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If $\frac{d^n y}{d x^n}=y_n$ and $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$, then $4 x y_2+2 y_1=$
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The correct answer is:
$\mathrm{y}$
$y=e^{\sqrt{x}}+e^{-\sqrt{x}}$
$y_1=e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}$ $-e^{-\sqrt{x}}\left(\frac{1}{2 \sqrt{x}}\right)$
$y_1=\frac{1}{2 \sqrt{x}}\left(e^{\sqrt{x}}-e^{-\sqrt{x}}\right)$ ...(i)
$y_2=\frac{1}{2 \sqrt{x}} \quad( e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}$ $-\left(-e^{-\sqrt{x}}\right)\left(\frac{1}{2 \sqrt{x}}\right))$
$+\frac{\left(e^{\sqrt{x}}-e^{-\sqrt{x}}\right)}{2}$ $.\left(-\frac{1}{2}\right) \times \frac{1}{x^{3 / 2}}$ ...(ii)
from (ii)
$2 y_1=\frac{1}{\sqrt{x}}\left(e^{\sqrt{x}}-e^{-\sqrt{x}}\right)$ ...(iii)
$4 x y_2=e^{\sqrt{x}}+e^{-\sqrt{x}}$ $-\left(\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{\sqrt{x}}\right)$ ...(iv)
Adding (iii) \& (iv),
$2y_1+4 x y_2=e^{\sqrt{x}}+e^{-\sqrt{x}}$
$=y$
$y_1=e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}$ $-e^{-\sqrt{x}}\left(\frac{1}{2 \sqrt{x}}\right)$
$y_1=\frac{1}{2 \sqrt{x}}\left(e^{\sqrt{x}}-e^{-\sqrt{x}}\right)$ ...(i)
$y_2=\frac{1}{2 \sqrt{x}} \quad( e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}$ $-\left(-e^{-\sqrt{x}}\right)\left(\frac{1}{2 \sqrt{x}}\right))$
$+\frac{\left(e^{\sqrt{x}}-e^{-\sqrt{x}}\right)}{2}$ $.\left(-\frac{1}{2}\right) \times \frac{1}{x^{3 / 2}}$ ...(ii)
from (ii)
$2 y_1=\frac{1}{\sqrt{x}}\left(e^{\sqrt{x}}-e^{-\sqrt{x}}\right)$ ...(iii)
$4 x y_2=e^{\sqrt{x}}+e^{-\sqrt{x}}$ $-\left(\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{\sqrt{x}}\right)$ ...(iv)
Adding (iii) \& (iv),
$2y_1+4 x y_2=e^{\sqrt{x}}+e^{-\sqrt{x}}$
$=y$
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