If y^{1/4} + y^{-1/4} = 2x,and (x^2 - 1) frac | vedclass

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(A) Given $y^{1/4} + y^{-1/4} = 2x$. Let $u = y^{1/4}$,then $u + \frac{1}{u} = 2x$,which implies $u^2 - 2xu + 1 = 0$. Solving for $u$,we get $u = x \p

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) Given $y^{1/4} + y^{-1/4} = 2x$. Let $u = y^{1/4}$,then $u + \frac{1}{u} = 2x$,which implies $u^2 - 2xu + 1 = 0$. Solving for $u$,we get $u = x \pm \sqrt{x^2 - 1}$.Thus,$y^{1/4} = x \pm \sqrt{x^2 - 1}$,so $y = (x \pm \sqrt{x^2 - 1})^4$.Differentiating with respect to $x$,$\frac{dy}{dx} = 4(x \pm \sqrt{x^2 - 1})^3 \left(1 \pm \frac{x}{\sqrt{x^2 - 1}}ight) = 4(x \pm \sqrt{x^2 - 1})^3 \left(\frac{\sqrt{x^2 - 1} \pm x}{\sqrt{x^2 - 1}}ight)$.Since $y^{1/4} = x \pm \sqrt{x^2 - 1}$,we have $\frac{dy}{dx} = 4(y^{3/4}) \left(\frac{\pm(x \pm \sqrt{x^2 - 1})}{\sqrt{x^2 - 1}}ight) = \frac{4y}{\sqrt{x^2 - 1}}$.Squaring both sides,$(x^2 - 1) \left(\frac{dy}{dx}ight)^2 = 16y^2$.Differentiating again with respect to $x$,$(x^2 - 1) \cdot 2 \frac{dy}{dx} \cdot \frac{d^2y}{dx^2} + 2x \left(\frac{dy}{dx}ight)^2 = 32y \frac{dy}{dx}$.Dividing by $2 \frac{dy}{dx}$ (assuming $\frac{dy}{dx} 
eq 0$),we get $(x^2 - 1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} - 16y = 0$.Comparing this with $(x^2 - 1) \frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0$,we find $\alpha = 1$ and $\beta = -16$.Therefore,$|\alpha - \beta| = |1 - (-16)| = |1 + 16| = 17$.

If $y^{1/4} + y^{-1/4} = 2x$,and $(x^2 - 1) \frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0$,then $|\alpha - \beta|$ is equal to ...... .

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