Question

If y = $(x+\sqrt{x^2-1})m$ then prove that $(1-x^2)y_2-x{y}_1+m^{2}y=0$

Answer 1

We have,

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

Differentiation this equation with respect to x and we get,

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

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

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

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

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

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

Squaring both side and we get,

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

Again differentiating and we get,

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

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

$(1-x^2)y_2-x{y}_1+m^{2}y=0$

Hence proved.