Question

If $y=\frac{1}{1+x+x^2+x^3}$, then $y_2\left(0\right)=$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

Given, $y=\frac{1}{(1+x)(1+x^2)}$
$\Rightarrow (1+x^2)y=\frac{1}{1+x}$
$\Rightarrow 2xy+(1+x^2)y^{{}^{\prime }}=\frac{-1}{(1+x{)}^2}$
$\Rightarrow 2y+2x{y}^{{}^{\prime }}+2x{y}^{{}^{\prime }}(1+x^2)y^{{}^{\prime \prime }}=\frac{2}{(1+x^2{)}^3}$
$\Rightarrow 2+y^{{}^{\prime \prime }}=2$

$\Rightarrow y^{{}^{\prime \prime }}=0$