20091029, 01:46  #1 
Sep 2004
13·41 Posts 
DifEQ theorem ?
Is there a theorem that says something like you need as many initial conditions as the order of your Dif Eq? And how would you prove it. Its kind of along the same lines with me wondering on one of the first order ones how you can say y=e^(rt). It always works out but how do you know it will before it does, to say that y will equal that.

20091029, 02:33  #2  
Dec 2008
7^{2}×17 Posts 
Quote:
The answer to your question thus follows (with little cosmetics on your part) from this simple notion. Last fiddled with by flouran on 20091029 at 02:34 

20091029, 04:27  #3 
"Kyle"
Feb 2005
Somewhere near M50..sshh!
2×3×149 Posts 
Think about it. If you don't have any initial conditions then you can't solve for your constants c1, c2, c3 etc. You don't need a proof for this as it is common sense. As for your other question... some smart individuals figured out the solution forms for many first and second order (homogeneous or not, linear or not, etc) a long time ago. Although I have had differential equations... I cannot not remember the proofs though I am sure they are an easy google search away.

20091029, 05:53  #4 
Sep 2004
13×41 Posts 
I guess a better question would be why is there as many c1 c2 c3 constants up to c sub n. It makes sense that that would be the max, but not that it is also the min.

20091029, 09:51  #5  
Nov 2003
1D24_{16} Posts 
Quote:
Sigh. The fact that you ask this question really shows that you lack fundamental background material. It is clear that you shouldn't even be studying DiffEqs. I will give a hint to your question: Do you know what the dimension of a vector space is? Have you taken linear algebra? Do you understand linear independence and what it means for functions to be linearly independent??? Quote:


20091029, 17:30  #6  
"William"
May 2003
New Haven
2^{3}·5·59 Posts 
Quote:
The second order equation can be recast as a two first order equations in two functions. These can often be arranged so that you can solve a first order equation in one function, then plug that into the other equation and solve another first order equation in one function. For these cases, it's obvious that you get two constants  one from each equation. 

20091029, 18:25  #7 
Sep 2004
13·41 Posts 
Thanks! I think I understand it now. I'm sorry I can't express myself clearly, but my answer seems to have a similar root as adding the plus c whenever you take an integral.

20091029, 21:43  #8  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:


20091030, 00:00  #9  
Dec 2008
7^{2}·17 Posts 
Quote:
Your analysis of the OP's mathematical background is word salad. Last fiddled with by flouran on 20091030 at 00:01 

20091030, 00:03  #10  
Dec 2008
7^{2}·17 Posts 
Quote:
However, it is helpful to know some Linear Algebra prior to taking a differential equations class (I am assuming the OP is learning ODE's right now). It ended up helping me in the long run. Last fiddled with by flouran on 20091030 at 00:07 

20091030, 01:42  #11 
Cranksta Rap Ayatollah
Jul 2003
1201_{8} Posts 

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