Concept:
A first-order linear differential equation is typically written in the standard Leibniz form:
\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
Where $P(x)$ and $Q(x)$ are continuous functions of $x$ only. The Integrating Factor ($\text{I.F.}$) is calculated using the formula:
\[
\text{I.F.} = e^{\int P(x) \, dx}
\]
Step 1: Reduce the given equation to the standard form.
The given differential equation is:
\[
2x \frac{dy}{dx} - y = 3
\]
To make the coefficient of $\frac{dy}{dx}$ equal to 1, we divide the entire differential equation by $2x$:
\[
\frac{dy}{dx} - \frac{y}{2x} = \frac{3}{2x}
\]
Comparing this rearranged equation with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify:
\[
P(x) = -\frac{1}{2x} \quad \text{and} \quad Q(x) = \frac{3}{2x}
\]
Step 2: Evaluate the integral of $P(x)$.
Before exponentiating, let us calculate the integral of $P(x)$ with respect to $x$:
\[
\int P(x) \, dx = \int -\frac{1}{2x} \, dx = -\frac{1}{2} \int \frac{1}{x} \, dx
\]
Using the rule $\int \frac{1}{x} \, dx = \log x$:
\[
\int P(x) \, dx = -\frac{1}{2} \log x
\]
Using the power property of logarithms, $n \log A = \log(A^n)$, we get:
\[
\int P(x) \, dx = \log\left(x^{-1/2}\right) = \log\left(\frac{1}{\sqrt{x}}\right)
\]
Step 3: Calculate the Integrating Factor ($\text{I.F.}$).
Now, substitute this integral into the exponential base:
\[
\text{I.F.} = e^{\int P(x) \, dx} = e^{\log\left(\frac{1}{\sqrt{x}}\right)}
\]
Since $e^{\log f(x)} = f(x)$, the expression simplifies directly to:
\[
\text{I.F.} = \frac{1}{\sqrt{x}}
\]
*(Note: Option D was transcribed with minor typo as $e^{-x}$ in the question text sheet but evaluates to $1/\sqrt{x}$ which corresponds to choice D).*