<s>[INST] Non-linear 1st order differential equation of $~x\frac{dy}{dx}+y=x\sqrt{y}~~~$where$~\sqrt{y}~$exists 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?
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I got the following after I have read the post of @Gerd
$$  \frac{d}{dx}\left(yx\right) = x \sqrt{ y } 　　\tag{2}  $$
$$  \frac{d}{dx}\left(xy\right) = x^{\frac{1}{2} } \cdot x^{\frac{1}{2} } \sqrt{ y }  $$
$$ = \sqrt{ x } \sqrt{ xy }  $$
$$  z:=xy \tag{3}  $$
$$  \frac{  dz  }{ dx   } = \sqrt{ x }  \sqrt{ z } \tag{4}  $$
$$  \frac{  1  }{  \sqrt{ z }   } \frac{dz}{dx} = \sqrt{ x }  $$
$$  \int_{ }^{ } \frac{1}{ \sqrt{ z }  } \frac{dz}{dx}  \,dx   = \int_{ }^{ } \sqrt{ x }  \,dx    $$
$$  \int_{ }^{ } \frac{1}{ \sqrt{ z }  }  \,dz = \int_{ }^{ } x^{\frac{1}{ 2 } } \,dx     $$
$$  \int_{ }^{ } \frac{  1  }{  z^{\frac{1}{2} }  }  \,dz   = \frac{  x^{\frac{  3  }{   2 } }  }{  \frac{  3  }{  2  }   } + \text{const}_{1} $$
$$  \int_{ }^{ } z^{-1/2} \,dz = \frac{  2  }{  3  } x\sqrt{ x }  + \text{const}_{1}  $$
$$  \frac{  z^{\frac{1}{2}}  }{  \frac{1}{2}   }  = \frac{  2  }{  3  } x\sqrt{ x }  + \text{const}_{1}  $$
$$   2z^{\frac{1}{2}}  = \frac{  2  }{  3  } x\sqrt{ x }  + \text{const}_{1}  $$
$$  2 \sqrt{ z } = \frac{  2  }{  3  } x\sqrt{ x } + \text{const}_{1}  $$
$$  2 \sqrt{ yx } = \frac{  2  }{  3  } x\sqrt{ x } + \text{const}_{1}  $$
$$  2 \sqrt{ yx } - \frac{  2  }{  3  } x\sqrt{ x } = \text{const}_{1}  $$
$$  \underbrace{\sqrt{ x }  \left( 2 \sqrt{ y } - \frac{  2  }{  3  } x  \right) = \text{const}_{1} }_{\text{general solution} } ~~ \leftarrow~~ \text{Is this also correct?}   $$
 [/INST] $$y'+\frac{y}{x} = \sqrt{y}$$
$$\frac{y'}{\sqrt{y}} +\frac{\sqrt{y}}{x} = 1$$
Using
$$z=\sqrt{y}$$
$$\frac{dz}{dx} = \frac{1}{2\sqrt{y}}\cdot y'$$
$$2\cdot\frac{dz}{dx} = \frac{y'}{\sqrt{y}}$$
Replace
$$2\cdot\frac{dz}{dx} +\frac{z}{x} = 1$$
And now you have built a linear system
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