Question: User: $x\,dy = y\,dx + \sqrt{xy}\,dx$
1. **State the problem:**
We are given the differential equation:
$$x\,dy = y\,dx + \sqrt{xy}\,dx$$
and we want to solve it.
2. **Rewrite the equation:**
Divide both sides by $dx$ (assuming $dx \neq 0$):
$$x \frac{dy}{dx} = y + \sqrt{xy}$$
3. **Isolate $\frac{dy}{dx}$:**
$$\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{xy}}{x} = \frac{y}{x} + \frac{\sqrt{x y}}{x}$$
Simplify the second term:
$$\frac{\sqrt{x y}}{x} = \frac{\sqrt{x} \sqrt{y}}{x} = \frac{\sqrt{y}}{\sqrt{x}}$$
So,
$$\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{y}}{\sqrt{x}}$$
4. **Substitution:**
Let $v = \sqrt{y}$, so $y = v^2$.
Then,
$$\frac{dy}{dx} = 2v \frac{dv}{dx}$$
Substitute into the equation:
$$2v \frac{dv}{dx} = \frac{v^2}{x} + \frac{v}{\sqrt{x}}$$
Divide both sides by $v$ (assuming $v \neq 0$):
$$2 \frac{dv}{dx} = \frac{v}{x} + \frac{1}{\sqrt{x}}$$
5. **Rewrite as:**
$$2 \frac{dv}{dx} - \frac{v}{x} = \frac{1}{\sqrt{x}}$$
or
$$\frac{dv}{dx} - \frac{v}{2x} = \frac{1}{2 \sqrt{x}}$$
6. **This is a linear first-order ODE:**
$$\frac{dv}{dx} + P(x) v = Q(x)$$
where
$$P(x) = -\frac{1}{2x}, \quad Q(x) = \frac{1}{2 \sqrt{x}}$$
7. **Find integrating factor $\mu(x)$:**
$$\mu(x) = e^{\int P(x) dx} = e^{\int -\frac{1}{2x} dx} = e^{-\frac{1}{2} \ln x} = x^{-\frac{1}{2}} = \frac{1}{\sqrt{x}}$$
8. **Multiply both sides by $\mu(x)$:**
$$\frac{1}{\sqrt{x}} \frac{dv}{dx} - \frac{v}{2 x^{3/2}} = \frac{1}{2 x}$$
This simplifies to:
$$\frac{d}{dx} \left( \frac{v}{\sqrt{x}} \right) = \frac{1}{2 x}$$
9. **Integrate both sides:**
$$\int \frac{d}{dx} \left( \frac{v}{\sqrt{x}} \right) dx = \int \frac{1}{2 x} dx$$
$$\frac{v}{\sqrt{x}} = \frac{1}{2} \ln |x| + C$$
10. **Solve for $v$:**
$$v = \sqrt{x} \left( \frac{1}{2} \ln |x| + C \right)$$
Recall $v = \sqrt{y}$, so:
$$\sqrt{y} = \sqrt{x} \left( \frac{1}{2} \ln |x| + C \right)$$
11. **Square both sides to find $y$:**
$$y = x \left( \frac{1}{2} \ln |x| + C \right)^2$$
**Final answer:**
$$\boxed{y = x \left( \frac{1}{2} \ln |x| + C \right)^2}$$