# Isotop Modeling

• Apr 17th 2014, 04:13 PM
Applestrudle
Isotop Modeling
http://i61.tinypic.com/2ekpyl0.jpg

I have:

$A=kte^{-at}$

since at any time there will be kt of A but its also decaying

$B = kt -kte^{-at}$

since at any moment in time kt will be the amount of substance so kt - (amount of A) = amount of B? (Thinking)

I don't know how to set up the differential equations at the start and I don't think your supposed to start with the solution!
• Apr 17th 2014, 07:46 PM
hollywood
Re: differential equations,
You are producing A at rate k and it is decaying at a rate $\alpha{A}$, so $\frac{dA}{dt} = k - \alpha{A}$. Your equation $A=kte^{-\alpha{t}}$ is incorrect since some of the A produced has decayed for a long time and some for a short time.

To solve this equation, substitute $A=Y+\frac{k}{\alpha}$. Everything else is pretty straightforward. Let us know if you have any trouble.

- Hollywood
• Apr 18th 2014, 02:33 AM
Applestrudle
Re: differential equations,
Thanks,

This is my solution:

$\frac{dA}{dt} = k-A\alpha$

$\frac{dB}{dt} = A\alpha$

$A(t) = \frac{k}{\alpha}$

I got the general solution using the integrating factor method

for part v)

$A(t) = \frac{k}{\alpha} - \frac{k}{\alpha}{e}^{-\alpha t}$

$A(0) > C$ C is some constant greater than k/a

$A(t) = \frac{k}{\alpha} + (C - \frac{k}{\alpha}{e}^{- \alpha t}$

for vi I subbed in the given B(t) into the differential equation I wrote for dB/dt

${c}_{1} = k$ and ${c}_{2} = -{A}_{0}$ (not sure about this)

then for B(t) I get

$B(t) = kt - {A}_{0}{e}^{- \alpha t} + {A}_{0}$