1. ## Exponential decay

have a problem with understanding how this integration took place:

dx/dt = -kx

x(t) = x0 exp(-kt)

the lecture i attended just said by integration of the first equation we obtain x(t)
and x0 is the initial avlue of x and exp is the exponential

2. Originally Posted by myenemy
have a problem with understanding how this integration took place:

dx/dt = -kx

x(t) = x0 exp(-kt)

the lecture i attended just said by integration of the first equation we obtain x(t)
and x0 is the initial avlue of x and exp is the exponential
Dear myenemy,

You have to use the seperation of variable method to solve this differential equation,

$\frac{dx}{dt}=-kx$

$\frac{dx}{x}=-k~dt$

$\int\frac{1}{x}dx=-\int{k~dt}$

Hope you can continue from here.