Question Number 104051 by bemath last updated on 19/Jul/20

(x^2 −1) (dy/dx) + 2y = (x+1)^2

Answered by bramlex last updated on 19/Jul/20

Commented bybramlex last updated on 19/Jul/20

typo integrating factor   u(x) = e^(∫(2/((x+1)(x−1)))dx) = ((x−1)/(x+1))  (d/dx)((((x−1)/(x+1))).y)= ∫(((x−1)/(x+1)))(((x+1)/(x−1)))dx  (((x−1)/(x+1)))y = ∫ dx   (((x−1)/(x+1)))y = x +C   ∴ y = ((x^2 +x)/(x−1)) +C(x−1)^(−1)  ■

Answered by Dwaipayan Shikari last updated on 19/Jul/20

IF=e^(2∫(1/(x^2 −1))) =(((x−1)/(x+1)))     {(dy/dx)+((2y)/(x^2 −1))=((x+1)/(x−1))  y.(((x−1)/(x+1)))=∫dx  y=((x(x+1))/(x−1))+C(((x+1)/(x−1)))

Answered by mathmax by abdo last updated on 19/Jul/20

(x^2 −1)y^′ +2y =(x+1)^2   h→(x^2 −1)y^′  +2y =0 ⇒(x^2 −1)y^′  =−2y ⇒(y^′ /y) =((−2)/((x^2 −1))) ⇒  ln∣y∣ =−2 ∫ (dx/(x^2 −1)) =−∫ ((1/(x−1))−(1/(x+1)))dx =ln∣((x+1)/(x−1))∣ +c ⇒  y(x) =k ∣((x+1)/(x−1))∣ solution on w={x /((x+1)/(x−1))>0} ⇒y =k×((x+1)/(x−1))  ⇒y^′  =k^′ ×((x+1)/(x−1)) +k×((x−1−(x+1))/((x−1)^2 )) =k^(′ ) (((x+1)/(x−1)))−((2k)/((x−1)^2 ))  e⇒(x^2 −1)k^′ (((x+1)/(x−1)))+(x^2 −1)×(((−2k))/((x−1)^2 )) +2k(((x+1)/(x−1))) =(x+1)^2  ⇒  (x^2 −1)k^′   =(x+1)^2  ⇒k^′  =(((x+1)^2 )/((x+1)(x−1))) =((x+1)/(x−1)) ⇒k(x) =∫ ((x+1)/(x−1))dx +c  =∫ ((x−1+2)/(x−1))dx +c =x+2ln∣x−1∣ +c ⇒  y(x) =((x+1)/(x−1)){x +2ln∣x−1∣ +c}