# Math Help - Decay of a Decay. Is the diff eqn correct

1. ## Decay of a Decay. Is the diff eqn correct

You have a decaying isotope $x=x_0e^{-kt}$ which decays in isotope $y$. Now $y$ decays in $z$ which is stable. Is the following correct

$\dfrac{dy}{dt}=kx-cy$

Also what is the equation of $z$?

Edit
I think $z=z_0+y_0-y+x_0-x$ which means $\dfrac{dz}{dt}=cy$ what you would expect

2. Here is what I have

$x=x_0e^{-kt}$

$\dfrac{dy}{dt}=kx-cy$

$\dfrac{dy}{dt}+cy=kx_0e^{-kt}$

$\dfrac{d}{dt}(e^{ct}y)=kx_0e^{(c-k)t}$

$e^{ct}y=\dfrac{kx_0}{c-k}e^{(c-k)t}+A$

$y=\dfrac{kx_0}{c-k}e^{-kt}+Ae^{-ct}$

Assuming $y_0=0=\dfrac{kx_0}{c-k}+A\therefore A=-\dfrac{kx_0}{c-k}$

$y=\dfrac{kx_0}{c-k}(e^{-kt}-e^{-ct})$

3. Originally Posted by fobos3
You have a decaying isotope $x=x_0e^{-kt}$ which decays in isotope $y$. Now $y$ decays in $z$ which is stable. Is the following correct

$\dfrac{dy}{dt}=kx-cy$

Also what is the equation of $z$?

Edit
I think $z=z_0+y_0-y+x_0-x$ which means $\dfrac{dz}{dt}=cy$ what you would expect
That looks OK

CB