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## $ydx = (e^{3x}+1)dy$

D.E.

$ydx = (e^{3x}+1)dy$

Since the equation has separable variables,

\begin{eqnarray*}

ydx &=& (e^{3x}+1)dy\\

\dfrac{dx}{e^{3x}+1} - \dfrac{dy}{y} &=& 0\\

\int \dfrac{e^{3x} dx}{e^{6x}+e^{3x}} - \int \dfrac{dy}{y} &=& \int 0\\

\dfrac{1}{3} (3x - \ln(e^{3x}+1)) - \ln y &=& C\\

3x - \ln(e^{3x}+1) - 3\ln y &=& C\\

\end{eqnarray*}

Rearranging, the answer is $\boxed{e^{3x}=Cy^3(e^{3x}+1)}$