# Decay of a Decay. Is the diff eqn correct

• Jul 26th 2010, 09:44 AM
fobos3
Decay of a Decay. Is the diff eqn correct
You have a decaying isotope \$\displaystyle x=x_0e^{-kt}\$ which decays in isotope \$\displaystyle y\$. Now \$\displaystyle y\$ decays in \$\displaystyle z\$ which is stable. Is the following correct

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

Also what is the equation of \$\displaystyle z\$?

Edit
I think \$\displaystyle z=z_0+y_0-y+x_0-x\$ which means \$\displaystyle \dfrac{dz}{dt}=cy\$ what you would expect
• Jul 28th 2010, 12:14 PM
fobos3
Here is what I have

\$\displaystyle x=x_0e^{-kt}\$

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

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

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

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

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

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

\$\displaystyle y=\dfrac{kx_0}{c-k}(e^{-kt}-e^{-ct})\$
• Jul 29th 2010, 02:40 AM
CaptainBlack
Quote:

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

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

Also what is the equation of \$\displaystyle z\$?

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

That looks OK

CB