Solve dy/dx=e^(3x+2y) by separation of variables

[ElijahRockers]

Homework Statement

solve dy/dx=e^(3x+2y) by separation of variables

The Attempt at a Solution

$\frac{dy}{dx}=e^{3x+2y}$

$\frac{dy}{dx}=e^{3x}e^{2y}$

$e^{-2y}dy=e^{3x}dx$

$\int e^{-2y}dy=\int e^{3x}dx$

$e^{-2y}=-\frac{2}{3}e^3x + C$

$-2y = ln(-\frac{2}{3}e^{3x}+C)$

$y=1\frac{1}{2}ln(-\frac{2}{3}e^{3x}+C)$

Just a little wondering mostly about the natural log thing. Can I take the natural log of the -ve function wrt x because of the constant?

 

[SammyS]

Homework Statement

solve dy/dx=e^(3x+2y) by separation of variables

The Attempt at a Solution

$\displaystyle \frac{dy}{dx}=e^{3x+2y}$

$\displaystyle \frac{dy}{dx}=e^{3x}e^{2y}$

$e^{-2y}dy=e^{3x}dx$

$\int e^{-2y}dy=\int e^{3x}dx$

$e^{-2y}=-\frac{2}{3}e^{3x} + C$

$-2y = ln(-\frac{2}{3}e^{3x}+C)$

$\displaystyle y=-\frac{1}{2}ln(-\frac{2}{3}e^{3x}+C)$

Just a little wondering mostly about the natural log thing. Can I take the natural log of the -ve function wrt x because of the constant?

(Fixed a couple of typos above.)

Of course this gives a "family" of solutions, depending upon the value of C.

For each value of C, the solution is real, only if $\displaystyle C-\frac{2}{3}e^{3x}>0\ \text{ i.e. } x<\frac{1}{3}\ln\left(\frac{3C}{2}\right) \ .$

 

[ElijahRockers]

so I don't need any absolute value brackets or anything? I am a little shaky when it comes to logs/abs value

 

[HallsofIvy]

No, you don't need absolute value here. If $e^x= y$, then ln(y)= x.

You are thinking of the integral, $\int (1/x)dx= ln|x|+ C$, but you are not doing that here. [itex]\int e^{ax}dx= (1/a)e^{ax}+ C without any absolute value.