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Sydney Sales
Sydney Sales's picture

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

Infinitesimal's picture


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

Since the equation has separable variables,

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\\

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

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