Question Number 71570 by TawaTawa last updated on 17/Oct/19

Solve: (2x + x^2 )y′′ − 2(1 + x)y′ + 2y = 0 Given y = x^2 is a solution.

Answered by mind is power last updated on 17/Oct/19

y=zx^2 y′=2xz+z′x^2 y′′=2z+4xz′+z′′x^2 (2x+x^2 )y′′−2(1+x)y′+2y=(2x+x^2 )(2z+4xz′+z′′x^2 )−2(1+x)(2xz+z′x^2 ) +2x^2 z=0 ⇒(2x^2 +2x^2 +4x−4x−4x^2 )z+(2x+x^2 )x^2 z′′+(8x^2 +4x^3 −2x^2 −2x^3 )z′=0 (2x^3 +x^4 )z′′+(6x^2 +2x^3 )z′=0 ⇔(x^2 +2x)z′′+2(3+x)z′=0..∀x≠0 Y=z′ (x^2 +2x)Y′+2(3+x)Y=0 ⇒((Y′)/Y)=((6+2x)/(x(x+2)))=(3/x)−(1/(x+2)) ⇒ln∣Y∣=3ln∣x∣−ln∣x+2∣ ln∣Y∣=ln∣(x^3 /(x+2))∣+c ⇒∣Y∣=k.∣(x^3 /(x+2))∣ Y=c.(x^3 /(x+2))....c=+_− k z′=c(x^3 /(x+2))=c(((x+2)(x^2 −2x+4)−8)/(x+2)) z′=c(x^2 −2x+4)−((8c)/(x+2)) ⇒z=c((x^3 /3)−x^2 +4x)−8cln∣x+2∣+k y=zx^2 =c((x^5 /3)−x^4 +4x^3 )+kx^2 −8cx^2 ln∣x+2∣ c,k constante solution is iner space of TWo Dimensiln S=vect((x^5 /3)−x^4 +4x^3 −8ln∣x+2∣,x^2 )

Commented byTawaTawa last updated on 17/Oct/19

Wow, God bless you sir

Commented bymind is power last updated on 17/Oct/19

y′re welcom