In connection with a thread posted earlier Anteeks was having problems with the equation: $\displaystyle y'-2y=3, y(0)=1$.

If you solve this with an integrating factor you get a factor of $\displaystyle e^{-2x}$.

$\displaystyle (e^{-2x}*y)'=3e^{-2x}$

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

$\displaystyle y=-\frac{3}{2}+ce^{2x}$

If you plug in the initial condition, $\displaystyle c=\frac{5}{2}$, 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:

$\displaystyle y'=2y+3$

$\displaystyle \frac{y'}{2y+3}=^{dx}$

$\displaystyle \ln({2y+3})=x+C$

$\displaystyle 2y+3=ce^x$

This does not satisfiy the original equation, so can't be right, but I don't see what I'm doing wrong.