If $y=\frac{\sin \mathrm{h}^{-1} x}{\sqrt{1+x^2}}$, then $\left(1+x^2\right) \mathrm{y}_2+3 \mathrm{xy}_1+\mathrm{y}=$

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Given, } y=\frac{\sinh ^{-1} x}{\sqrt{1+x^2}} \ & \sqrt{1+x^2} y=\sinh ^{-1} x \end{aligned} $ On differentiating w.r.t. $x$, we get $ \begin{aligned} & \sqrt{1+x^2} y_1+y \frac{1}{2 \sqrt{1+x^2}} \cdot 2 x=\frac{1}{\sqrt{1+x^2}} \ & \left(1+x^2ight) y_1+x y=1 \end{aligned} $ Again, on differentiating w.r.t. $x$, we get $ \begin{aligned} & \left(1+x^2ight) y_2+y_1 \cdot 2 x+x y_1+y=0 \ & \left(1+x^2ight) y_2+3 x y_1+y=0 \end{aligned} $

If $y=\frac{\sin \mathrm{h}^{-1} x}{\sqrt{1+x^2}}$, then $\left(1+x^2\right) \mathrm{y}_2+3…

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