How to find the particular solution of $dx = (2y-x)dy$ I need to find the particular solution of $dx = (2y-x)dy$ but I can't find a way to move x and dx on one side and y and dy on the other side. I know that this is a simple question but it seems that I just can't find the answer.
 A: $$dx = (2y-x)dy$$
Rewrite it as:
$$x'+x=2y$$
$$(xe^y)'=2ye^y$$
Integrate.

If you only need the particular solution of the DE, you can try this :
$$x_p=Ay+B$$
Where $A,B$ are constants.
A: This answer, which took me some time, is (in effect) a very inelegant trodding
on the path very similar to Aryadeva's answer.  Everything below presumes that
Aryadeva's elegant answer doesn't exist.

I agree with Gerry Myerson.  However, based on my bible, Calculus Vol 1,
2nd Ed. (1966 : Apostol), this is something of a trick question (at least
from the perspective of someone with an undergraduate level comprehension of
Real Analysis).
Apostol solves $y' + P(x)y = Q(x)$, where $f(a) = b$
by $f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t)e^{A(t)}dt,$
where $A(x) = \int_a^x P(t)dt.$
The original equation $dx = (2y - x)dy$ translates into 
$x' = (2y - x),$ where $x'$ represents $dx/dy.$
Consequently, I attack the problem by construing $x = g(y)$, rather than $y = f(x).$
Under this construance:
$x' + P(y)x = Q(y)$, where 
$P(y) = 1$ and $Q(y) = 2y.$
Here, $A(y) = \int_a^y 1dt = (y - a).$
Next, $\int_a^y Q(t)e^{A(t)}dt = \int_a^y 2te^{(t-a)}dt.$
Integration by parts resolves this to $(2t - 2) e^{(t - a)}$ 
evaluated at $t = y$ and $t = a$, which resolves to 
$(2y-2)e^{(y-a)} - (2a - 2).$
This yields 
$g(y) = be^{(a-y)} + e^{(a-y)} \times [(2y-2)e^{(y-a)} - (2a - 2)].$
With no initial value condition imposed, $b$ and $a$ are any arbitrary
constants.  Therefore, the above expression collapses to 
$g(y) = be^{(a-y)} + (2y - 2).$
I don't have the knowledge or sophistication to determine whether this must be
converted into the $y = f(x)$ format, and (if so) how to do that.
