If frac{d^n y}{d x^n}=y_n and y=e^{sqrt{x}}+e | vedclass

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(B) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.First derivative $y_1 = \frac{d}{dx}(e^{\sqrt{x}} + e^{-\sqrt{x}}) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x

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$y$

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(B) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.First derivative $y_1 = \frac{d}{dx}(e^{\sqrt{x}} + e^{-\sqrt{x}}) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} - e^{-\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}}$.Multiplying by $2\sqrt{x}$,we get $2\sqrt{x} y_1 = e^{\sqrt{x}} - e^{-\sqrt{x}}$.Differentiating both sides with respect to $x$:$2\sqrt{x} y_2 + 2 y_1 \cdot \frac{1}{2\sqrt{x}} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} + e^{-\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$.$2\sqrt{x} y_2 + \frac{y_1}{\sqrt{x}} = \frac{e^{\sqrt{x}} + e^{-\sqrt{x}}}{2\sqrt{x}} = \frac{y}{2\sqrt{x}}$.Multiplying the entire equation by $2\sqrt{x}$:$4 x y_2 + 2 y_1 = y$.

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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