If the solve the problem
Question:

If $y=\operatorname{cosec}^{-1} x, x>1$, then show that $x\left(x^{2}-1\right) \frac{d^{2} y}{d x^{2}}+\left(2 x^{2}-1\right) \frac{d y}{d x}=0$

Solution:

Formula: –

(i) $\frac{d y}{d x}=y_{1}$ and $\frac{d^{2} y}{d x^{2}}=y_{2}$

(ii) $\frac{d\left(\operatorname{cosec}^{-1} x\right)}{d x}=\frac{-1}{|x| \sqrt{x^{2}-1}}$

(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

(iv) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\text { wou })}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

Given: –

$Y=\operatorname{cosec}^{-1} x$

We know that

$\frac{\mathrm{d}\left(\operatorname{cosec}^{-1} \mathrm{x}\right)}{\mathrm{dx}}=\frac{-1}{|\mathrm{x}| \sqrt{\mathrm{x}^{2}-1}}$

Let $y=\operatorname{cosec}^{-1} x$

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

Since $x>1,|x|=x$

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

Differentiating the above function with respect to $x$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{x} \frac{2 \mathrm{x}}{2 \sqrt{\mathrm{x}^{2}-1}}+\sqrt{\mathrm{x}^{2}-1}}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{x}^{2}}{\sqrt{\mathrm{x}^{2}-1}}+\sqrt{\mathrm{x}^{2}-1}}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{x}^{2}+\mathrm{x}^{2}-1}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{\frac{3}{2}}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{2 \mathrm{x}^{2}-1}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{\frac{3}{2}}}$

Thus

$x\left(x^{2}-1\right) \frac{d^{2} y}{d x^{2}}=\frac{2 x^{2}-1}{x \sqrt{x^{2}-1}} \ldots \cdots \cdots$(2)

Similarly

$\Rightarrow\left[2 x^{2}-1\right] \frac{d y}{d x}=\frac{-2 x^{2}+1}{x \sqrt{x^{2}-1}}$

$\Rightarrow x\left(x^{2}-1\right) \frac{d^{2} y}{d x^{2}}+\left[2 x^{2}-1\right] \frac{d y}{d x}=\frac{2 x^{2}-1}{x \sqrt{x^{2}-1}}+\frac{-2 x^{2}+1}{x \sqrt{x^{2}-1}}=0$

Hence proved.

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