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I'm relatively new to differential equations and I recently read that if the functions $p(x)$ and $q(x)$ are continuous at $x=0$ then it is never possible for $x^2$ to be a solution to the equation $y''+p(x)y'+q(x)y = 0$

Given that this is still relatively elementary work as I've only just started looking at second order differential equations, I was wondering if anyone could explain this to me please as I can't see how it holds with the theorems I have.

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Hint Substituting $x^2$ into the equation $$y'' + p(x) y' + q(x) y$$ gives that $$(2) + p(x) (2x) + q(x) (x^2) = 0.$$ Now, what happens at $x = 0$?

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  • $\begingroup$ Ah, ok. I was expecting something a little more theorem based than this by the way the book talks but this makes complete sense. Thanks! $\endgroup$ – vs9734 Apr 29 '15 at 11:18
  • $\begingroup$ You're welcome, I hope you found it useful. Since $x^2$ is a specific function, in this case it's easiest simply to substitute. Note that we only used a particular property of $x^2$ and so our argument proves a more general result: A general second-order linear ODE $y'' + py' + qy = 0$ with coefficient functions continuous at $x = x_0$ has no solution that vanishes precisely to second order at $x = x_0$. In fact, the same sort of argument shows that this still holds if we replace second order with $k$th order in both uses of the term. $\endgroup$ – Travis Willse Apr 29 '15 at 11:49
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Subsititute $y=x^2$ in $y''+p(x)y'+q(x)y$ and you'll get

$$2+2p(x)x+q(x)x^2$$

Now, what is the limit of this for $x\to 0$ assuming continuity of $p(x)$ and $q(x)$?

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    $\begingroup$ We only need $p$ and $q$ to be defined at $x=0$. $\endgroup$ – lhf Apr 29 '15 at 11:21

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