Why is a Constant added to front?

Question

I made the differential equation : $$dQ = (-1/100)2Q dt$$

I separate it and get: $\int_a^b x (dQ/Q) = \int_a^b x (-2/100)dt$

this leads me to: $\log(|Q|) = (-t/50) + C$

I simplify that to $Q = e^{-t/50}$

My TI-Nspire differential equation solver, however, gives me: $Q = Ce^{-t/50}$

I'm confused as to why the calculator is multiple my answer with a constant and which one is the correct answer.

Answer 1

log(|Q|)=(−t/50)+C

$e^{log(|Q|} = e^{\frac{-t}{50}+C}$

$Q = e^{\frac{-t}{50}} e^{C}$

Now you need to recognized that $e^{C}$ is just another constant, so we substitute $e^{C}$ with just C.

You could say that $C_1 = e^{C}$ and use $C_1$, but most people say use C and gloss over this distinction.

$Q = C e^{\frac{-t}{50}}$

Answer 2

$\ln(x)=t+c$ implies that $x=e^{t+c}=e^ce^t=ke^t$ where $k=e^c$.

Answer 3

In your second to last step $ln(|q|) = (-t/50) + C$ you get $e^{\ln|q|}=e^{-t/50+C}$ so $|q|=e^{-t/50}e^C$ which can be stated as $q=\pm Ce^{-t/50}$ or $q= Ce^{-t/50}$.

Answer 4

Even though $\log(|Q|) = (-t/50) + C$ implies $|Q| = e^{-t/50+c} \iff Q= \pm e^{-t/50+c}=\pm e^{-t/50}e^c$, but $e^c$ is a constant raised to a constant, which is (...you guessed right) a constant! So $Q= \pm C e^{-t/50}$ where $C=e^c$