# 2nd order DE violating theroem 2?

#### accountkiller

1. The problem statement, all variables and given/known data
y1 = x2 and y2 = x3 are two different solutions of x2y'' - 4xy' + 6y = 0, both satisfying the initial conditions y(0) = 0 = y'(0). Explain why these facts don't contradict Theorem 2 (with respect to the guaranteed uniqueness).

2. Relevant equations

3. The attempt at a solution
I have no attempt because I don't understand how it doesn't violate the theorem?

#### HallsofIvy

Since I have no idea exactly what "theorem 2" is in your text book (you say "with respect to the guarenteed uniqueness" but there are many theorems about uniqueness) so I can't say what, if any, part of the hypotheses does not hold.

Please state exactly what "theorem 2" says.

#### accountkiller

Oh wow, I'm sorry, hadn't even realized that. Theorem 2 is the 2nd order uniqueness - existence theorem - how if there is a solution on interval I satisfying the initial condition, it is the one and only solution to that DE.

#### HallsofIvy

So you are refusing to state exactly what "theorem 2" says? Not much point in continuing this, is there?

#### gomunkul51

Note that the "solution" to your ODE is this:

y(x) = A*x^2 + B*x^3

*A, B are constants.

and this is the only possible solution.

(as it is a ordinary linear homogeneous equation, its solution is made from any constant multiplicands of the 2 linearly interdependent solution to that ODE.)

Then, there is only a single pair of constants A and B that satisfy a certain IVP.

#### accountkiller

HallsofIvy, I gave you a quick description of the theorem, thinking that you would go "Ah, yes, that one" and be able to help me. Did you want exact wording? I'm sorry for the miscommunication.

gomunkul51, thanks for your reply, it was well-worded and clear.

I figured it out, though - the theorem requires continuity and once my equation is put into 'standard form,' I got numbers divided by x.. which obviously means discontinuity where x = 0, so the theorem does not apply here, so it does not violate it.

Thanks!

#### Mark44

Mentor
HallsofIvy, I gave you a quick description of the theorem, thinking that you would go "Ah, yes, that one" and be able to help me. Did you want exact wording? I'm sorry for the miscommunication.

gomunkul51, thanks for your reply, it was well-worded and clear.

I figured it out, though - the theorem requires continuity and once my equation is put into 'standard form,' I got numbers divided by x.. which obviously means discontinuity where x = 0, so the theorem does not apply here, so it does not violate it.
Yes, that's the key - putting the diff. eqn. in standard form, making it y'' - (4/x)y' + (6/x^2)y = 0, together with the fact that both initial conditions are at x = 0.

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