Final Answer
$y=\frac{-\ln\left(-\frac{2}{3}e^{3x}+C_1\right)}{2}$
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Step-by-step Solution
Specify the solving method
1
Apply the property of the product of two powers of the same base in reverse: $a^{m+n}=a^m\cdot a^n$
$\frac{dy}{dx}=e^{3x}e^{2y}$
2
Group the terms of the differential equation. Move the terms of the $y$ variable to the left side, and the terms of the $x$ variable to the right side of the equality
$\frac{1}{e^{2y}}dy=e^{3x}dx$
3
Integrate both sides of the differential equation, the left side with respect to $y$, and the right side with respect to $x$
$\int\frac{1}{e^{2y}}dy=\int e^{3x}dx$
4
Solve the integral $\int\frac{1}{e^{2y}}dy$ and replace the result in the differential equation
$\frac{-1}{2e^{2y}}=\int e^{3x}dx$
5
Solve the integral $\int e^{3x}dx$ and replace the result in the differential equation
$\frac{-1}{2e^{2y}}=\frac{1}{3}e^{3x}+C_0$
6
Find the explicit solution to the differential equation. We need to isolate the variable $y$
$y=\frac{-\ln\left(-\frac{2}{3}e^{3x}+C_1\right)}{2}$
Final Answer
$y=\frac{-\ln\left(-\frac{2}{3}e^{3x}+C_1\right)}{2}$