I think this is the first time when I handle of **non-linear** differential equation.

$$ \underbrace{x \frac{dy}{dx} + y = x\sqrt{ y } }\_{x > 0} \tag{1} $$

$$ \frac{dy}{dx} + \frac{1}{ x } y = \sqrt{ y } $$

$$ \underbrace{\frac{d}{dx}\left(yx\right)}\_{\text{LHS of eqn1} } = x \sqrt{ y } ~~ \leftarrow~~ \text{What can I do for next?} $$

Which website is suitable for this problem?

**ADD**

I got the following after I have read the post of @Gerd

$$ \frac{d}{dx}\left(yx\right) = x \sqrt{ y } 　　\tag{2} $$

$$ \frac{d}{dx}\left(xy\right) = x^{\frac{1}{2} } \cdot x^{\frac{1}{2} } \sqrt{ y } $$

$$ = \sqrt{ x } \sqrt{ xy } $$

$$ z:=xy \tag{3} $$

$$ \frac{ dz }{ dx } = \sqrt{ x } \sqrt{ z } \tag{4} $$

$$ \frac{ 1 }{ \sqrt{ z } } \frac{dz}{dx} = \sqrt{ x } $$

$$ \int\_{ }^{ } \frac{1}{ \sqrt{ z } } \frac{dz}{dx} \,dx = \int\_{ }^{ } \sqrt{ x } \,dx $$

$$ \int\_{ }^{ } \frac{1}{ \sqrt{ z } } \,dz = \int\_{ }^{ } x^{\frac{1}{ 2 } } \,dx $$

$$ \int\_{ }^{ } \frac{ 1 }{ z^{\frac{1}{2} } } \,dz = \frac{ x^{\frac{ 3 }{ 2 } } }{ \frac{ 3 }{ 2 } } + \text{const}\_{1} $$

$$ \int\_{ }^{ } z^{-1/2} \,dz = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}\_{1} $$

$$ \frac{ z^{\frac{1}{2}} }{ \frac{1}{2} } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}\_{1} $$

$$ 2z^{\frac{1}{2}} = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}\_{1} $$

$$ 2 \sqrt{ z } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}\_{1} $$

$$ 2 \sqrt{ yx } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}\_{1} $$

$$ 2 \sqrt{ yx } - \frac{ 2 }{ 3 } x\sqrt{ x } = \text{const}\_{1} $$

$$ \underbrace{\sqrt{ x } \left( 2 \sqrt{ y } - \frac{ 2 }{ 3 } x \right) = \text{const}\_{1} }\_{\text{general solution} } ~~ \leftarrow~~ \text{Is this also correct?} $$