2nd order differential equations

[kingwinner]

1) Assume that p and q are continuous on some open interval I, and that y₁ and y₂ are solutions of y'' + p(t)y' + q(t)y = 0 on I.
a) Prove that if {y₁, y₂} form a fundamental set of solutions on I, then they can't have a common inflection point in I, unless p and q are both 0 at this point.
b) If 0 E I, prove that y(t)=t³ cannot be a solution of the differential equation on I.

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1a) My first question: "A unless B", is this equivalent to "A if and only if not B". That is, do I have to prove both directions for question 1a simply by seeing the word "unless"?

Now my thoughts about this problem:
Inflection point at c => f ''(c)=0 or f ''(c) does not exist.
But in this case, since we're given that y₁ and y₂ are solutions of the differential equation, this means that y₁ and y₂ must be twice differentiable, so we can reject the possiblity f ''(c) does not exist.
Now we have: inflection point at c => f ''(c)=0
Fundamental set of solutions means Wronskian is nonzero.
But now I am having a lot of trouble deriving one from the another...


1b) No clue...need some hints...


Thanks for your help!

 

[kingwinner]

For part b, since we are given that p, q are continuous, I think this may be related to existence-uniqueness theorem...but I can't figure out how to use it...

 

[kingwinner]

1b) y(t)=t^3
=>y'(t)=3t^2
=>y''(t)=6t

=>y'(0)=0, y''(0)=0

But the initial value problem (IVP) y'' + p(t)y' + q(t)y = 0, y'(0)=0, y''(0)=0 has the solution y≡0 by inspection, and by existence-uniqueness theorem, y≡0 must be the only solution to the IVP. So y(t)=t^3 cannot be a solution to the ODE on I if 0 E I.

I've tried my best. Is this a correct proof?