DifEQ theorem ?

Joshua2 · 2009-10-29, 01:46

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.

flouran · 2009-10-29, 02:33

Quote:

Originally Posted by Joshua2

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.

By definition, f is a solution to an n-th order differential equation on an interval if f has at least n derivatives on that interval and satisfies the equation.

The answer to your question thus follows (with little cosmetics on your part) from this simple notion.

Primeinator · 2009-10-29, 04:27

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.

Joshua2 · 2009-10-29, 05:53

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.

R.D. Silverman · 2009-10-29, 09:51

Quote:

Originally Posted by Joshua2

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.

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:

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.

Word salad. Gibberish.

wblipp · 2009-10-29, 17:30

Quote:

Originally Posted by Joshua2

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.

Do you understand why there is one constant for the first order equation?

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.

Joshua2 · 2009-10-29, 18:25

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.

Orgasmic Troll · 2009-10-29, 21:43

Quote:

Originally Posted by R.D. Silverman

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???

Bob, you're being an a-hole. DiffEq is often offered without Linear Algebra as a pre-requisite. Whether you agree with math departments doing that, don't fault the kid for the system. I went through the same system and wondered similar things when I was taking classes, and I eventually ended up not being an idiot.

flouran · 2009-10-30, 00:00

Quote:

Originally Posted by R.D. Silverman

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.

Your analysis is not grounded in a lot of substantive evidence. You are essentially stating that the OP lacks "fundamental background material" (if I may quote you) because of ONE question that he asked. That surely is not conclusive. Furthermore, you do not even know the OP personally to accurately assess his mathematical abilities. I suggest you help the OP out rather than barrage him with empty insults (which is not impressive either, since anyone can do that).

Quote:

Originally Posted by R.D. Silverman

Word salad. Gibberish.

Your analysis of the OP's mathematical background is word salad.

flouran · 2009-10-30, 00:03

Quote:

Originally Posted by Orgasmic Troll

DiffEq is often offered without Linear Algebra as a pre-requisite. Whether you agree with math departments doing that, don't fault the kid for the system. I went through the same system and wondered similar things when I was taking classes, and I eventually ended up not being an idiot.

True.
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.

Orgasmic Troll · 2009-10-30, 01:42

Quote:

Originally Posted by flouran

True.
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.

I don't dispute that at all, it's just irrelevant to my point.

2009-10-29, 01:46	#1
Joshua2 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.

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2009-10-29, 04:27	#3
Primeinator "Kyle" Feb 2005 Somewhere near M52.. 2²·227 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.

2009-10-29, 05:53	#4
Joshua2 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.

2009-10-29, 18:25	#7
Joshua2 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.