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I'am trying to solve $\frac{\mathrm{dy} }{\mathrm{d} x}= e^{2x+3y}$

I use the law of exponent to obtain $\frac{\mathrm{dy} }{\mathrm{d} x}= e^{2x}e^{3y}$

I send the $dx$ to the other side and integrate both sides after seperating the variables.$\int \frac{dy}{e^{3y}} = \int(e^{2x})dx$

I know the right hand side is equal to $\frac{e^{2x}}{2} + c$.How about the left hand side?

2 Answers 2

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Same idea as on the "$x$" side. You are integrating $e^{-3y}dy$, and get $\frac{1}{-3}e^{-3y}$. So we end up with $\frac{1}{-3}e^{-3y}=\frac{1}{2}e^{2x}+C.$ This can be simplified in various ways. In this case, you can get an explicit formula for $y$ in terms of $x$.

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That is $\int e^{-3y}dy$ which is equal to $\frac{1}{-3}e^{-3y}$. Note that $a^b=\frac{1}{a^{-b}}$

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    Well done! +++++2013-03-01