First year calculus differential equations question

[wisvuze]

Homework Statement

Suppose \alpha satisfies

1) a_{n} \alpha^{n} + .. + a_{0} = 0
Suppose that polynomial splits into n real linear factors ( it has n real roots )
Then, any solution of

a_{n} y^{n} + .. + a_{1}y' + a_{0} = 0

is of the form
y(x ) = c_{1}e^{\alpha_{1}x} + .. + c_{n}e^{\alpha_{n}x}
where the c's are real numbers, and the alphas are the distinct roots of the polynomial 1)

The Attempt at a Solution

Nooo clue. I verified that the thing is a solution at all, then tried to use induction on the degree of n, but then realized that I would get a solution for any differential equation ever so I scrapped that method.

Any ideas? thank you
and the class has no differential equations / linear algebra coverage. Everything is in elementary calculus/theory

 

[Dick]

Evaluate
<br /> a_{n} y^{n} + .. + a_{1}y&#039; + a_{0}<br />
where y=c1*e^(alpha1*x). Do you see why you get zero?

 

[wisvuze]

hm, you're right so is the statement incorrect?

 

[Dick]

hm, you're right so is the statement incorrect?

Why? What am I right about? What did you find? Aren't you supposed to get zero?

 

[wisvuze]

Why? What am I right about? What did you find? Aren't you supposed to get zero?

never mind, I got zero because I can factor out all the e's and the rest of it is a linear combination of the alpha_1 's .. which is equal to zero so multiplying that gives me zero

 

[wisvuze]

I proved the general solution to the equation a_1 y' + a_0 y = 0 , can I use this for nth derivative equations?

 

[Dick]

I proved the general solution to the equation a_1 y' + a_0 y = 0 , can I use this for nth derivative equations?

You were right on the first response, except that it isn't a linear function of alpha_1, it's polynomial in alpha_1. But that means the exponential expression solves the equation, yes? How close does that put you to being done? I'm not sure that showing that that is ONLY solution family falls within the domain Calc I. Unless I'm underestimating Calc I.

 

[wisvuze]

You were right on the first response, except that it isn't a linear function of alpha_1, it's polynomial in alpha_1. But that means the exponential expression solves the equation, yes? How close does that put you to being done? I'm not sure that showing that that is ONLY solution family falls within the domain Calc I. Unless I'm underestimating Calc I.

I have Calc I and Calc II under my belt ( and linear algebra , but these courses don't officially yet )
The original thing is a polynomial in x where alpha_1 is a solution

I noticed that I can say something along the lines of: if I have an nth order equation ( they act like polynomials ) , I can have at most n linearly indepedent solutions, then every solution will be a linear combination of these? If I can prove this then I am done. I haven't really thought about it, but it doesn't seem like it should be so hard to prove?

 

[Dick]

I have Calc I and Calc II under my belt ( and linear algebra , but these courses don't officially yet )
The original thing is a polynomial in x where alpha_1 is a solution

I noticed that I can say something along the lines of: if I have an nth order equation ( they act like polynomials ) , I can have at most n linearly indepedent solutions, then every solution will be a linear combination of these? If I can prove this then I am done. I haven't really thought about it, but it doesn't seem like it should be so hard to prove?

Ok, you've got that right. You have n linearly independent solutions since the alphas are distinct. And all of the alphas are solutions. Nothing special about alpha1. So that combination must be the most general solution, yes?

 

[wisvuze]

Ok, you've got that right. You have n linearly independent solutions since the alphas are distinct. And all of the alphas are solutions. Nothing special about alpha1. So that combination must be the most general solution, yes?

Yeah, but I don't know how I should prove that I'll have at most n linearly independent solutions. I think that the differentiation operation composition can be defined as a multiplication operation for an algebra then the proof that there are at most n distinct roots to the equation a_n y^(n) + .. + y a_0 will be analogous to how you factorize polynomials over R, but I am not allowed to say this in a calculus course

 

[Dick]

Yeah, but I don't know how I should prove that I'll have at most n linearly independent solutions. I think that the differentiation operation composition can be defined as a multiplication operation for an algebra then the proof that there are at most n distinct roots to the equation a_n y^(n) + .. + y a_0 will be analogous to how you factorize polynomials over R, but I am not allowed to say this in a calculus course

You are already given that the polynomial splits into n real factors. If they mean (I hope) that those n factors are distinct and you know i) that an nth order linear ode has exactly n linearly independent solution and ii) that the exponentials are linearly independent then you are done.

 

[wisvuze]

You are already given that the polynomial splits into n real factors. If they mean (I hope) that those n factors are distinct and you know i) that an nth order linear ode has exactly n linearly independent solution and ii) that the exponentials are linearly independent then you are done.

How do I prove that nth order linear ODEs have n linearly independent solutions? (I don't know that , I believe it, but i have never taken an ODE course )

 

[Dick]

How do I prove that nth order linear ODEs have n linearly independent solutions? (I don't know that , I believe it, but i have never taken an ODE course )

That's why I'm wondering what parts of this question actually NEED to be proved, and which ones you can just quote theorems on. The proof there are n linearly independent solutions comes from existence and uniqueness for the solutions given initial conditions like y(0)=y0, y'(0)=y1, ... y^(n-1)(0)=y_n-1. Since your functions are linearly independent you can always find a combination of them to solve this. Therefore all solution are combinations of them. But I don't see any reason to repeat the argument for this particular problem. It's a standard result you can look up in ODE texts.

 

[wisvuze]

Thanks, I will consult ODE books
thanks :)