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 $\boxed{e^{3x}=Cy^3(e^{3x}+1)}$
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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 $\boxed{e^{3x}=Cy^3(e^{3x}+1)}$
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