Diff Eqn: dx = r dt - kx dt ... Can you solve it?

Question

The concept of this equation is that the amount of a substance in a container is x. The amount is being increased at a rate r per unit time, and is being decreased at a rate that is proportional to the amount in the container (-kx).

It seems like a simple differential equation, but I don't remember how to solve it.
Thanks for your help.

Answer 1

The differential equation has two parts, rate in and rate out. The rate in is of course r. And then the rate out is -kx...That gives you the differential equation:
dx/dt = r - kx

The simplest way to solve this equation is probably by separating the variables. Take (r - kx) to be the x term. Divide both sides by it to get it on the dx side. Then multiply both sides by dt to get dt on the right side (Note: there aren't really any t terms so you just end up integrating 1.)...
(dx/dt)/(r-kx) = (r-kx)/(r-kx) [divided by r-kx]
(dx/dt)/(r-kx) = 1
(dt)(dx/dt)/(r-kx) = 1 (dt) [multiplied both sides by dt.]
dx/(r - kx) = dt

Now do the integral of the right side and the left side. The integral of (1)dt is simply t + c.
dx/(r - kx) = dt
integral[ dx/(r - kx) ] = t + c

The integral on the left side requires only a little bit more work. Do a u substitution.
Let u = r - kx (the entire denominator)
Then du = -k dx
So (-1/k)du = dx
Plug it in and solve...

integral[ (-1/k)du/u ] = t + c
(-1/k) ln| u | = t+c
(-1/k)ln | r - kx | = t +c [substituted for u.]
ln | r - kx | = -kt +c [ multiplied both sides by -k (the constant c absorbs the constant k)
e^ln | r - kx | = e^(-kt +c) = e^(-kt)e^(c) [exponentiated both sides]
r-kx = Ce^(-kt) [the e^ln cancels. The e^c just becomes another constant.]
-kx = Ce^(-kt) - r [subtracted r from both sides]
x = Ce^(-kt) + (r/k) [divided both sides by -k]

Hopefully this is right unless I made a mistake somewhere...but hopefully it will maybe get you on the right track. Good luck!

Answer 2

Try separation of variables: dx = (r-kx) dt
(1/(r-kx))dx = dt
Integral(1/(r-kx)) = t + C
Integrate (it's a natural log function) then solve for x (you'll get an exponential function)

Answer 3

dx / (r - kx) = dt
(-k).dx / (-k).(r - kx) = dt
(-1/k).log(r - kx) = t + C
log (r - kx) = - k(t + C)
r - kx = e^(-k(t + C)
kx = r - e^(-k(t + C)
x = (1/k).(r - e^(-k(t + C))