The general solution of $\large\frac{dy}{dx}$$=2xe^{x^2-y}$ is: $$(A)\;e^{x^2-y}=c \quad (B)\;e^{-y}+e^{x^2}=c \quad(C)\;e^y=e^{x^2}+c \quad (D)\;e^{x^2}+y=c$$

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