Question

If $y^{1/m}+y^{-1/m}=2x,$ then the value of $(x^2-1)y_2+x{y}_1,$ is

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

Answer 1

Answer: (A)

Given, $y^{1/m}+y^{-1/m}=2x$
$\Rightarrow$ $y^{1/m}+\frac{1}{y^{1/m}}=2x$
$\Rightarrow$ ${(y^{1/m})}^2+1=2x{y}^{1/m}$
$\Rightarrow$ ${(y^{1/m})}^2-2x{y}^{1/m}+1=0,$ which is a quadratic equation.
So, $y^{1/m}=\frac{2x\pm \sqrt{4x^2-4}}{2}$
$\Rightarrow$ $y^{1/m}=x\pm \sqrt{x^2-1}$
$\Rightarrow$ $y={[x\pm \sqrt{x^2-1}]}^m$ ...(i)
$\therefore$ $\frac{dy}{dx}=y_1$
$=m{[x\pm \sqrt{x^2-1}]}^{m-1}$
$\left[1\pm \frac{1}{2\sqrt{x^2-1}}.2x\right]$
$\Rightarrow$ $y_1=m{[x\pm \sqrt{x^2-1}]}^{m-1}\left[\frac{\sqrt{x^2-1}\pm x}{\sqrt{x^2-1}}\right]$
$\Rightarrow$ $y_1=\frac{m{[x\pm \sqrt{x^2-1}]}^m}{\sqrt{x^2-1}}$
$\Rightarrow$ $y_1=\frac{my}{\sqrt{x^2-1}}$ [from Eq.(i)] $\Rightarrow$
$\sqrt{x^2-1}.y_1=my$
On squaring both sides, we get
$(x^2-1)y_1^2=m^2{y}^2$
On differentiating both sides, we get
$(x^2-1)2y_1{y}_2+2x{y}_1^2=2m^{2}y{y}_1$
$\Rightarrow$ $y_1{y}_2(x^2-1)+x{y}_1^2=m^{2}y{y}_1$
$\Rightarrow$ $(x^2-1)y_2+x{y}_1=m^{2}y$