Differential equation $\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$ I am not sure which type of differential equation this falls into:
$$\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$$ any hints? P.S. I first tried reornazing it so I have $y'$ alone, and hoping that I would get a homogeneous equation, but no such luck.
 A: Divide both side to the $x^{ 2 }$ $$\left( x^{ 2 }+xy \right) y'=x\sqrt { x^{ 2 }-y^{ 2 } } +xy+y^{ 2 }\\ \left( 1+\frac { y }{ x }  \right) { y }^{ \prime  }=\sqrt { 1-\frac { { y }^{ 2 } }{ { x }^{ 2 } }  } +\frac { y }{ x } +\frac { { y }^{ 2 } }{ { x }^{ 2 } } \\ y=zx\\ { y }^{ \prime  }={ z }^{ \prime  }x+z\\ \left( 1+z \right) \left( { z }^{ \prime  }x+z \right) =\sqrt { 1-{ z }^{ 2 } } +z+{ z }^{ 2 }\\ { z }^{ \prime  }x+z+z{ z }^{ \prime  }x+{ z }^{ 2 }=\sqrt { 1-{ z }^{ 2 } } +z+{ z }^{ 2 }\\ { z }^{ \prime  }x\left( 1+z \right) =\sqrt { 1-{ z }^{ 2 } } \\ \int { \frac { \left( 1+z \right) dz }{ \sqrt { 1-{ z }^{ 2 } }  }  } =\int { \frac { dx }{ x }  } $$
substitute now :$z=\sin { \theta  } $
$$\int { \frac { 1+\sin { \theta  }  }{ \cos { \theta  }  } \cos { \theta  } d\theta  } =\ln { Cx } \\ \theta -\cos { \theta  } =\ln { Cx } \\ \arcsin { z-\sqrt { 1-{ z }^{ 2 } }  } =\ln { Cx } $$
so the final aswer is :

$$\arcsin { \frac { y }{ x }  } -\frac { \sqrt { { x }^{ 2 }-{ y }^{ 2 } }  }{ x } =\ln { Cx } $$

A: Divide $x^2$ from both sides and then let $z=\dfrac{y}x$.
Note that $y'=z+xz'$.
The new equation becomes $(1+z)y'=\sqrt{1-z^2}+z+z^2$.
$y'=\sqrt{\dfrac{1-z}{1+z}}+z$
$x\dfrac{\mathrm dz}{\mathrm dx}=\sqrt{\dfrac{1-z}{1+z}}$
$\dfrac{\mathrm dx}{x}=\sqrt{\dfrac{1+z}{1-z}}\ \mathrm dz$
$\ln(x)=\sqrt{\dfrac1{1-z}}\left((z-1)\sqrt{z+1}+2\sqrt{1-z}\sin^{-1}\left(\dfrac{\sqrt{1+z}}{\sqrt2}\right)\right)+C$
And... it gets too complicated.
