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Question:

If $\mathrm{y}=\left(\cot ^{-1} \mathrm{x}\right)^{2}$, prove that $\mathrm{y}_{2}\left(\mathrm{x}^{2}+1\right)^{2}+2 \mathrm{x}\left(\mathrm{x}^{2}+1\right) \mathrm{y}_{1}=2$

Solution:

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

(ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cot ^{-1} \mathrm{x}=\frac{-1}{1+\mathrm{x}^{2}}$

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

(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: –

$y=\left(\cot ^{-1} x\right)^{2}$

differentiating w.r.t $x$

$\frac{d y}{d x}=y_{1}=2 \cot ^{-1} x \cdot\left[\frac{-1}{1+x^{2}}\right]$

$\Rightarrow y_{1}=\frac{-2 \cot ^{-1} x}{1+x^{2}}$

Differentiating w.r.t $x$

$\Rightarrow\left(1+x^{2}\right) y_{2}+2 x y_{1}=2\left(\frac{1}{1+x^{2}}\right)$

$\Rightarrow\left(1+x^{2}\right)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2$

Hence proved