- #1

Bill Foster

- 338

- 0

## Homework Statement

I'm pretty rusty at these. But given the following:

[tex]\frac{dN_a}{dt}=-\frac{N_a}{\tau_a}[/tex]

[tex]\frac{dN_b}{dt}=\frac{N_a}{\tau_a}-\frac{N_b}{\tau_b}[/tex]

## The Attempt at a Solution

The first one, naturally, is easy [tex]N_a(t)=N_a(0)e^\frac{-t}{\tau_a}[/tex]

The second one is giving me a little trouble.

I tried it as such:

[tex]\frac{dN_b}{dt}=\frac{N_a\tau_b-N_b\tau_a}{\tau_a \tau_b}[/tex]

[tex]\frac{dN_b}{N_a\tau_b-N_b\tau_a}=\frac{dt}{\tau_a \tau_b}[/tex]

Let [tex]x=N_a\tau_b-N_b\tau_a[/tex]

[tex]dx=-dN_b\tau_a[/tex]

[tex]\frac{dx}{x\tau_a}=\frac{dt}{\tau_a \tau_b}[/tex]

[tex]\frac{dx}{x}=\frac{dt}{\tau_b}[/tex]

[tex]x(t)=x_0e^\frac{-t}{\tau_b}[/tex]

[tex]N_a\tau_b-N_b\tau_a=x_0e^\frac{-t}{\tau_b}[/tex]

[tex]N_b\tau_a=N_a\tau_b-x_0e^\frac{-t}{\tau_b}[/tex]

[tex]N_b(t)=\frac{N_a\tau_b-x_0e^\frac{-t}{\tau_b}}{\tau_a}[/tex]

Now, at [tex]t=0[/tex], [tex]N_b(t=0)=N_{b0}[/tex]

So [tex]x_0=N_a\tau_b-N_{b0}\tau_a[/tex]

[tex]N_b(t)=\frac{N_a\tau_b-(N_a\tau_b-N_{b0}\tau_a)e^\frac{-t}{\tau_b}}{\tau_a}[/tex]

But I don't think that's right because it blows up. It's supposed to be a

**decay**problem.

Did I solve it correctly?