Solve the differential equation left[frac{e^{ | vedclass

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(A) Given the differential equation: $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1$Rearranging the equation to t

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$y e^{2 \sqrt{x}} = 2 \sqrt{x} + C$

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(A) Given the differential equation: $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}ight] \frac{d x}{d y}=1$Rearranging the equation to the form $\frac{dy}{dx} + Py = Q$:$\frac{dx}{dy} = \frac{\sqrt{x}}{e^{-2 \sqrt{x}} - y}$$\frac{dy}{dx} = \frac{e^{-2 \sqrt{x}} - y}{\sqrt{x}} = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}$Thus,$\frac{dy}{dx} + \frac{1}{\sqrt{x}} y = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{\sqrt{x}}$ and $Q = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$.Calculating the Integrating Factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int \frac{1}{\sqrt{x}} dx} = e^{2 \sqrt{x}}$The general solution is given by $y(I.F.) = \int (Q 	imes I.F.) dx + C$:$y e^{2 \sqrt{x}} = \int \left( \frac{e^{-2 \sqrt{x}}}{\sqrt{x}} 	imes e^{2 \sqrt{x}} ight) dx + C$$y e^{2 \sqrt{x}} = \int \frac{1}{\sqrt{x}} dx + C$$y e^{2 \sqrt{x}} = 2 \sqrt{x} + C$

Solve the differential equation $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{dy}=1$ where $x \neq 0$.

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