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