Way to make this homogoneous ODE seperable? Is my algebra correct, turning this:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4y-3x}{2x-y}$$
Into this:
I split into the difference of the two fractions,
then factored x out of the left fraction, and factored y
out of the right fraction, getting:

$$\frac{4\frac{y}{x}}{x(1-\frac{y}{x})} - \frac{3\frac{x}{y}}{2\frac{x}{y}-1}$$
Now I can substitute $v = \frac{y}{x}, v^{-1}= \frac{x}{y}$, right?
Then separate? Then integrate?

I solved it maybe it will help you:$$\frac { dy }{ dx } =\frac { 4y-3x }{ 2x-y } =\frac { 4\frac { y }{ x } -3 }{ 2-\frac { y }{ x }  } \\ \frac { y }{ x } =t$$
$$ y=xt\\ y^{ \prime  }=t+x{ t }^{ \prime  }\\ t+x{ t }^{ \prime  }=\frac { 4t-3 }{ 2-t } $$ $$ x{ t }^{ \prime  }=\frac { { t }^{ 2 }+2t-3 }{ 2-t } =\frac { \left( t+3 \right) \left( t-1 \right)  }{ 2-t } \\ \int { \frac { 2-t }{ \left( t+3 \right) \left( t-1 \right)  } dt } =\int { \frac { dx }{ x }  } \\ \frac { 2-t }{ \left( t+3 \right) \left( t-1 \right)  } =\frac { A }{ t+3 } +\frac { B }{ t-1 } $$ $$2-t=A\left( t-1 \right) +B\left( t+3 \right) =\left( A+B \right) t+3B-A$$
$A=-\frac { 5 }{ 4 } $,$B=\frac { 1 }{ 4 } $
$$ -\frac { 5 }{ 4 } \int { \frac { dt }{ t+3 }  } +\frac { 1 }{ 4 } \int { \frac { dt }{ t-1 } = } \int { \frac { dx }{ x }  } $$
$$-\frac { 5 }{ 4 } \ln { \left| t+3 \right|  } +\frac { 1 }{ 4 } \ln { \left| t-1 \right| =\ln { Cx }  } $$ $$\ln { \sqrt [ 4 ]{ \left| \frac { t-1 }{ \left( t+3 \right) ^{ 5 } }  \right|  }  } =\ln { Cx } $$ $$ \frac { t-1 }{ \left( t+3 \right) ^{ 5 } } =Cx^{ 4 }$$ $$ \frac { \frac { y }{ x } -1 }{ { \left( \frac { y }{ x } +3 \right)  }^{ 5 } } =Cx^{ 4 }$$
