# Homework Help: Differential equation

1. Feb 14, 2009

### Bill Foster

1. The problem statement, all variables and given/known data

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

$$\frac{dN_a}{dt}=-\frac{N_a}{\tau_a}$$
$$\frac{dN_b}{dt}=\frac{N_a}{\tau_a}-\frac{N_b}{\tau_b}$$

3. The attempt at a solution

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

The second one is giving me a little trouble.

I tried it as such:

$$\frac{dN_b}{dt}=\frac{N_a\tau_b-N_b\tau_a}{\tau_a \tau_b}$$

$$\frac{dN_b}{N_a\tau_b-N_b\tau_a}=\frac{dt}{\tau_a \tau_b}$$

Let $$x=N_a\tau_b-N_b\tau_a$$

$$dx=-dN_b\tau_a$$

$$\frac{dx}{x\tau_a}=\frac{dt}{\tau_a \tau_b}$$

$$\frac{dx}{x}=\frac{dt}{\tau_b}$$

$$x(t)=x_0e^\frac{-t}{\tau_b}$$

$$N_a\tau_b-N_b\tau_a=x_0e^\frac{-t}{\tau_b}$$

$$N_b\tau_a=N_a\tau_b-x_0e^\frac{-t}{\tau_b}$$

$$N_b(t)=\frac{N_a\tau_b-x_0e^\frac{-t}{\tau_b}}{\tau_a}$$

Now, at $$t=0$$, $$N_b(t=0)=N_{b0}$$

So $$x_0=N_a\tau_b-N_{b0}\tau_a$$

$$N_b(t)=\frac{N_a\tau_b-(N_a\tau_b-N_{b0}\tau_a)e^\frac{-t}{\tau_b}}{\tau_a}$$

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?

2. Feb 14, 2009

### Bill Foster

I believe that is right.

Now I'm trying to solve it using the Euler method...

$$N_b(t+\Delta{t})=N_b(t)-\Delta{t}*\frac{dN_b}{dt}$$

I get: $$N_b(t+\Delta{t})=N_b(t)-\Delta{t}*(\frac{N_a}{\tau_a}-\frac{N_b}{\tau_b})$$

And THAT blows up.

3. Feb 14, 2009

### HallsofIvy

Instead use the solution to the first equation to write the equation as
$$\frac{dN_b}{dt}=\frac{N_a(0)}{\tau_a}e^\frac{-t}{\tau_a}-\frac{N_b}{\tau_b}$$
$$\frac{dN_b}{dt}+ \frac{N_b}{\tau_b}= \frac{N_a(0)}{\tau_a}e^\frac{-t}{\tau_a}$$

That's a linear non-homogeneous differential equation with constant coefficients. Its characteristic equation is $$r+ \frac{r}{\tau_b}= 0$$ and you can look for a specific solution of the entire equation of the form $$Ae^\frac{-t}{\tau_a}$$.

You will find that has solutions with decreasing exponentials.

Last edited by a moderator: Feb 14, 2009
4. Feb 14, 2009

### Bill Foster

Finally got it right. I was right with my analytical solution, except for some sign errors. And it checks out with the Euler method solution. Here's a perl script that I wrote to test and illustrate:

Code (Text):

#!/usr/bin/perl

$dt=0.01;$t=0;
$ta=1;$tb=2;
$Na0=100;$Nb0=0;
$Nae=$Na0;
$Nbe=$Nb0;
$i=0;$imax=20;

for($i=0;$i<=$imax/$dt;$i++) {$t=$i*$dt;
$Na=$Na0*exp(-$t/$ta);
$Nb=-($Na*$tb-($Na0*$tb+$Nb0*$ta)*exp(-$t/$tb))/$ta;
if($i > 0) {$Nae=$Nae-$dt*$Nae/$ta;
$Nbe=$Nbe+($Nae/$ta-$Nbe/$tb)*$dt; } printf("%f %f %f %f %f\n",$t,$Na,$Nae,$Nb,$Nbe);
}

5. Feb 18, 2009

### Bill Foster

I guess it wasn't right. If I use different values for the decay constants my analytic solutions differs from the Euler solution.

I don't know how to solve that.

6. Feb 18, 2009

### djeitnstine

The solution to the characteristic equation posted is the argument of an exponential $$e^{\frac{1}{\tau_{b}}}$$. This added with the particular solution of the form $$Ae^{\frac{-t}{\tau_{a}}}$$ will give you your general solution.

Solving for the particular. Simply take the solution of the form $$Ae^{\frac{-t}{\tau_{a}}}$$, differentiate it, add them together and solve for the coefficients.

I'll get you started:

$$N = Ae^{\frac{-t}{\tau_a}}$$
$$N'= \frac{-1}{\tau_{a}}Ae^{\frac{-t}{\tau_{a}}}$$

Sum them $$N'+\frac{N}{\tau_{b}} = \frac{-1}{\tau_{a}}Ae^{\frac{-t}{\tau_{a}}}+\frac{1}{\tau_{b}}Ae^{\frac{-t}{\tau_{a}}}$$

Solve for coefficients $$A(\frac{-1}{\tau_{a}}+\frac{1}{\tau_{b}})= \frac{N_{a}(0)}{\tau_{b}}$$ and

Last edited: Feb 18, 2009
7. Feb 18, 2009

### HallsofIvy

What course is this for then? It is now a single non-homogeneous linear equation with constant coefficients. Much simpler than the original problem and much simpler than what you were trying to do.

