To solve the differential equation
\[ y' = \frac{4}{x} y + x \sqrt{y}, \]
we can rearrange it as
\[ \frac{dy}{dx} - \frac{4}{x} y = x \sqrt{y}. \]
This is a first-order linear differential equation. We can rewrite it in the standard form:
\[ \frac{dy}{dx} - \frac{4}{x} y - x \sqrt{y} = 0. \]
This equation is not directly linear due to the \( \sqrt{y} \) term. We can try a substitution to simplify the equation. Let's use the substitution:
\[ y = v^2, \]
where \( v \) is a function of \( x \). Then we have:
\[ y' = 2v \frac{dv}{dx}, \]
which leads us to substitute into the original differential equation:
\[ 2v \frac{dv}{dx} = \frac{4}{x} v^2 + x v. \]
We can simplify this as:
\[ 2v \frac{dv}{dx} - \frac{4}{x} v^2 - x v = 0. \]
Dividing through by \( v \) (assuming \( v \neq 0 \)) gives:
\[ 2 \frac{dv}{dx} = \frac{4}{x} v + x. \]
Now we can separate the differential equation:
\[ \frac{dv}{dx} = \frac{2}{x} v + \frac{x}{2}. \]
This is again a linear first-order differential equation. We can find an integrating factor, which is
\[ \mu(x) = e^{\int \frac{2}{x} dx} = e^{2 \ln |x|} = |x|^2. \]
Multiplying the entire equation by the integrating factor:
\[ |x|^2 \frac{dv}{dx} - 4|x|v = \frac{x^3}{2}. \]
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx} \left( |x|^2 v \right) = \frac{x^3}{2}. \]
Integrating both sides with respect to \( x \):
\[ |x|^2 v = \frac{x^4}{8} + C, \]
where \( C \) is the integration constant. We can write this expression for \( v \):
\[ v = \frac{x^4}{8|x|^2} + \frac{C}{|x|^2}. \]
Now substituting back for \( y \) gives:
\[ y = v^2 = \left( \frac{x^4}{8|x|^2} + \frac{C}{|x|^2} \right)^2. \]
This would be the general solution for the differential equation.
Thus, the solution of the differential equation is:
\[ y = \left( \frac{x^2}{8} + \frac{C}{x^2} \right)^2, \]
where \( C \) is a constant.
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