First Order Differential Equation involving cotangent I am trying to solve the following:
\begin{align*}
y' & = 2 + \cot(y-2x) \\ 
\end{align*} 
I think it may be possible to get it into a linear form
\begin{align*} 
y' + P(x)y = Q(x) \\
\end{align*} 
but the cot is a function of both y and x, which is causing confusion for me. Is there a better method for solving?
edit: so I performed the recommended substitution of $u = y - 2x$ and found a solution of 
\begin{align*}
(y-2x)\operatorname{arccot}(y-2x) + \frac{1}{2}\ln(1+(y-2x)^2) = x + C
\end{align*}
Now the question is: If an initial value of
\begin{align*}
y(1) = 2
\end{align*}
is given, how would I find a specific solution?
 A: Let $ u=y-2x$ and your equation $$\begin{align*}
y' & = 2 + cot(y-2x) \\ 
\end{align*}$$
transforms to $$ u' = \cot u $$ which is separable.
A: $$\begin{align*}
y' & = 2 + \cot(y-2x) \\ 
\end{align*}$$
$$y' -2 =  \cot(y-2x)  $$
We have a derivative ...
$$(y' -2) \tan(y-2x) =1 $$
$$(y' -2) \frac {\sin(y-2x) }{\cos(y-2x)}=1 $$
$$-(y' -2) \frac {\sin(y-2x) }{\cos(y-2x)}=-1 $$
$$\ln |\cos(y-2x)|)'=-1$$ 
Integrate
$$\implies  \cos(y-2x)=Ke^{-x} $$
$$\boxed{ y= 2x+\arccos(Ke^{-x})}$$
For the specific solution we have $\cos(y-2x)=Ke^{-x}$
$$ y(1)=2  \implies 1=Ke^{-1} \implies K=e$$
Therefore
$$ y(x)= 2x+\arccos(e^{1-x})$$
A: let $z=y-2x$ then differenciate respect x and you'll obtain $z'=y'-2$ and replace
$$z'+2=2+\cot(2x)\\z'=\cot(z)\\\int\tan(z)dz=\int dx\\\ln(\sec(z))=x+C\\\sec(z)=Ce^{x}\\C=e^{x}\cos(z)$$
$$C=e^{x}\cos(y-2x)\\\text{General Solution}$$
Now for the condition $y(1)=2$ replace in General Solution
$$C=e\cos(2-2(1))\\C=e\cos(0)=e$$
$$e=e^{x}\cos(y-2x)\\\text{then that means}$$
$$e^{x-1}=\sec(x-2y)\\\text{Particular Solution}$$
if you need to isolate $y$ apply $\operatorname{arcsec}$ both sides
$$\operatorname{arcsec}(e^{x-1})=x-2y\\y=-\frac{1}{2}\left(\operatorname{arcsec}(e^{x-1})-x\right)\\y=\frac{1}{2}\left(x-\operatorname{arcsec}(e^{x-1})\right)$$
