Differential equation- (haven't had any courses on diff. equations yet

[Lisa...]

Differential equation- please help (haven't had any courses on diff. equations yet :(

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!

 

[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.

 

[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 ?!

 

[siddharth]

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

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

 

[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) C^n-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

 

[Lisa...]

is that correct?

 

[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.

 

[Lisa...]

How about now?

 

[Lisa...]

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

 

[siddharth]

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

 

[Lisa...]

But now that I have

- 1/(n-1) C^n-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) C^n-1 = A

ln C= A

 

[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)C₀^n-1 = A and ln C₀ = A

 

[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 ;)

 

[siddharth]

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

 

[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.