equations

2 posts / 0 new
Last post
Sydney Sales
Sydney Sales's picture
equations

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

Infinitesimal
Infinitesimal's picture

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

 
 
Subscribe to MATHalino on