If the solve the problem

Formula: -

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

(ii) $\frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}=\frac{1}{\mathrm{x}}$

(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: -

$y=\left[\log \left(x+\sqrt{1+x^{2}}\right)\right\}^{2}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d\left[\log \left(x+\sqrt{1+x^{2}}\right)\right\}^{2}}{d x}$

Using formula(ii)

$\Rightarrow \frac{d y}{d x}=2 \log \left(x+\sqrt{1+x^{2}}\right) \cdot \frac{1}{\left(x+\sqrt{1+x^{2}}\right)} \cdot\left(1+\frac{2 x}{2 \sqrt{1+x^{2}}}\right)$

Using formula(i)

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

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

Squaring both sides

$\left(y_{1}\right)^{2}=\frac{4}{1+x^{2}}\left[\log \left(x+\sqrt{1+x^{2}}\right)\right.$

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

$\Rightarrow\left(1+x^{2}\right) y_{2} y_{1}+2 x\left(y_{1}\right)^{2}=4 y_{1}$

Using formual(iii)

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

Hence proved