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The equation given in the problem is:

\[
2xy' - 3y = \sqrt{x}
\]

We are asked to find the solutions of the equation on the interval \((0, +\infty)\), and the possible solutions are provided in the form of four different function choices: A, B, C, and D. Let's analyze this step by step.

### Step 1: Solve the Differential Equation

The differential equation is:

\[
2xy' - 3y = \sqrt{x}
\]

First, let's rewrite it in a more standard form:

\[
y' - \frac{3}{2x}y = \frac{\sqrt{x}}{2x}
\]

This is a linear first-order differential equation of the form \( y' + P(x)y = Q(x) \), where:

- \( P(x) = -\frac{3}{2x} \)
- \( Q(x) = \frac{\sqrt{x}}{2x} \)

### Step 2: Solve Using an Integrating Factor

The integrating factor \( \mu(x) \) is given by:

\[
\mu(x) = e^{\int P(x) dx} = e^{\int -\frac{3}{2x} dx} = e^{-\frac{3}{2} \ln |x|} = x^{-\frac{3}{2}}
\]

Now, multiply the entire equation by the integrating factor:

\[
x^{-\frac{3}{2}} y' - x^{-\frac{3}{2}} \frac{3}{2x} y = x^{-\frac{3}{2}} \frac{\sqrt{x}}{2x}
\]

The left-hand side simplifies to:

\[
\frac{d}{dx}\left( x^{-\frac{3}{2}} y \right) = \frac{1}{2} x^{-2}
\]

### Step 3: Integrate Both Sides

Now integrate both sides with respect to \(x\):

\[
x^{-\frac{3}{2}} y = \int \frac{1}{2} x^{-2} dx
\]

The integral on the right-hand side is:

\[
\int \frac{1}{2} x^{-2} dx = -\frac{1}{2x} + C
\]

So, we have:

\[
x^{-\frac{3}{2}} y = -\frac{1}{2x} + C
\]

### Step 4: Solve for \(y\)

Multiplying both sides by \(x^{\frac{3}{2}}\), we get:

\[
y = -\frac{\sqrt{x}}{2} + C x^{\frac{3}{2}}
\]

### Step 5: Compare with the Given Solutions

The general solution we found is:

\[
y = -\frac{\sqrt{x}}{2} + C x^{\frac{3}{2}}
\]

This matches option A:

\[
A. \quad y(x) = -\frac{\sqrt{x}}{2} + \lambda x^{\frac{3}{2}}, \quad \lambda \in \mathbb{R}
\]

Thus, the correct answer is **A**.

---

Would you like further details on this process or a deeper explanation of any step? Here are some related questions to explore:

1. How do you find the integrating factor for a first-order linear differential equation?
2. What is the significance of an integrating factor in solving differential equations?
3. How do we determine the general solution for non-homogeneous linear differential equations?
4. Can this method be applied to other forms of first-order linear differential equations?
5. What are some other examples where integrating factors simplify the solving process?

**Tip**: Always check if a given differential equation can be simplified into a standard form before solving. This reduces complexity and avoids errors.

On considère l'équation 2xy' - 3y = √x. Les solutions de l'équation sur ]0,+∞[ sont les fonctions de la forme : A. x → -√x/2 + λx^(3/2), λ ∈ ℝ; B. x → √x/2√2 + λx^(3/2), λ ∈ ℝ; C. x → -√x/2 + λx, λ ∈ ℝ; D. x → -√x/2 + λx^(1/2), λ ∈ ℝ.

Discover how to solve the first-order linear differential equation 2xy' - 3y = √x using the integrating factor method. The solution involves detailed steps including transforming the equation and integrating. Suitable for calculus students.

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