Understanding exponential decay

Question

Say I have a variable $x$ that decays over time $t$ as follows:

$$ \frac{dx}{dt} = \frac{-x}{\tau}. $$

Solving for $x$, I get

\begin{align} x &= \frac{-1}{\tau}\int x dt\\ &=e^{-t/\tau}. \end{align}

My question is so simple I'm embarrassed to be asking it:

Can someone walk me through why

$$x=\frac{-1}{\tau}\int x dt$$

implies

$$x=e^{-t/\tau}?$$

I know this is correct, and know I've thought through the solution in the past, but now have no idea how to think through this correctly.

Answer 1

I find that using the Laplace transform is the easiest way for me to solve this. (That, I now remember, is what I used in the past.)

\begin{align} \frac{dx}{dt} &= \frac{-x}{\tau}\\ -x(0) + sx_L &= \frac{-x_L}{\tau}\\ x_L(s+1/\tau)&=x(0)\\ x_L &= \frac{x(0)}{s + 1/\tau}\\ x&=x(0)e^{-t/\tau}. \end{align}

Note the presence of the initial value of $x$ in the final answer, something I had overlooked in my question.

Answer 2

You want to integrate via separation of variables. So,

$\dfrac{dx}{dt}=-\dfrac{x}{\tau}$

$\dfrac{dx}{x}=-\dfrac{dt}{\tau}\quad\Rightarrow\quad \int\dfrac{dx}{x}=-\int\dfrac{dt}{\tau} \quad \Rightarrow \quad \ln(x)=-\dfrac{t}{\tau}$

Thus, $e^{-\frac{t}{\tau}}=x$

Answer 3

$$\frac{\mathrm{d}x }{\mathrm{d} t} = \frac{-x}{\tau }$$ Seperate Variables $$\frac{\mathrm{d}x }x=\frac{\mathrm{d}t }{-\tau}$$ Integrate both sides $$\int \frac{\mathrm{d}x }x=\int \frac{\mathrm{d}t }{-\tau}$$ Antiderivative: $$\ln \left | x \right |= \frac{-t}{\tau} + c_{1}$$ Solve for Variable $$\left | x \right |=e^{(-t/\tau)+c_{1}}$$ $$\left | x \right |=e^{(c_{1})}e^{(-t/\tau)}$$ $$\left | x \right |=c_{2}e^{(-t/\tau)}$$ $$x=\pm c_{2}e^{(-t/\tau)}$$