If 2x = y^{1/5} + y^{-1/5} and (x^2 - 1) frac | vedclass

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(B) Given: $y^{1/5} + y^{-1/5} = 2x$. Differentiating with respect to $x$: $\frac{1}{5} y^{-4/5} \frac{dy}{dx} - \frac{1}{5} y^{-6/5} \frac{dy}{dx} =

Vedclass · Accepted Answer

$-24$

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(B) Given: $y^{1/5} + y^{-1/5} = 2x$. Differentiating with respect to $x$: $\frac{1}{5} y^{-4/5} \frac{dy}{dx} - \frac{1}{5} y^{-6/5} \frac{dy}{dx} = 2$. $\frac{1}{5} y^{-1} (y^{1/5} - y^{-1/5}) \frac{dy}{dx} = 2$. $(y^{1/5} - y^{-1/5}) \frac{dy}{dx} = 10y$. We know that $(y^{1/5} - y^{-1/5})^2 = (y^{1/5} + y^{-1/5})^2 - 4 = (2x)^2 - 4 = 4(x^2 - 1)$. Thus,$y^{1/5} - y^{-1/5} = 2\sqrt{x^2 - 1}$. Substituting this value: $2\sqrt{x^2 - 1} \frac{dy}{dx} = 10y$,which simplifies to $\sqrt{x^2 - 1} \frac{dy}{dx} = 5y$. Differentiating again with respect to $x$: $\frac{x}{\sqrt{x^2 - 1}} \frac{dy}{dx} + \sqrt{x^2 - 1} \frac{d^2y}{dx^2} = 5 \frac{dy}{dx}$. Multiplying by $\sqrt{x^2 - 1}$: $x \frac{dy}{dx} + (x^2 - 1) \frac{d^2y}{dx^2} = 5 \sqrt{x^2 - 1} \frac{dy}{dx}$. Since $\sqrt{x^2 - 1} \frac{dy}{dx} = 5y$,we get $x \frac{dy}{dx} + (x^2 - 1) \frac{d^2y}{dx^2} = 5(5y) = 25y$. Rearranging: $(x^2 - 1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} - 25y = 0$. Comparing with $(x^2 - 1) \frac{d^2y}{dx^2} + \lambda x \frac{dy}{dx} + ky = 0$,we get $\lambda = 1$ and $k = -25$. Therefore,$\lambda + k = 1 - 25 = -24$.

If $2x = y^{1/5} + y^{-1/5}$ and $(x^2 - 1) \frac{d^2y}{dx^2} + \lambda x \frac{dy}{dx} + ky = 0$,then $\lambda + k$ is equal to

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