Question

e^(2x-y) dx + e^(y-2x) dy =0

Answer 1

Let us solve the differential equation $e^{2x-y} dx + e^{y-2x}dy =0,$ which is equivalent to $e^{2x}e^{-y} dx + e^{y}e^{-2x}dy =0.$ Let us multiply both parts by $e^ye^{2x}.$ Then we get $e^{4x}dx+e^{2y}dy=0.$ It follows that $\int e^{4x}dx+\int e^{2y}dy=C,$ and hence $\frac{1}{4}e^{4x}+\frac{1}{2}e^{2y}=C.$ We conclude that the general solution can be written in the form $e^{4x}+2e^{2y}=C.$