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

$y={(x+\sqrt{x^2-1})}^m$

$y_1=m{(x+\sqrt{x^2-1})}^{m-1}.(1+\frac{1}{2\sqrt{x^2-1}}\times 2x)$

$y_1=\frac{m{(x+\sqrt{x^2-1})}^m}{x+\sqrt{x^2-1}}\frac{(x+\sqrt{x^2-1})}{\sqrt{x^2-1}}$

$y_1=\frac{my}{\sqrt{x^2-1}}\Rightarrow y_1\sqrt{x^2-1}=my$

${y_1}^2(x^2-1)=m^2{y}^2$

${y_1}^2\left(2x\right)+(x^2-1)2y_1{y}_2=m^{2}2y{y}_1$

$2x{y}_1+(x^2-1)y_2=m^{2}y$