Question

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

Answer 1

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

Differentiating w.r.t $x$, $2=\left[\right(\frac{1}{m})\times y^{\frac{1}{m}–1}-\frac{1}{m}\times y^{\frac{-1}{m}–1}]\left(\frac{dy}{dx}\right)$ --------- (2)

$2my=[y^{\frac{1}{m}}-y^{\frac{-1}{m}}]\frac{dy}{dx}$

Squaring this, $4m^2{y}^2={\left(\frac{dy}{dx}\right)}^2[y^{\frac{2}{m}}+y^{\frac{-2}{m}}-2]$ --------- (3)

Squaring the (1),

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

$4x^2=y^{\frac{2}{m}}+y^{\frac{-2}{m}}+2$

$y^{\frac{2}{m}}+y^{\frac{-2}{m}}=4x²-2$ ---- (4)

Substituting this in (1) above,

$4m^2{y}^2={\left(\frac{dy}{dx}\right)}^2(4x^2-4)$

Dividing by 4, $m^2{y}^2={\left(\frac{dy}{dx}\right)}^2(x^2-1)$

Again differentiating the above,

$2m^{2}y\left(\frac{dy}{dx}\right)=2\left(\frac{dy}{dx}\right)\left(\frac{d^{2}y}{d{x}^2}\right)(x^2-1)+2x{\left(\frac{dy}{dx}\right)}^2$

Dividing throughout by $2\times \frac{dy}{dx}$,

$m^{2}y=(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}$

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

Or, $(x^2-1)\left(y_2\right)+x{y}_1-\left(m^2\right)y=0$ --- Proved