Question Number 102323 by Learner101 last updated on 08/Jul/20

Solve this linear Equation (dy/dx) + y cos x = (1/2) sin x

Answered by bemath last updated on 08/Jul/20

IF u(x)=e^(∫cos x dx) = e^(sin x) y(x)=((∫ (1/2)sin x e^(sin x) dx +C)/e^(sin x) ) y(x)= (((1/2)∫sin xe^(sin x) dx +C)/e^(sin x) ) y(x)= e^(−sin x) {(1/2)∫sin x e^(sin x) +C}

Commented bybemath last updated on 08/Jul/20

yes sir.

Answered by mathmax by abdo last updated on 08/Jul/20

y^′ +y cosx =((sinx)/2) (he)→y^′ =−ycosx ⇒(y^′ /y) =−cosx ⇒ln∣y∣ =−sinx +c ⇒y =k e^(−sinx) lagrange method →y^′ =k^′ e^(−sinx) −cosx k e^(−sinx) e ⇒k^′ e^(−sinx) −kcosx e^(−sinx) +kcosx e^(−sinx) =(1/2)sinx ⇒ k^′ e^(−sinx) =(1/2)sinx ⇒ k^′ =(1/2)sinx e^(sinx) ⇒ k =(1/2)∫^x sint e^(sint) dt +c ⇒ y(x) =((1/2)∫^x sint e^(sint) dt +c)e^(−sinx) =c e^(−sinx ) +(e^(−sinx) /2) ∫^x sint e^(sint) dt