How can I find an ODE equation from $dy/dx$ What is the ODE satisfied by $y=y(x)$ 
given that $$\frac{dy}{dx} = \frac{-x-2y}{y-2x}$$
I understand that I need to get it in some form of $\int \cdots \;dy = \int \cdots \; dx$, but am not sure how to go about it.

\begin{align}
\frac{dy}{dx} = \frac{ -x-2y }{ y-2x } \\ 
\frac{ dy }{ dx } =\frac{ -1-2\frac{ y }{ x }  }{ \frac{ y }{ x } -2 } \\ \frac{ y }{ x } = t \to \quad y=xt \quad \Rightarrow \quad dy=t+x \frac{ dt }{ dx } \\ 
t+x\frac{ dt }{ dx } =\frac{ -1-2t }{ t-2 } \\ 
x\frac{ dt }{ dx } =\frac{ -1-2t }{ t-2 } -t=\frac{ -1-{ t }^{ 2 } }{ t-2 } \\
\int { \frac{ 2-t }{ t^{ 2 }+1 } dt } =\int { \frac{ dx }{ x }  } \\ 
\int { \left( \frac{ 2 }{ { t }^{ 2 }+1 } -\frac{ t }{ { t }^{ 2 }+1 }  \right) dt= } \int { \frac { dx }{ x }  } \\ 2\int { \frac { 1 }{ { t }^{ 2 }+1 } dt-\frac { 1 }{ 2 } \int { \frac { d\left( { t }^{ 2 }+1 \right)  }{ { t }^{ 2 }+1 }  } =\int { \frac { dx }{ x }  }  } \\ 2\arctan { \left( t \right) -\frac { 1 }{ 2 }  } \ln { \left( 1+{ t }^{ 2 } \right) =\ln { \left| x \right| +C }  } \\ \arctan { \left( t \right) =\frac { 1 }{ 2 } \ln { \left| x\sqrt { 1+{ t }^{ 2 } }  \right| +C }  } \\ \arctan { \left( \frac { y }{ x }  \right) =\frac { 1 }{ 2 } \ln { \left| x\sqrt { 1+{ \left( \frac { y }{ x }  \right)  }^{ 2 } }  \right| +C }  } 
\end{align}
