Question

If $y^{1/m}=[x+\sqrt{1+x^2}]$, then $(1+x^2)y_2+x{y}_1$ is equal to

• A. $m^{2}y$
• B. $m{y}^2$
• C. $m^2{y}^2$
• D. None of these

Answer 1

Answer: (A)

$m^{2}y$

We have, $y^{1/m}=[x+\sqrt{1+x^2}]\Rightarrow y={[x+\sqrt{1+x^2}]}^m$
$\Rightarrow \frac{dy}{dx}=m{[x+\sqrt{1+x^2}]}^{m-1}(1+\frac{x}{\sqrt{x^2+1}})$
$=m\frac{{[x+\sqrt{1+x^2}]}^m}{\sqrt{x^2+1}}$
On differentiating w.r.t $x$ we get
$\frac{dy}{dx}=\frac{my}{\sqrt{1+x^2}}$
$\Rightarrow y_1^2(1+x^2)=m^2{y}^2$
again differentiating we get
$2y_1{y}_2(1+x^2)+2x{y}_1^2=2m^{2}y{y}_1$
$\Rightarrow y_2(1+x^2)+x{y}_1=m^{2}y$