Question

lf $y=(x+\sqrt{1+x^2}{)}^m$, then $(1+x^2)y_2+x{y}_1-m^{2}y=$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$0$

$y=(x+\sqrt{1+x^2}{)}^m$
differentiating y with respect to x
$y_1=m(x+\sqrt{1+x^2}{)}^{m-1}(1+\frac{x}{\sqrt{1+x^2}})$
$⟹y_1=m(x+\sqrt{1+x^2}{)}^{m-1}\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}$
$⟹y_1=m(x+\sqrt{1+x^2}{)}^m\ast \frac{1}{\sqrt{1+x^2}}=\frac{my}{\sqrt{1+x^2}}$
$⟹y_1\sqrt{1+x^2}=my$
differentiating both side with respect to x
$y_2\sqrt{1+x^2}+\frac{x{y}_1}{\sqrt{1+x^2}}=m{y}_1$
$⟹y_2(1+x^2)+x{y}_1=m{y}_1\sqrt{1+x^2}=m^{2}y$
$⟹y_2(1+x^2)+x{y}_1-m^{2}y=0$