8. Feb 18, 2009

### Bill Foster

Computational Physics.

The problem is to approximate the original system of differential equations using the Euler method, which I can do.

I am now trying to solve it analytically so I can compare it with the Euler solution.

9. Feb 18, 2009

### djeitnstine

Take a look at how I solved it...

10. Feb 23, 2009

### Bill Foster

Shouldn't it be : $$N = Ae^{\frac{-t}{\tau_b}}$$?

11. Feb 23, 2009

### djeitnstine

No. Halls of Ivy showed you this result,

$$\frac{dN_b}{dt}=\frac{N_a(0)}{\tau_a}e^\frac{-t}{\tau_a}-\frac{N_b}{\tau_b}$$

Which in turn becomes

$$\frac{dN_b}{dt}+ \frac{N_b}{\tau_b}= \frac{N_a(0)}{\tau_a}e^\frac{-t}{\tau_a}$$

using this result, we get a non-homogeneous $$2^{nd}$$ order differentual equation with a particular solution $$Ae^{\frac{-t}{\tau_{a}}}$$

So we get a particular solution $$N_{p} = Ae^{\frac{-t}{\tau_{a}}}$$

then, $$N'=\frac{-1}{\tau_{a}}Ae^{\frac{-t}{\tau_{a}}}$$

We sum the 2 with your coefficient: $$N'+\frac{N}{\tau_{b}} = \frac{-1}{\tau_{a}}Ae^{\frac{-t}{\tau_{a}}}+\frac{1}{\tau_{b}}Ae^{\frac{-t}{\tau_{a}}}$$

Of course we solve for the coefficient A: $$A(\frac{-1}{\tau_{a}}+\frac{1}{\tau_{b}})= \frac{N_{a}(0)}{\tau_{b}}$$

Lets get a decent looking fraction first $$A(\frac{\tau_{a} - \tau_{b}}{\tau_{a} \tau_{b}})= \frac{N_{a}(0)}{\tau_{b}}$$

$$A = \frac{N_{a}(0)\tau_{a} \tau_{b}}{\tau_{b} (\tau_{a} - \tau_{b})}$$

$$N_{p}=\frac{N_{a}(0) \tau_{a} \tau_{b} e^{\frac{-t}{\tau_{a}}}}{\tau_{b} (\tau_{a} - \tau_{b})}$$
We have only solved the particular equation. Now we have to add it to the homogeneous equation.

The characteristic is as Ivy showed is $$r+ \frac{1}{\tau_b}= 0$$ Thus your $$N_{h}$$ homogeneous is $$N_{h}= C_{1}e^{\frac{-t}{\tau_{b}}$$

Now summing them to get a general solution...$$N_{g}=N_{h}+N_{p}$$ gives

$$N_{g}= C_{1} e^{\frac{-t}{\tau_{b}}} + \frac{N_{a}(0) \tau_{a} \tau_{b} }{\tau_{b} (\tau_{a} - \tau_{b})}e^{\frac{-t}{\tau_{a}}}$$

Now use your initial conditions to find $$C_1$$

Last edited: Feb 23, 2009
12. Feb 25, 2009

### Bill Foster

Well, that's quite a bit different than what I just got:

$$N_b(t)=[N_b(0)-\tau_aN_a(0)]e^{-\frac{t}{\tau_b}}-\tau_aN_a(0)e^{-\frac{t}{\tau_a}}$$

Hmmm

13. Feb 25, 2009

### Bill Foster

There's a problem with this. If $$\tau_a = \tau_b$$, it blows up.

14. Feb 25, 2009

### djeitnstine

Yea cus division by 0 doesn't usually go well in math =\

15. Feb 25, 2009

### djeitnstine

Oh man I got mixed up with all these tau sub a's and tau sub b's the final answer is
$$N_{g}= C_{1} e^{\frac{-t}{\tau_{b}}} + \frac{N_{a}(0) \tau_{a} \tau_{b} }{\tau_{a} (\tau_{a} - \tau_{b})}e^{\frac{-t}{\tau_{a}}}$$

(its the little $$\tau_{b}$$ before the bracket in the denominator that's wrong)

16. Feb 26, 2009

### Bill Foster

Final solution that matches the Euler solution:

$$N_b(t)=(N_b(0)-\frac{\tau_b}{\tau_a-\tau_b}N_a(0))e^{-\frac{t}{\tau_b}}+\frac{\tau_b}{\tau_a-\tau_b}N_a(0))e^{-\frac{t}{\tau_A}}$$

But what can I do about that condition when $$\tau_a=\tau_b$$?

The function should not blow up.

17. Feb 26, 2009

### djeitnstine

Check that condition on your original DE, then compare that result with the actual answers. Basically check to see if any other solutions were lost.