2nd order DE violating theroem 2?

[accountkiller]

Homework Statement

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).

Homework Equations

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 textbook (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]

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.