Separation of variables, can't get y out of exponent

[find_the_fun]

Solve the DE by using separation of variables
\(\displaystyle \frac{dy}{dx} = e^{3x+2y}\)

Break up \(\displaystyle e^{3x+2y} = e^{3x}e^{2y}\) Move x's and y's to their own side of the equation.
\(\displaystyle \frac{1}{e^{2y}} dy = e^{3x} dx\)
Integrate both sides of the equation to get \(\displaystyle \frac{-e^{2y}}{2x}=\frac{e^{3x}}{3}+C\)

I don't know how to isolate the y; I don't know how to get it down from the exponent.

 

[topsquark]

Solve the DE by using separation of variables
\(\displaystyle \frac{dy}{dx} = e^{3x+2y}\)

Break up \(\displaystyle e^{3x+2y} = e^{3x}e^{2y}\) Move x's and y's to their own side of the equation.
\(\displaystyle \frac{1}{e^{2y}} dy = e^{3x} dx\)
Integrate both sides of the equation to get \(\displaystyle \frac{-e^{2y}}{2x}=\frac{e^{3x}}{3}+C\)

I don't know how to isolate the y; I don't know how to get it down from the exponent.

I'm going to presume that the x on the LHS is a typo. Otherwise I have no idea where it came from.

Simplifying a bit we have:
\(\displaystyle e^{-2y} = -\frac{2}{3}e^{3x} - 2C\)

Your turn: Take ln of both sides. And no, it doesn't simplify beyond this, unless C = 0, which is a matter for the boundary conditions, which you don't have.

-Dan