First Order ODE/dy = k y dx/Proof 2

Proof
Write the differential equation as:


 * $y' - k y = 0$

Taking Laplace transforms:


 * $\laptrans {y' - k y} = \laptrans 0$

From Laplace Transform of Constant Mapping, we have:


 * $\laptrans 0 = 0$

We also have:

So:


 * $\paren {s - k} \laptrans y = \map y 0$

Giving:


 * $\laptrans y = \dfrac {\map y 0} {s - k}$

So:

Setting $C = \map y 0$ gives the result.