1. Apr 30, 2006

### Lisa...

Hi!

I need to solve this differential equation:

$$\frac{dC}{dt} = -k C ^n$$
with C(0)= C0
for n= 0, 1, 2, ....,

Then I need to plot C(t) as a function of t in a way in order to obtain a straight line.

(This is used for reaction kinetics)

The only problem is that I haven't had a course on differential equations yet (it is planned for next month), yet I have to solve this problem before tomorrow...

Could anyone of you be able to help me? I'd be very very grateful!

2. Apr 30, 2006

### quasi426

Use the method of separation by parts, its usually the first topic taught in a Diff. Eqn. course. Try grouping C^n with dC and k with dT. Then integrate, not forgetting to add the constant of integration, this is where the intial conditions will come to play.

3. Apr 30, 2006

### Lisa...

Thanks, I find:

$$\frac{dC}{dt} = -k C ^n$$

$$= \frac{dC}{C^n} = - k dt$$

integrating gives:

$$\frac{1}{ln C} C^n = -kt + A$$

with $$A = \frac{1}{ln C} C^n$$ so

$$\frac{1}{ln C} C^n = -kt + \frac{1}{ln C} C^n$$ ??

But how do I find C as a function of t ?!

Last edited: Apr 30, 2006
4. Apr 30, 2006

### siddharth

$$\frac{1}{ln C} C^n$$
is not correct.

What's
$$\int \frac{dC}{C^n}$$ when $n \neq 1$.

5. Apr 30, 2006

### Lisa...

Oh I see.... I've considered C n to be of the form g x (an exponential function).

Though it should be - 1/(n-1) Cn-1 if n is NOT 1 and if n=1 (cause n most certainly can be 1) it should be of the form: ln C

Last edited: Apr 30, 2006
6. Apr 30, 2006

### Lisa...

is that correct?

7. Apr 30, 2006

### siddharth

-n/Cn+1 is not right

An easy way to check is to differentiate your answer. For example, if you differentiate -n/Cn+1, you get n(n+1)/Cn+2 which isn't what you need.

Last edited: Apr 30, 2006
8. Apr 30, 2006

### Lisa...

Last edited: Apr 30, 2006
9. Apr 30, 2006

### Lisa...

I guess I'm mixing integration & differentiation up again .... just blame it on being tired :P

10. Apr 30, 2006

### siddharth

Yes, it's right now. That's why I don't do my homework late in the night when I'm tired

11. Apr 30, 2006

### Lisa...

But now that I have

- 1/(n-1) Cn-1 = -kt + A if n is NOT 1
ln C = -kt +A if n =1

how on earth would I get a function for C(t)?! + I don't get how I can put the initial condition C(0)= C0 into the equation.... if I choose t=0, it gives me respectively:

- 1/(n-1) Cn-1 = A

ln C= A

Last edited: Apr 30, 2006
12. Apr 30, 2006

### siddharth

You need to plot C(t) as a function of t in a way in order to obtain a straight line, right?

So your answer is right there! When you plot a graph of $$\frac{-1}{(n-1)C^{n-1}}$$ (on the y-axis) vs $t$ (on the x-axis) or ln C vs t when n=1, you have a straight line

To use the initial condition, put C(0) = C0, not C.
So, you relation will be - 1/(n-1)C0n-1 = A and ln C0 = A

Last edited: Apr 30, 2006
13. Apr 30, 2006

### Lisa...

OMG that's too easy :D Thanks a lot for helping me out!

Sorry for my poor planning btw, but yesterday was Queens Day, so I could not resist going to town ;)

14. Apr 30, 2006

### siddharth

You're welcome. But, I thought the Dutch celebrate it on the 30th (ie, today)?

15. Apr 30, 2006

### Lisa...

Very good of you, that's true!
But the 30th (today) is on a Sunday and because the queen is Christian and there's a Christian holiday on this Sunday (it has something to do with thanking the Lord for having our crops or something like that) she decided to make an exception and celebrate it a day earlier.

Last edited: Apr 30, 2006