Question Number 105118 by john santu last updated on 26/Jul/20

x^2 y′′+xy′−y = (1/(x+1))

Answered by john santu last updated on 26/Jul/20

⇒ x^2 y′′+2xy′−xy′ − y = (1/(x+1)) (d/dx)[ x^2 y′−xy ] = (d/dx)[ ln (x+1) ] ∫ (d/dx)[ x^2 y′−xy ] = ∫ (d/dx)[ ln(x+1)] x^2 y′−xy = ln (x+1)+c ((y′)/x) − (y/x^2 ) = ((ln (x+1)+c)/x^3 ) (d/dx)[ (y/x) ] = ((ln (x+1)+c)/x^3 ) ⇔ y = x { ∫ ((ln (x+1)+c)/x^3 ) dx } y = x { −(c/(2x^2 )) + ∫x^(−3) ln (1+x)dx} let J=∫x^(−3) ln (1+x) dx by parts → { ((u=ln (1+x)→du=(1/(1+x))dx)),((v = −(x^(−2) /2))) :} J = −((ln (1+x))/(2x^2 ))+∫(dx/(2x^2 (1+x))) (1/(x^2 (x+1))) = (A/x)+(B/x^2 )+(C/(x+1)) 1 = Ax(x+1)+B(x+1)+Cx^2 x=0→1=B x=−1→1=C x=1→1=2A+2B+C 1 =2A+3 →A=−1 J= −((ln (x+1))/(2x^2 ))+(1/2)∫((1/(x+1))+(1/x^2 )−(1/x))dx J= −((ln (x+1))/(2x^2 ))+(1/2){ln (x+1)−(1/x)−ln x} ∴ y = −(((ln C(x+1))/(2x)))+(x/2){ln (((x+1)/x))−(1/x)}

Answered by OlafThorendsen last updated on 26/Jul/20

(x^2 y′′+2xy′)−(xy′+y) = (1/(x+1)) (x^2 y′)′−(xy)′ = (1/(x+1)) x^2 y′−xy = ln∣1+x∣+C_1 ((xy′−y)/x^2 ) = (1/x^3 )ln∣1+x∣+(C_1 /x^3 ) (d/dx)((y/x)) = (1/x^3 )ln∣1+x∣+(C_1 /x^3 ) (y/x) = −(1/(2x^2 ))ln∣1+x∣+∫(1/(2x^2 )).(1/(1+x))dx+(C_2 /x^2 )+C_3 ∫(dx/(x^2 (x+1))) = ∫((1/x^2 )−(1/x)+(1/(1+x)))dx = −(1/x)+ln∣((1+x)/x)∣ (y/x) = −(1/(2x^2 ))ln∣1+x∣−(1/(2x))+(1/2)ln∣((1+x)/x)∣+(C_2 /x^2 )+C_3 y = −(1/(2x))ln∣1+x∣+(1/2)xln∣((1+x)/x)∣+(C_2 /x)+C_3 x−(1/2)