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