1. ## Doing something wrong

In connection with a thread posted earlier Anteeks was having problems with the equation: $y'-2y=3, y(0)=1$.
If you solve this with an integrating factor you get a factor of $e^{-2x}$.
$(e^{-2x}*y)'=3e^{-2x}$
$e^{-2x}y=-\frac{3}{2}e^{-2x}+C$
$y=-\frac{3}{2}+ce^{2x}$
If you plug in the initial condition, $c=\frac{5}{2}$, and this satisfies the original equation.
However, Mr Fantastic pointed out that the equation was seperable, and indeed it is. But when I solve it that way:
$y'=2y+3$
$\frac{y'}{2y+3}=^{dx}$
$\ln({2y+3})=x+C$
$2y+3=ce^x$
This does not satisfiy the original equation, so can't be right, but I don't see what I'm doing wrong.

$\displaystyle\int\frac{dy}{2y+3}=\ln|2y+3|,$ which is incorrect.

You have to do a $u=2y+3, du/2=dy$ substitution thus:

$\displaystyle\int\frac{dy}{2y+3}=\frac{1}{2}\int\f rac{du}{u}=\frac{1}{2}\,\ln|u|=\frac{1}{2}\,\ln|2y +3|.$

How does that change things?

3. Thanks.

4. You're welcome. Does that enable you to finish the problem?