Question

lf $y=(x^2-1{)}^n$ and $y_n$ denotes the $n^{th}$ derivative of $y$, then $(x^2-1)y_{n+2}+2x{y}_{n+1}=$

• A. $(n^2+1)y_n$
• B. $(n^2-1)y_n$
• C. $n(n^2+1)y_n$
• D. $n(n+1)y_n$

Answer 1

The correct option is D $n(n+1)y_n$

$y=(x^2-1{)}^n$
$y_1=2xn(x^2-1{)}^{n-1}$
$y_2=2n(x^2-1{)}^{n-1}+4x^2(x^2-1{)}^{n-2}n(n-1)$
$(x^2-1)y_2=2n(x^2-1{)}^n+4x^2(x^2-1{)}^{n-1}n(n-1)$
$(x^2-1)y_2=2ny+(n-1)2x{y}_1$
$(x^2-1)y_3+2x{y}_2=2n{y}_1+2(n-1)y_1+(n-1)2x{y}_2$
$(x^2-1)y_3=2(n+(n-1\left)\right)y_1+(n-2)2x{y}_2$
$(x^2-1)y_4=2(n+(n-1)+n-2)y_2+2(n-3)x{y}_3$
.
.
$(x^2-1)y_{n+2}=2(n+(n-1)+(n-2)+(n-3)+\cdots +(n-(n-1))+(n-n\left)\right)y_n-2x{y}_{n+1}$
$(x^2-1)y_{n+2}=2(n+(n-1)+(n-2)+(n-3)+\cdots +1+0)y_n-2x{y}_{n+1}$
$(x^2-1)y_{n+2}=n(n+1)y_n-2x{y}_{n+1}$
Hence, option D.