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Solution:

Formula: -

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

(ii) $\frac{\mathrm{d}(\cot x)}{d x}=-\operatorname{cosec}^{2} x$

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

(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: -

$Y=\cot x$

Differentiating w.r.t. $x$

$\frac{d y}{d x}=\frac{d(\cot x)}{d x}$

Using formula (ii)

$\Rightarrow \frac{d y}{d x}=-\operatorname{cosec}^{2} x$

Differentiating w.r.t $\mathrm{x}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-[2 \operatorname{cosecx}(-\operatorname{cosecxcotx})]$

Using formual (iii)

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2 \operatorname{cosec}^{2} \mathrm{x} \cot \mathrm{x}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-2 \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \mathrm{y}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0$

Hence proved.