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