Why does $\frac{dy}{dx} \propto y$ imply $x_1 - x_2$ is constant for $\frac{y_2}{y_1}$ constant where $(x_1, y_1)$ and $(x_2, y_2)$ can vary?

Question

The time taken for the activity of a radioactive sample to halve is called the half-life. This is a constant.

More generally, given the curve $y=Ae^{-bx}$, given $y_1$ and $y_2 = ky_1$, then $\frac{y_2}{y_1}$ is always the same number for all choices of $y_1$.

I can justify this by writing $y_1 = Ae^{-bx_1}$ and $y_2 = Ae^{-bx_2}$ and then using $y_2 = ky_1$ to get $x_1 - x_2 = \frac{1}{b} \ln k$, which is a constant.

But is there anyway to justify this starting from the fact that we have a curve that satisfies $\frac{dy}{dx} \propto y$?

Answer 1

$\frac{dy}{dx} \propto y \Rightarrow \frac{dy}{dx} = ky \ $ ($k$ is a constant)

Now you're provided with 2 arbitrary points $(x_1,y_1)$ and $(x_2,y_2)$.

So,

$\begin{align}&\int_{y_1}^{y_2}\frac{dy}{y} = k\int_{x_1}^{x_2} dx \\\Rightarrow \ & \ln\frac{y_2}{y_1} = k(x_2- x_1) \text{ or } \boxed{x_1 - x_2 = \frac1k\ln\frac{y_1}{y_2}}\end{align}$

Now, $k$ and $\frac{y_1}{y_2}$ are constats. So, $\boxed{x_1- x_2}$ is also constant