If y=≤ft(x+√(1+x^2))^m then ≤ft(1+x^2) y_2+x y_1-m^2 y=

$y_1=m\left[x+\sqrt{1+x^2}\right]^{m-1} . (\left.1+\frac{1 . 2 x}{2 \sqrt{1+x^2}}\right)$

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

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

Differentiating w.r.t. x,

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

Canceling 2y₁,

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

Hence (1) is correct answer.