# Math Help - 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 $Qo g$ initially $(t = 0)$, find the amount present at any time $t$.

I started off using the restricted growth model of

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

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 $Qo g$ initially $(t = 0)$, find the amount present at any time $t$.

I started off using the restricted growth model of

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

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

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

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

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

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

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

3. After making the following substitution:

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

we get the following simple ODE:

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

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 $Qo g$ initially $(t = 0)$, find the amount present at any time $t$.

I started off using the restricted growth model of

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

You apparently have the wrong equation. If Q' is directly proportional to Q then
$Q' = -kQ$
(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
$Q' = -kQ$
(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,

$dQ/dt =-kQ$

$Q(0) = Qo$

6. Originally Posted by topsquark
You apparently have the wrong equation. If Q' is directly proportional to Q then
$Q' = -kQ$
(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