If $y^{1 / m}=\left[x+\sqrt{1+x^2}\right]$, then $\left(1+x^2\right) y_2+x y_1$ is equal to

$m^2~y$

We have,

$y^{1 / m}=\left\{x+\sqrt{1+x^2}\right\}$

$\Rightarrow y=\left\{x+\sqrt{1+x^2}\right\}^m$

$\Rightarrow \frac{d y}{d x}=m\left\{x+\sqrt{1+x^2}\right\}^{m-1}\left\{1+\frac{x}{\sqrt{x^2+1}}\right\}=m \frac{\left\{x \sqrt{1+x^2}\right\}^m}{\sqrt{1+x^2}}$

$\Rightarrow \frac{d y}{d x}=\frac{m y}{\sqrt{1+x^2}}$

$\Rightarrow y_1^2\left(1+x^2\right)=m^2 y^2$

$\Rightarrow 2 y_1 y_2\left(1+x^2\right)+2 x y_1^2=2 m^2 y y_1$

$\Rightarrow y_2\left(1+x^2\right)+y_1=m^2 y$