Question Number 103412 by abdomsup last updated on 14/Jul/20

solve (1+x^2 )^2 y^(′′)  +2x(1+x^2 )y^′  +2=0

Answered by OlafThorendsen last updated on 15/Jul/20

(1+x^2 )^2 y^(′′)  +2x(1+x^2 )y^′  +2 = 0  y′ = (u/(1+x^2 ))  ⇒ y′′ = (((1+x^2 )u′−2xu)/((1+x^2 )^2 ))  (1+x^2 )u′−2xu+2xu+2 = 0  u′ = −(2/(1+x^2 ))  ⇒u = −2Arctanx+C_1   y′ = −2((Arctanx)/(1+x^2 ))+(C_1 /(1+x^2 ))  ⇒ y = −Arctan^2 x+C_1 Arctanx+C_2

Commented bymathmax by abdo last updated on 15/Jul/20

thank you sir