1 is a root of the characteristic equation, i1 is not, so we use y p = Acosx Bsinx Then y0 p = −Acosx Bsinx and y00 p = −Acosx − Bsinx In the differential equation these Show That Y E X A Cos X B Sin X Is The Solution Of The Differential Equation D 2y Dx 2 2 Dy Dx 2y 0 Sarthaks Econnect Largest Online Education Community If y=e^-x(acosx+bsinx)