The given differential equation is $$ \frac{dy}{dx} = e^{3x} \cdot e^{-y} $$ By separating the variables $x$ and $y$, we get: $$ e^{y} \, dy = e^{3x} \, dx $$
Integrate both sides of the equation: $$ \int e^{y} \, dy = \int e^{3x} \, dx $$ The integral of $e^{y}$ is $e^{y}$ and the integral of $e^{3x}$ is $\frac{e^{3x}}{3}$. Adding the constant of integration $C$: $$ e^{y} = \frac{e^{3x}}{3} + C $$
To match the given options, multiply the entire equation by 3: $$ 3e^{y} = e^{3x} + 3C $$ Since $3C$ is still an arbitrary constant, we can replace it with $C$: $$ 3e^{y} = e^{3x} + C $$
Final Answer: $3e^{y}=e^{3x}+C$
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