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?
 A: The usual way of doing such a question is to divide the top and bottom of the fraction by $x$ yielding $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4\frac{y}{x} - 3}{2 - \frac{y}{x}}$$ and then use the substitition $v = \frac{y}{x} \implies y = vx$ so that $\frac{\mathrm{d}y}{\mathrm{d}x} = v + x\frac{\mathrm{d}v}{\mathrm{d}x}$ to get $$v + x\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{4v}{2-v} - \frac{3}{2-v}$$ can you take it from there? If not, you can continue by subtracting $v$ from both sides, giving you $$x \frac{\mathrm{d}v}{\mathrm{d}x} = - \frac{(v+3)(v-1)}{v-2} \implies \int \frac{2-v}{(v-1)(v+3)} \, \mathrm{d}v = \ln |x| + c$$ where partial fraction decomposition on the rational function of $v$ yields $$\frac{2-v}{(v-1)(v+3)} = \frac{1}{4(v-1)} - \frac{5}{4(v+3)}$$ Hence we get $$\frac{1}{4} \ln |v-1| - \frac{5}{4}\ln |v+3| = \ln |x| + c$$
A: Your equation belongs to a class of homogeneous ODE
$$f(x,y)dx+ g(x,y)dy=0 $$
where $f(x,y)=x(-3+\frac{y}{x})$ and $g(x,y)=x(-2+\frac{y}{x})$
The typical way to solve this is : let $u=\frac{y}{x}$ hence $y=ux$
therefore $dy=udx +xdu$
plug these into your ODE, simplify, you will get a first order ODE in term of $u$
and you can solve it using the method of separation.
A: 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 } $
$$ -\frac { 5 }{ 4 } \int { \frac { dt }{ t+3 }  } +\frac { 1 }{ 4 } \int { \frac { dt }{ t-1 } = } \int { \frac { dx }{ x }  } $$
$$-\frac { 5 }{ 4 } \ln { \left| t+3 \right|  } +\frac { 1 }{ 4 } \ln { \left| t-1 \right| =\ln { Cx }  } $$ $$\ln { \sqrt [ 4 ]{ \left| \frac { t-1 }{ \left( t+3 \right) ^{ 5 } }  \right|  }  } =\ln { Cx } $$ $$ \frac { t-1 }{ \left( t+3 \right) ^{ 5 } } =Cx^{ 4 }$$ $$ \frac { \frac { y }{ x } -1 }{ { \left( \frac { y }{ x } +3 \right)  }^{ 5 } } =Cx^{ 4 }$$
