Integration with Natural Logarithms I was looking at a problem in calculus 1 that said: "solve the differential equation $\frac{dy}{dt} = (y-15)(t+20)$ when $y(0)=3$"
So I separated the variables and took the integral of both sides to be left with 
$ln|y-15|=\frac{1}{2}t^2+20t+c$
Here, if I were to plug in the numbers I was given you would get
$ln|-12|=c$
so then c must be equal to $ln12$
However, my professor said that what you had to do was rewrite the solution in exponential form so you would get
$y=e^{\frac{1}{2}t^2+20t+c}+15$
which then would simplify to 
$y=ce^{\frac{1}{2}t^2+20t}+15$
so when you solve, you would get
$3=c+15$ so then
$c=-12$
Yet, being able to reduce $y=e^{\frac{1}{2}t^2+20t+c}+15$ 
to 
$y=ce^{\frac{1}{2}t^2+20t}+15$
requires you to be able to compute $e^c$, which you cannot, since, according to the equation, the result of $e^c$ is $-12$ something that is seemingly impossible. I wanted to know if there is a way to get around the inability to get a negative number from an exponent in the same way we get around the inability to take the square root of a negative number by using an imaginary number.
 A: You have confused yourself by using poor notation. You can’t reduce $e^{t+c}$ to $ce^{t}$ because the equation $e^{t+c}=ce^t$ isn’t true. The true equation is $e^{t+c}=de^t$, where $c$ and $d$ relate according to $e^c=d$. When you follow this through, you’ll find that your problem has disappeared and a new one has appeared: you find that $-12=12$! This can be handled with branch-cuts.
This shows a limitation of separation of variables: your original solution doesn't find the whole solution set.
A: Reducing to exponential form you will first get
$$|y-15|=e^{\frac{1}{2}t^2+20t+c}$$
and so
$$y-15=\pm e^{\frac{1}{2}t^2+20t+c}$$
and
$$y=\pm e^ce^{\frac{1}{2}t^2+20t}+15\ .$$
So the constant in front of the exponential can be either positive or negative.  It can also be zero (though not with the initial condition $y(0)=3$), as you can check that $y=15$ is a solution of the DE.  If you take $y=15$ and go through your working, you can see why the separation of variables method fails to find this solution.
