## Solving a Differential Equation By Regrouping Terms

\[(2x-y)dx+(2y-x)dx=0\]

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 \[x^2+y^2-xy=1\]

.