If $y^{1/m} + y^{-1/m} = 2x$, then the value of $(x^2 - 1)\frac{d^2y}{dx^2} + x\frac{dy}{dx}$ is:

Given: $y^{1/m} + y^{-1/m} = 2x$

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

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Let $u = y^{1/m}$

Then $y^{-1/m} = \frac{1}{u}$

The equation becomes:

$u + \frac{1}{u} = 2x$

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Multiply both sides by $u$:

$u^2 + 1 = 2xu$

$u^2 - 2xu + 1 = 0$

Using the quadratic formula:

$u = \frac{2x \pm \sqrt{4x^2 - 4}}{2}$

$u = \frac{2x \pm 2\sqrt{x^2-1}}{2}$

$u = x \pm \sqrt{x^2 - 1}$

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Since $u = y^{1/m}$:

$y^{1/m} = x \pm \sqrt{x^2 - 1}$

Raising both sides to power $m$:

$y = \left(x \pm \sqrt{x^2 - 1}\right)^m$

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Using chain rule:

$\frac{dy}{dx} = m\left(x \pm \sqrt{x^2 - 1}\right)^{m-1} \cdot \frac{d}{dx}(x \pm \sqrt{x^2 - 1})$

$\frac{d}{dx}(x \pm \sqrt{x^2 - 1}) = 1 \pm \frac{x}{\sqrt{x^2-1}}$

$= \frac{\sqrt{x^2-1} \pm x}{\sqrt{x^2-1}}$

$= \frac{x \pm \sqrt{x^2-1}}{\sqrt{x^2-1}}$

Therefore:

$\frac{dy}{dx} = m\left(x \pm \sqrt{x^2 - 1}\right)^{m-1} \cdot \frac{x \pm \sqrt{x^2-1}}{\sqrt{x^2-1}}$

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

Since $y = \left(x \pm \sqrt{x^2 - 1}\right)^m$:

$\frac{dy}{dx} = \frac{my}{\sqrt{x^2-1}}$

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Using quotient rule on $\frac{dy}{dx} = \frac{my}{\sqrt{x^2-1}}$:

$\frac{d^2y}{dx^2} = \frac{m\frac{dy}{dx} \cdot \sqrt{x^2-1} - my \cdot \frac{x}{\sqrt{x^2-1}}}{x^2-1}$

$= \frac{m(x^2-1)\frac{dy}{dx} - mxy}{(x^2-1)^{3/2}}$

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

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

Substituting $\frac{dy}{dx} = \frac{my}{\sqrt{x^2-1}}$:

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

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

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

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

$= m^2y$

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