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What is the general solution to the differential equation: $\cfrac{dy}{dx}=e^{3y-x}$?

$3e^{3y}={e^{x}}+C$

$\cfrac{1}{3e^{3y}}=\cfrac{1}{e^{x}}+C$

$e^{3y}=e^{x}+C$

$\cfrac{1}{e^{3y}}=\cfrac{1}{e^{x}}+C$