$y-xy' =2(x+yy'), y(1) =1 $ is.. 
Question: $y-xy' =2(x+yy'), y(1) =1 $ is..
  

Firstly, I do not even know what the question is asking for.
Secondly, Why does this question put $y(1)=1$? Isn't it obvious or is there any special meaning to it.
Surprisingly, the answer to it is also very unique.
The answer is: $\tan^{-1} \frac{y}{x} + \log (x^2 + y^2) = \frac{\pi}{4} + \log 2$.
Also, another thing is that this question was solved by trigonometry, which is confusing me further.
P.S.: Please do not downvote as I completely do not understand the question even a bit. Thats why I am asking it here.

How the equation can be solved
The equation can be rewritten as
$$
y' = \frac{y-2x}{2y+x}.
$$
Introduce $u = y/x$ so that $y = xu$ and $y' = xu' + u.$ This makes the equation take the form
$$
xu' + u = \frac{xu-2x}{2xu+x} = \frac{u-2}{2u+1}
$$
i.e.
$$
xu' = \frac{u-2}{2u+1} - u = \frac{(u-2)-u(2u+1)}{2u+1} = -2 \frac{u^2+1}{2u+1}.
$$
or, separated,
$$
\frac{2u+1}{u^2+1}u' = -\frac{2}{x}.
$$
The left hand side can be split into two terms:
$$
\frac{2u+1}{u^2+1}u' = \frac{2uu'}{u^2+1} + \frac{u'}{u^2+1} = \left( \ln(u^2+1) + \arctan(u) \right)'.
$$
Thus,
$$
\left( \ln(u^2+1) + \arctan(u) \right)' = -\frac{2}{x} = \left( -2 \ln(x) \right)' = \left( \ln\frac{1}{x^2} \right)'
$$
so
$$
\ln(u^2+1) + \arctan(u) = \ln\frac{1}{x^2} + C,
$$
which after backsubstitution of $u=y/x$ gives
$$
\ln((y/x)^2+1) + \arctan(y/x) = \ln\frac{1}{x^2} + C
$$
or
$$
\ln(x^2+y^2) + \arctan(y/x) = C.
$$
We now get $C$ from inserting $(x,y)=(1,1)$:
$$
C = \ln(1^2+1^2) + \arctan(1/1) = \ln 2 + \pi/4.
$$
