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Let's solve the differential equation step by step. The given equation is:

\[
x^2 y' + 2xy = 4x \cos(2x) \sqrt{y}
\]

### Step 1: Simplifying the equation
We can divide both sides by \(x^2\) to simplify:

\[
y' + \frac{2}{x} y = \frac{4 \cos(2x) \sqrt{y}}{x}
\]

Now, let’s make a substitution to reduce the complexity. Define \( z = \sqrt{y} \), so \( y = z^2 \) and \( y' = 2z z' \). Substituting this into the equation:

\[
2z z' + \frac{2}{x} z^2 = \frac{4 \cos(2x) z}{x}
\]

### Step 2: Simplifying further
We can divide the entire equation by \(z\) (assuming \(z \neq 0\)):

\[
2z' + \frac{2}{x} z = \frac{4 \cos(2x)}{x}
\]

### Step 3: Solving this equation
This is a linear first-order differential equation in \(z\). We can solve it using the integrating factor method.

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

\[
I(x) = e^{\int \frac{2}{x} dx} = e^{2 \ln x} = x^2
\]

### Step 4: Apply the integrating factor
Multiplying both sides of the equation by \(x^2\):

\[
x^2 (2z') + 2x z = 4x \cos(2x)
\]

This simplifies to:

\[
\frac{d}{dx}(x^2 z) = 4x \cos(2x)
\]

### Step 5: Integrating both sides
Integrating both sides with respect to \(x\):

\[
x^2 z = \int 4x \cos(2x) dx
\]

Using integration by parts for the right-hand side:

\[
\int 4x \cos(2x) dx = 2x \sin(2x) + \cos(2x) + C
\]

Thus, the solution becomes:

\[
x^2 z = 2x \sin(2x) + \cos(2x) + C
\]

### Step 6: Solving for \(z\)
Dividing both sides by \(x^2\):

\[
z = \frac{2 \sin(2x)}{x} + \frac{\cos(2x)}{x^2} + \frac{C}{x^2}
\]

### Step 7: Substituting back \(z = \sqrt{y}\)
Finally, since \(z = \sqrt{y}\), we square both sides:

\[
y = \left( \frac{2 \sin(2x)}{x} + \frac{\cos(2x)}{x^2} + \frac{C}{x^2} \right)^2
\]

### Conclusion:
After evaluating the options, the correct answer seems to be:

\[
\boxed{(b) \ y^{1/2} = \frac{\sin(2x) + C}{x}}
\]

Would you like further clarification or a deeper breakdown of any step? Here are some follow-up questions:

1. What are integrating factors and how do they help solve differential equations?
2. Can we always substitute \(z = \sqrt{y}\) in such problems?
3. Why do we assume \(z \neq 0\) when dividing the equation?
4. How does the method of integration by parts work?
5. What are some general techniques for solving first-order linear differential equations?

**Tip:** When solving linear differential equations, always check if an integrating factor can simplify the equation efficiently.

The general solution of x²y' + 2xy = 4x cos(2x)√y is:

Discover the general solution for the differential equation x²y' + 2xy = 4x cos(2x)√y using integrating factors and substitution methods. This problem involves key concepts such as first-order linear differential equations and is suitable for undergraduate students.

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