# Can sin(x^2) be a solution

1. Oct 17, 2012

Can $y = sin(x^{2})$ be a solution on an interval $(\alpha,β) \subseteq ℝ[/itext] containing x= 0 of an equation y'' + p(x)y' +q(x)y = 0 where p(x) and q(x) are continuous on [itex](\alpha,β)[/itext]. I'm not sure how to start this. What I tried to do is use x=0 and create an initial value problem so you have y(0) = 1. 2. Oct 17, 2012 ### dextercioby Why would you do that ? You may want to read the problem once more. What does it mean for a function to be a solution of an ODE ? 3. Oct 17, 2012 ### th3chemist Does it not mean that if you differentiate the solution you would get that function? 4. Oct 17, 2012 ### dextercioby It means that if you plug the confirmed/alleged solution back in the ODE, you get 0=0. That's what you should do. 5. Oct 17, 2012 ### th3chemist I did try doing that though. I don't know what p and q are so that's what is also confusing me. And there's also the trouble of having x=0 which is also confusing me :(. 6. Oct 17, 2012 ### th3chemist If I set up the initial condition of y(0) = 1, how could I even solve this equation? I did get y = sin(x^2) y' = 2xcos(x^2) y'' = 2cos(x^2) - 4x^2 sin(x^2) if I input these values into y'' + p(x)y' + y = 0 how can I tell if it's a solution that fits? :( 7. Oct 17, 2012 ### dextercioby I fail to see the relevance of the initial condition, since sin (0^2) =0 and not 1. 8. Oct 17, 2012 ### th3chemist yeah, I mean 0 sorry. was thinking of cos (0). 9. Oct 18, 2012 ### th3chemist Anyone have any other ideas? :( I feel like I'm overcontemplating things. 10. Oct 18, 2012 ### X89codered89X Is the question you gave precisely what how the question was given to you? ( I assume Homework?) I don't believe it's asking you to solve the equation, rather prove that it's possible for y to be a solution for any q and p. e.g., [itex]q(x), p(x),$ are continous in some neigborhood $(\alpha,\beta)$ and $y(x)$ is $C^2 \; \forall x \in (\alpha,\beta) \Rightarrow Ly = 0$ where the operator $L$ is defined as $L = (\frac{d^2}{dx^2} + p(x)\frac{d}{dx}+q(x))$.

I'm just guessing, in fact, I think I mixed up the inference. Oh well, good luck! Was anything about smoothness of functions as solutions to ode's mentioned in class? It must have. I say this because this is a time-varying diff eq., which adds an order of magnitude of complexity from time-invariant diff eq.'s. Normally, you have to check for singular points over the interval $(\alpha, \beta)$,but the problem gave you this assumption for free. And $sin(x^2)$ certainly has at least 2 derivatives. Look in your notes =D.

EDITED: for LaTeX...

Last edited: Oct 18, 2012
11. Oct 18, 2012

### th3chemist

yes! I think that is what it asking.

I'm not sure how about to go and prove it :(.

12. Oct 19, 2012

### HallsofIvy

Staff Emeritus
Yes, the problem asks if $sin(x^2)$ can be a solution for such an equation. That is, is it possible to find p and q that such that $sin(x^2)$ is a solution. When you set $y= sin(x^2)$, find its first and second derivatives, what do you get? Can you find p and q that will make that equation true?
(I suspect there will be many such p and q. Try to pick some simple ones.

13. Oct 19, 2012

### th3chemist

I got
y = sin(x^2)
y' = 2xcos(x^2)
y'' = 2cos(x^2) - 4x^2 sin(x^2)

Would I sub this into the equation and just solve for p and q?

14. Oct 19, 2012

### X89codered89X

you wouldn't be 'solving' p and q, since you'd have infinite smooth p and q that could work. you'd just 'pick' any p and q that happen to work.

15. Oct 19, 2012

### Citan Uzuki

On problems like these, I like to apply some metamathematics. Since we naturally assume that a smooth function can be a solution of a particular type of ODE unless proven otherwise, it follows that if the answer to this problem is "yes, y=sin(x^2) can be a solution of y'' + py' + q = 0," then the problem is uninteresting and would never have been asked. Therefore we deduce that y=sin(x^2) cannot be a solution and that we need to start looking for proofs that this is impossible. And since we are given the stipulation that the interval must contain zero, the proof will probably come from looking at the behavior of y'' + py' + q at or near zero. So ask yourself, what happens when we plug x=0 into that expression (regardless of what p and q are)?

16. Oct 22, 2012

### Vargo

Hi all,

Citan Uzuki is right to doubt the easy answer here. The correct answer is "no". Consider the facts about existence/uniqueness of solutions of 2nd order linear ODE with continuous coefficient functions p and q.

Last edited: Oct 22, 2012
17. Oct 22, 2012

### Vargo

Also, my guess is that this is a textbook question? Like Boyce and DiPrima section 3.2 maybe? If you at least give the section title in your question, then people would be more likely to pick up on the thrust of the question. I am actually teaching this class this semester and saw that question in the book, so it was already in my mind when I read your question.

18. Oct 22, 2012

### willem2

This is actually pretty easy, just compute y'''(0), y'(0) and y(0) and put them in the equation and see what happens.

19. Oct 23, 2012

### X89codered89X

Ridiculous I didn't see it.

20. Oct 23, 2012

### Vargo

The thing is, that technique for solution works and is simple. But the "real reason" is more basic. Suppose you take sin(x^4). Then that technique does not work. However, the same conclusion is still true. You can't have a solution whose first and second derivatives are simultaneously 0 (no matter what the second derivative does there).