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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?

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up vote 3 down vote accepted

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! +++++ –  amWhy Mar 1 '13 at 1:03
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