# Thread: Differential Eq stuck again

1. ## Differential Eq stuck again

A radioactive substance decays at a rate directly proportional to the amount present. If the substance is present in the amount of $\displaystyle Qo g$ initially $\displaystyle (t = 0)$, find the amount present at any time $\displaystyle t$.

I started off using the restricted growth model of

$\displaystyle dQ/dt = kQ(C-Q)$

That is about as far as I made it.

2. Originally Posted by vodka
A radioactive substance decays at a rate directly proportional to the amount present. If the substance is present in the amount of $\displaystyle Qo g$ initially $\displaystyle (t = 0)$, find the amount present at any time $\displaystyle t$.

I started off using the restricted growth model of

$\displaystyle dQ/dt = kQ(C-Q)$

That is about as far as I made it.
Well I can help you integrate that:

$\displaystyle \frac{dQ}{dt} = kQ(C-Q) \Rightarrow \int_{Q_0}^{Q} \frac{dQ}{Q(C-Q)} = \int_0^t k\, dt$

Now $\displaystyle \frac1{Q(C-Q)} = \frac1{C}\left(\frac1{Q} + \frac1{C-Q}\right)$,

$\displaystyle \int_{Q_0}^{Q} \frac{dQ}{Q(C-Q)} = \int_0^t k\, dt \Rightarrow \frac1{C}\left(\int_{Q_0}^{Q} \frac{dQ}{Q} + \int_{Q_0}^{Q} \frac{dQ}{C-Q}\right) = kt$

$\displaystyle \frac1{C}\left(\ln\left(\frac{Q}{Q_0}\right) - \ln\left(\frac{C - Q}{C - Q_0}\right) \right) = kt$

$\displaystyle \left(\frac{Q(C - Q_0)}{Q_0(C - Q)}\right) = e^{Ckt}$

$\displaystyle Q = \frac{CQ_0}{Q_0 + (C- Q_0)e^{-Ckt}}$

3. After making the following substitution:

$\displaystyle z = \frac{1} {Q}$

we get the following simple ODE:

$\displaystyle \frac{{dz}} {{dt}} + kcz - k = 0$

I'm sure you'll be able to take it from here...

4. Originally Posted by vodka
A radioactive substance decays at a rate directly proportional to the amount present. If the substance is present in the amount of $\displaystyle Qo g$ initially $\displaystyle (t = 0)$, find the amount present at any time $\displaystyle t$.

I started off using the restricted growth model of

$\displaystyle dQ/dt = kQ(C-Q)$

That is about as far as I made it.
You apparently have the wrong equation. If Q' is directly proportional to Q then
$\displaystyle Q' = -kQ$
(negative because the amount present is getting smaller.)

I'm guessing you will have no trouble solving this equation, but if you do simply say so.

-Dan

5. Originally Posted by topsquark
You apparently have the wrong equation. If Q' is directly proportional to Q then
$\displaystyle Q' = -kQ$
(negative because the amount present is getting smaller.)

I'm guessing you will have no trouble solving this equation, but if you do simply say so.

-Dan

Cheers again thats what I was looking for,

$\displaystyle dQ/dt =-kQ$

$\displaystyle Q(0) = Qo$

6. Originally Posted by topsquark
You apparently have the wrong equation. If Q' is directly proportional to Q then
$\displaystyle Q' = -kQ$
(negative because the amount present is getting smaller.)
-Dan
I thought so too... But I thought probably the laws have changed over the past 4 years when I did not touch physics*

*Actually I just dont want to admit that I have got rusty