First Order Differential Equation involving cotangent

Question

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?

Answer 1

Let $ u=y-2x$ and your equation $$\begin{align*} y' & = 2 + cot(y-2x) \\ \end{align*}$$

transforms to $$ u' = \cot u $$ which is separable.

Answer 2

$$\begin{align*} y' & = 2 + \cot(y-2x) \\ \end{align*}$$ $$y' -2 = \cot(y-2x) $$ 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 $$ y(x)= 2x+\arccos(e^{1-x})$$

Answer 3

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 $$\operatorname{arcsec}(e^{x-1})=x-2y\\y=-\frac{1}{2}\left(\operatorname{arcsec}(e^{x-1})-x\right)\\y=\frac{1}{2}\left(x-\operatorname{arcsec}(e^{x-1})\right)$$