# Thread: trouble with the constant term of a separable DE

1. ## trouble with the constant term of a separable DE

I have a really simple separable DE from an input output problem and I'm having trouble with the constant. (I think)

$\displaystyle \frac{dx}{dt}=0.6-\frac{3x(t)}{500}$ and $\displaystyle x(0)=0$

$\displaystyle \frac{dx}{dt}=\frac{300-3x(t)}{500}$

$\displaystyle \frac{dx}{300-3x(t)}=\frac{dt}{500}$

$\displaystyle \int\frac{1}{300-3x(t)}dx(t)=\int\frac{1}{500}dt$

$\displaystyle \int\frac{1}{300-3x(t)}dx(t)=\frac{t}{500}+c$ let $\displaystyle u=300-3x(t)$ then $\displaystyle du=0-3dx(t)$

$\displaystyle \frac{-1}{3}\int\frac{1}{u}du(t)=ln|u|=ln|300-3x(t)|$

$\displaystyle \frac{-1}{3}ln|300-3x(t)|=\frac{t}{500}+c$

$\displaystyle ln|300-3x(t)|=-3\frac{t}{500}+c$

So I'm guessing that my problem starts here. I know I have to exponentiate both sides to solve for the dependant variable. And I'm assuming that after exponentiation the equation looks like this;

$\displaystyle 300-3x(t)=e^{-3\frac{t}{500}+c}$

$\displaystyle x(t)=100-\frac{-1}{3}e^{-3\frac{t}{500}+c}$

Even if I were to evaluate the problem with the initial value provided this doesn't seem to match the provided answer, $\displaystyle x(t)=100(1-e^{-3t/500})$, at all.

2. ## Re: trouble with the constant term of a separable DE

Originally Posted by bkbowser
$\displaystyle 300-3x(t)=e^{-3\frac{t}{500}+c}$

$\displaystyle x(t)=100-\frac{-1}{3}e^{-3\frac{t}{500}+c}$

Even if I were to evaluate the problem with the initial value provided this doesn't seem to match the provided answer, $\displaystyle x(t)=100(1-e^{-3t/500})$, at all.
You have too many negative signs in the second equation. Let's try it out with the negative sign corrected:

$\displaystyle x(t)=100-\frac{1}{3}e^{-3t/500+c}$

$\displaystyle x(0) = 100 - \dfrac{1}{3}e^{0+c} = 0$

$\displaystyle e^c = 300$

Going back to the original equation:

$\displaystyle x(t)=100-\frac{1}{3}e^{-3t/500+c} = 100 - \frac{1}{3}e^{-3t/500}e^c = 100-100e^{-3t/500}$

Looks like you got the right answer to me.