Question Number 104265 by mohammad17 last updated on 20/Jul/20

(dy/dx)−y=sinx

Answered by bemath last updated on 20/Jul/20

Homogenous solution   λ−1=0 ⇒λ=1 ⇒y_h  = C_1 e^x   particular solution   y_p =Acos x+Bsin x  y_p ′=−Asin x+Bcos x  comparing coefficient   (−Asin  x+Bcos  x)−(Acos x  + Bsin x) = sin x  ⇒(B−A)cos x+(−B−A)sin x  = sin x   B=A ∧ −B−A=1 ⇒A=−(1/2)  y_p  = −(1/2)cos x−(1/2)sin x  general solution   ∴y = Ce^x −(1/2)(cos x+sin x)

Commented bymohammad17 last updated on 20/Jul/20

    sory sir how this i think this y=((∫e^(−x) sinxdx)/e^(−x) )

Commented bymohammad17 last updated on 20/Jul/20

Commented bymohammad17 last updated on 20/Jul/20

is the solution its right?

Commented bybemath last updated on 20/Jul/20

yes sir

Commented bymohammad17 last updated on 20/Jul/20

thank you

Answered by Dwaipayan Shikari last updated on 20/Jul/20

IF=e^(∫−1dx) =e^(−x)   y.e^(−x) =∫e^(−x) sinxdx=I  ye^(−x) =−sinxe^(−x) +∫e^(−x) cosxdx  ye^(−x) =−sinx e^(−x) −cosx e^(−x) −∫sinx e^(−x) =I  ye^(−x) =−(1/2)e^(−x) (sinx+cosx)+C  y=((−1)/2)(sinx+cosx)+Ce^x

Answered by Ar Brandon last updated on 20/Jul/20

{(dy/dx)−y=sinx}∙e^(−x)   ⇒e^(−x) (dy/dx)−ye^(−x) =e^(−x) sinx  ⇒((d(ye^(−x) ))/dx)=e^(−x) sinx  ⇒ye^(−x) =∫e^(−x) sinxdx                 =sinx∫e^(−x) dx−∫{((dsinx)/dx)∫e^(−x) dx}dx                 =−e^(−x) sinx+∫e^(−x) cosxdx                 =−e^(−x) sinx+{cosx∫e^(−x) dx−∫{((dcosx)/dx)∫e^(−x) dx}dx}                 =−e^(−x) sinx−e^(−x) cosx−∫e^(−x) sinxdx                 =((−(sinx+cosx)e^(−x) )/2)+C  ⇒y=−(((sinx+cosx))/2)+Ce^x

Answered by mathmax by abdo last updated on 20/Jul/20

y^′ −y =sinx  h→r−1 =0 ⇒r =1 ⇒y_h =Ke^x   lagrange method →y^′  =k^′  e^x  +ke^x   e⇒k^′  e^x  +ke^x −ke^x  =sinx ⇒k^′  =e^(−x) sinx ⇒k =∫ e^(−x) sinx dx  =Im(∫ e^(−x+ix) dx) =Im(∫ e^((−1+i)x) dx)  but   ∫ e^((−1+i)x) dx =(1/(−1+i)) e^((−1+i)x)  +c =−(1/(1−i)) e^(−x) (cosx +sinx)  =−((1+i)/2) e^(−x) {cosx +isinx} =−(e^(−x) /2){cosx+isinx+icosx−sinx} ⇒  Im(∫...) =−(e^(−x) /2){cosx +sinx} ⇒the general solution is  (−(e^(−x) /2){cosx +sinx} +c)e^x  =−(1/2)(cosx +sinx) +ce^x