ODE Homogeneous Equation Algebra Assignment Statement
Given 
\begin{equation}
\frac{dy}{dx} = \frac{(y/x)-4}{1-(y/x)}
\end{equation}
Let $v = y/x$ a dependent variable such that $y(x)=x\ v(x)$
Express $dy/dx$ in terms of $x$ , $v$ and $dv/dx$ , then replace $y$ and $dy/dx$ in Equation.1 by the expression found which involves $v$ and $dv/dx$. Show the resulting differential equation is 
$$ x\frac{dv}{dx} = \frac{v^{2}-4}{1-v}$$
Attempt at Solution
First and foremost, since $y(x) = x v(x)$ then ,
\begin{align*}
   \frac{dv}{dx} &= \frac{d}{dx}\left(\frac{y(x)}{x}\right) \\
    &=\frac{dy/dx}{x}-\frac{y(x)}{x^{2}}\\
    x^{2}\frac{dv}{dx} &= x(dy/dx) - y(x)\\
    \implies &\frac{dy}{dx} = x\frac{dv}{dx} + \frac{y(x)}{x} 
\end{align*}
Here i'm then assuming that $y(x)/x = v \iff y/x = v$. Then I have an expression of $dy/dx$ involving $x$ , $v$ and $dv/dx$.
I then replace $y$ and $dy/dx$ in Equation 1 by this result as I'm asked to : 
$$ x\frac{dv}{dx} + v = \frac{\left(\frac{dv}{dx}+\frac{v}{x}\right)-4}{1-\left(\frac{dv}{dx}+\frac{v}{x}\right)}$$.
So from here I must show that: 
$$ \frac{\left(\frac{dv}{dx}+\frac{v}{x}\right)-4}{1-\left(\frac{dv}{dx}+\frac{v}{x}\right)} + v =\frac{v^{2}-4}{1-v}$$
After many pages of algebra I'm stuck , the farthest I've got is 
$$ \frac{x-4x^{2}-vx^{2}+v^{2}}{x^{2}-v}$$
Can someone help me ?
Thank you very much.
 A: Why try to compute $v'$ in terms of $y'$ when the expression provided begins with $y'$? We should get
$$y'=(xv)'=xv'+v$$
After this, when substituting expressions in, we should have $v=y/x$ so that
$$xv'+v=\frac{v-4}{1-v}$$
which, after a bit of algebra, becomes
$$xv'=\frac{v^2-4}{1-v}$$
as desired.
A: If $y=xv$ then by the product rule, $\dfrac{dy}{dx} = x\dfrac{dv}{dx} + v.$
\begin{align}
\frac{dy}{dx} = {} & \frac{v-4}{1-v} \\[8pt]
x \frac{dv}{dx} +v = {} & \frac{v-4}{1-v} \\[8pt]
x\frac{dv}{dx} = {} & \frac{v-4}{1-v} -v \\[8pt]
= {} & \frac{v^2 - 4}{1-v}.
\end{align}
After that you can separate variables:
$$
\frac{1-v}{v^2-4} \, dv = \frac{dx} x.
$$
Then integrate both sides, using partial fractions on the left side.
A: You wrote:
\begin{align*}
   \frac{dv}{dx} &= \frac{d}{dx}\left(\frac{y(x)}{x}\right) \\
    &=\frac{dy/dx}{x}-\frac{y(x)}{x^{2}}\\
    x^{2}\frac{dv}{dx} &= x(dy/dx) - y(x)\\
    \implies &\frac{dy}{dx} = x\frac{dv}{dx} + \frac{y(x)}{x} 
\end{align*}
These lines are correct but you should conclude that:
$$\frac{dy}{dx} = x\frac{dv}{dx} + \frac{y(x)}{x} $$
$$ \implies \frac{dy}{dx} = x\frac{dv}{dx} + v$$
Then transform your original equation into:
$$ \implies x\frac{dv}{dx} + v = \frac{v-4}{1-v}$$
Which becomes a separable DE.
