Question

help to solve this problem using Variation of parameters

thanks

y"+y = sen x

Answer 1

Given ve+ y = sin x →(1), here R(x) = sin x. Generalform is +P(x) —+Q(x)y= R(x) Differential equation in operator form is (D2 +1)y=sin x, where D Auxiliary equation is m* +1=0 m =-1=m=+/-1 = ti, the roots are complex If m= a tib, then C.F. = (, cosbx+c sin bx), here a = 0, b = 1. :: Complementary function,C.F. = y. Eccos x+c sinx. Let u = cos X, V = sin x. By method of variation of parameters: Let Particular integral,P.1. = A(x)u(x)+B(x)v(x) Wronskian,W(uv)=wv'- v' = cos*x+sinx = 1 VR sin x sinx where A(x)=-1; dx=- W(u,v) 1 --ſsin” xdx = -2( |(1-cos2x)dx 2x) -- 1 1 X- sin 2x x+ sin 2x 2 2 2 COS X sin x B(x)=\ with you = 1 = }{sin 2x dx = |:; }((x))* })dx= (f(x))*** +c, n-1 » +1 =-- cos 2x 4 = P.1. = A(x)u(x)+(x)v(x) 1 -6**+ an 21 ]cos x cos 2x din y - cox+(an 2x 600 x = coa 2x sin ) 1 1 =-xcosx++(sin x) Genral solution is 1 y(x) = C.F.+P.1. = 4 cos x+c sin x-xcox+ (sin x) = cos x+| x-xcosx + (sinx)= 4 cos x + 1 *+(s++)sin x-_rcos x 1 #y(x)=Ccos x+C. sin x-srcos x 2