Question Number 100337 by I want to learn more last updated on 26/Jun/20

Solve:   (1 +  x^2 )y′′  −  4xy′  +  6y   =  0

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

let determine a solution develppsble at integr serie y =Σ_(n=0) ^(∞ ) a_n x^n   ⇒y^′  =Σ_(n=1) ^∞  na_n x^(n−1)  ⇒y^(′′)  =Σ_(n=2) ^∞  n(n−1)a_n x^(n−2)   e⇒(1+x^2 )Σ_(n=2) ^∞  n(n−1)a_n x^(n−2)  −4x Σ_(n=1) ^∞  na_n x^(n−1)  +6Σ_(n=0) ^∞  a_n x^(n )  =0 ⇒  Σ_(n=2) ^∞  n(n−1)a_n x^(n−2)  +Σ_(n=2) ^∞  n(n−1)a_n x^n  −4Σ_(n=1) ^∞  na_n x^n  +6 Σ_(n=0) ^∞  a_n x^n  =0 ⇒  Σ_(n=0) ^∞  (n+2)(n+1)a_(n+2) x^n  +Σ_(n=1) ^∞ (n^2 a_n −na_n −4na_n  +6a_n )x^n  +6a_0 =0 ⇒   +Σ_(n=1) ^∞  {(n+1)(n+2)a_(n+2) +(n^2 −5n +6)a_n }x^n  +6a_0  +2a_2 =0 ⇒  ⇒ { (((n+1)(n+2)a_(n+2)  +(n^2 −5n+6)a_n =0)),((3a_0  +a_2 =0)) :}  ⇒ { ((a_(n+2) =−((n^2 −5n+6)/(n^2  +3n +2))a_n )),((a_2 =−3a_0 )) :}  and from this relation we calculate a_(2n)  and a_(2n+1) ....

Commented byI want to learn more last updated on 26/Jun/20

Thanks sir

Answered by maths mind last updated on 26/Jun/20

y=ax^2 +bx+c solution  (1+x^2 )2a−8ax^2 −4bx+6ax^2 +6bx+6c=0⇒  b=0  a=−3c⇒y=−3x^2 +1  y=(−3x^2 +1)z⇒  y′=−6xz+z′(−3x^2 +1)⇒y′′=−12xz′−6z+z′′(−3x^2 +1)  (1+x^2 )(−3x^2 +1)z′′−12x(1+x^2 )z′−6(1+x^2 )z−4x(−6xz+z′(−3x^2 +1))  +6(−3x^2 +1)z=0  (1+x^2 )(−3x^2 +1)z′′−16xz′=0  ((z′′)/(z′))=((16x)/((1+x^2 )(1−3x^2 )))  ln∣z′∣=∫((16x)/((1+x^2 )(1−3x^2 )))dx=2log(((1+x^2 )/(1−3x^2 )))  ⇒z′=k(((1+x^2 )^2 )/((1−3x^2 )^2 ))  z=k∫(((1+x^2 )^2 )/((1−3x^2 )^2 ))dx=k((x(x^2 −3))/(9x^2 −3))+c  y=k((x(x^2 −3))/(−3))+c(1−3x^2 )

Commented byI want to learn more last updated on 28/Jun/20

Thanks sir