Step-by-Step Solution

The given differential equation is \((\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}})\frac{dx}{dy}=1\). We can rewrite this as:

\(\frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\)

Rearrange the equation to the standard form of a first-order linear differential equation: \(\frac{dy}{dx} + P(x)y = Q(x)\)

\(\frac{dy}{dx} + \frac{1}{\sqrt{x}}y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)

Here, \(P(x) = \frac{1}{\sqrt{x}}\) and \(Q(x) = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)

The integrating factor (IF) is given by \(e^{\int P(x) dx}\)

So, \(IF = e^{\int \frac{1}{\sqrt{x}} dx}\)

Let's evaluate the integral: \(\int \frac{1}{\sqrt{x}} dx = \int x^{-1/2} dx = \frac{x^{1/2}}{1/2} + C = 2\sqrt{x} + C\)

Therefore, the integrating factor is \(e^{2\sqrt{x}}\)

Step-by-Step Solution

Given the differential equation: \((\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}})\frac{dx}{dy}=1\)

Rewrite the equation as: \(\frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\)

Rearrange the equation into the standard form: \(\frac{dy}{dx} + \frac{y}{\sqrt{x}} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)

Here, \(P(x) = \frac{1}{\sqrt{x}}\) and \(Q(x) = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)

The integrating factor (IF) is given by: \(IF = e^{\int P(x) dx}\)

So, \(IF = e^{\int \frac{1}{\sqrt{x}} dx}\)

Evaluate the integral: \(\int \frac{1}{\sqrt{x}} dx = \int x^{-1/2} dx = 2x^{1/2} + C = 2\sqrt{x} + C\)

Therefore, the integrating factor is: \(e^{2\sqrt{x}}\)