Solving differential equation ${\rm d}y/{\rm d}x = (4x - y)^2$ Let $u = 4x - y$
${\rm d}u = 4{\rm d}x - {\rm d}y$
$4{\rm d}x - {\rm d}u = {\rm d}y$
$4{\rm d}x - {\rm d}u = (4x - y)^2$
$4{\rm d}x - {\rm d}u = u^2$
I was stuck at $-{\rm d}u = (u - 4){\rm d}x$
by separation of variables (SOV), so what's next?
 A: $$u=4x-y \implies \frac{du}{dx}=4-\frac{dy}{dx},$$ the the ODE becomes
$$\frac{du}{dx}=4-u^2 \implies \int \frac{du}{4-u^2}= \int dx \implies \frac{1}{2} \tanh^{-1}(u/2)=x+C.$$
A: To solve
$$dy/dx = (4x-y)^2$$
is to find a function $y(x)$, that satisfies this equation.
Substitute
$$u=4x-y$$
It follows
$$\Rightarrow du/dx = 4 - dy/dx \Leftrightarrow dy/dx = 4-du/dx  $$
Plug in the $u$ and the $du/dx$ in the original equation
$$\begin{align} & dy/dx = 4-du/dx = u^2 = (4x-y)^2 \\ \Rightarrow & \ du/dx = 4-u^2 \\  \Rightarrow & \ dx/du = 1/(4-u^2) \\  \Rightarrow &  \ dx = du/(4-u^2)\end{align}$$
It follows
$$\begin{align}x &= \int du/(4-u^2) +C\\
&= (1/4) \cdot  \int du/(1-(u/2)^2) + C\\
&= (2/4) \cdot  \int da/(1-a^2) + C && a = u/2 \ \ \text{and} \ \ da/du = 1/2\\ 
&= (1/2) \cdot \text{arctan}(a) + C \\ 
&= (1/2) \cdot \text{arctan}\left(u/2\right) + C \\ 
&= (1/2) \cdot \text{arctan}\left((4x-y)/2\right) + C && u = 4x-y \\ 
\end{align}$$
This means that
$$ x = (1/2) \cdot \text{arctan}\left((4x-y)/2\right) + C$$
which has to be solved for $y$. Maybe this way:
Substract C from both sides
$$ x - C = (1/2) \cdot \text{arctan}\left((4x-y)/2\right) $$
Multiply by $2$
$$ 2(x - C) =  \text{arctan}\left((4x-y)/2\right) $$
Take $\tan$ on both sides
$$ \tan(2(x - C)) =  (4x-y)/2 $$
Multiply by $2$
$$ 2\tan(2(x - C)) =  4x-y $$
Rearrange
$$ y =  4x-2\tan(2(x - C)) $$
Not so sure if $ y =  4x-2\tan(2(x - C)) $ satisfies $dy/dx = (4x-y)^2$ ... ?
