Question Number 101846 by bemath last updated on 05/Jul/20

Answered by bramlex last updated on 05/Jul/20

⇔ (dy/dx) −2x^(−1) y = x^2 cos x integrating factor u = e^(∫−2x^(−1) dx) u = e ^(−2 ln(x)) = x^(−2) ⇔ y = ((∫ x^(−2) .x^2 cos x dx)/x^(−2) ) ⇔ y = x^2 { sin x + C } y = x^2 sin x + Cx^2 (Bram−□ )

Answered by mathmax by abdo last updated on 05/Jul/20

(1/x)y^′ −(2/x^2 )y =xcosx ⇒xy^′ −2y =x^3 cosx (x>0) he →xy^′ −2y =0 ⇒xy^′ =2y ⇒(y^′ /y) =(2/x) ⇒ln∣y∣ =2lnx +c ⇒ y =k x^2 lsgrange method →y^′ =k^′ x^2 +2kx e ⇒k^′ x^3 +2kx^2 −2kx^2 =x^3 cosx ⇒k^′ =cosx ⇒k =sinx +λ ⇒ y(x) =(sinx +λ)x^2 =λx^2 +x^2 sinx