Question Number 105163 by bemath last updated on 26/Jul/20

x^2 y′′+xy′−y=(1/(1+x^2 ))

Answered by bramlex last updated on 27/Jul/20

we solve homogenous equation  This is an Euler−Cauchy   y=x^n → { ((y′=nx^(n−1) )),((y′′=(n^2 −n)x^(n−2) )) :}  ⇒x^2 (n^2 −n)x^(n−2) +x(nx^(n−1) )−x^n =0  n^2 −n+n−1=0 →n = ±1  y_h  = C_1 x+(C_2 /x)  let y_1  = x ; y_2 = x^(−1)   → { ((y_1 ^′  = 1)),((y_2 ′=−x^(−2) )) :} ⇒ W(y_1 ,y_2 )= determinant (((x       x^(−1) )),((1    −x^(−2) )))=−2x^(−1)   u_1 = −∫((x^(−1) .(1/(x^2 (x^2 +1))))/(−2x^(−1) )) dx   u_1 =∫(1/2)((1/x^2 )−(1/(1+x^2 ))) dx  u_1 = −(1/(2x))−(1/2)tan^(−1) (x)  u_2 = ∫((x.(1/(x^2 (1+x^2 ))))/(−2x^(−1) )) dx  u_2 = ∫(( −1)/(2(1+x^2 ))) dx = −(1/2)tan^(−1) (x)  particular solution   y_p  = y_1 u_1 +y_2 u_2   y_p  = x(−(1/(2x))−(1/2)tan^(−1) (x))+(1/x)(−(1/2)tan^(−1) (x))  y_p = −(1/2)−(x/2) tan^(−1) (x) −(1/(2x))tan^(−1) (x)  General solution   y_G  = C_1 x+(C_2 /x)−(x/2) tan^(−1) (x)−(1/(2x))tan^(−1) (x)−(1/2)  ★▼◊

Answered by OlafThorendsen last updated on 26/Jul/20

(x^2 y′′+2xy′)−(xy′+y) = (1/(1+x^2 ))  (x^2 y′)′−(xy)′ = (1/(1+x^2 ))  x^2 y′−xy = arctanx+C_1   ((xy′−y)/x^2 ) = (1/x^3 )arctanx+(C_1 /x^3 )  (d/dx)((y/x)) = (1/x^3 )arctanx+(C_1 /x^3 )  (y/x) = −(1/(2x^2 ))arctanx+∫(dx/(2x^2 (1+x^2 )))+(C_2 /x^2 )+C_3   (1/(x^2 (1+x^2 ))) = (1/x^2 )−(1/(1+x^2 ))  (y/x) = −(1/(2x^2 ))arctanx−(1/(2x))−(1/2)arctanx+(C_2 /x^2 )+C_3   y = −(1/(2x))arctanx−(1/2)xarctanx+(C_2 /x)+C_3 x−(1/2)

Answered by mathmax by abdo last updated on 26/Jul/20

x^2 y^(′′)  +xy^′ −y =(1/(1+x^2 ))  h→x^2 y^((2))  +xy^((1)) −y =0 let y =x^m  ⇒y^′  =mx^(m−1)  and y^((2))  =m(m−1)x^(m−2)   e⇒m(m−1)x^m +mx^m −x^m  =0 ⇒(m^2 −m+m−1)x^m  =0 ⇒  m^2 −1=0 ⇒m =+^− 1 ⇒y =αx +(β/x) =αu_1  +β u_2   W(u_1  ,u_2 ) = determinant (((x          (1/x))),((1          −(1/x^2 ))))=−(1/x)−(1/x) =−(2/x) ≠0  W_1 = determinant (((o            (1/x))),(((1/(1+x^2 ))      −(1/x^2 ))))=−(1/(x(1+x^2 )))  W_2 = determinant (((x            0)),((1             (1/(1+x^2 )))))=(x/(1+x^2 ))  v_1 =∫ (w_1 /w)dx =−∫  (1/(x(1+x^2 )))×(((−x)/2))dx =(1/2) ∫  (dx/((1+x^2 )))=(1/2)arctanx  =∫ (w_2 /w)dx =∫  (x/(1+x^2 ))×(−(x/2))dx =−(1/2)∫ (x^2 /(1+x^2 ))dx  =−(1/2)∫((1+x^2 −1)/(1+x^2 ))dx =−(x/2) +(1/2)arctanx ⇒  y_p =u_1 v_1  +u_2 v_2 =(x/2)arctanx+(1/x)(−(x/2) +(1/2) arctanx)  =(x/2) arctanx−(1/2) +(1/(2x)) arctanx =(1/2)(x+(1/x))arctanx−(1/2)  the general solution is   y =αx +βx^(−1)  +(1/2)(x+(1/x))arctanx −(1/2)