Question Number 102163 by mathmax by abdo last updated on 07/Jul/20

solve  y^(′′)  −2xy^′   =xe^(−x^2 )

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

let y^′  =z  so (e)⇒z^′ −2xz =xe^(−x^2 )   (he)→z^′ =2xz ⇒(z^′ /z) =2x ⇒ln∣z∣ =x^2  +c ⇒ z = ke^x^2    lagrange method →z^′  =k^′  e^x^2   +2xk e^x^2   so  e ⇒k^′  e^x^2   +2xk e^x^2   −2xk e^x^2   =xe^(−x^2 )  ⇒k^′  =x e^(−2x^2  )  ⇒  k =∫ x e^(−2x^2 )  dx +λ =−(1/4)e^(−2x^2 )  +λ ⇒z(x) =(−(1/4)e^(−2x^2 )  +λ)e^x^2    =−(1/4) e^(−x^2 )  +λe^x^2    wehave y^′  =z ⇒y(x) =∫^x z(t)dt =−(1/4) ∫^x  e^(−t^2 ) dt +λ ∫^x  e^t^2   dt +C