## Solving a Differential Equation By Regrouping Terms

The differential equation
$(2x-y)dx+(2y-x)dx=0$
with the curve passing through
$(2,1)$
cannot be solved by separation of variables, use of integrating factors, parts, substitution,...
If we expand the brackets and rearrange however we get
$2xdx+2ydy-(ydx+xdy)=0$

We can write this as
$d(x^2+y^2)-d(xy)=0$
.
Integration now gives
$x^2+y^2-xy=c$
.
The curve passes through
$(2,1)$
so
$2^2+1^2-2 \times 2 \times 1=1=c$
and the solution is
$x^2+y^2-xy=1$
.