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Is the solution of ${1\over x}(y''-y)=0$ simply $y(x)=a\cosh{x}+b\sinh{x}$? Or is there something fishy to do with the $1\over x$?

What are the solutions s.t. ${1\over x}y(x)$ is bounded as $x\to 0$? I am guessing $y(x)=b\sinh{x}$? And the ones bounded as $x\to \infty$?


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"Oops, error" is not very clear. If there's an error, please remove it or mark it clearly. – joriki Nov 13 '11 at 17:36
@joriki: the "oops error" bit is just to bring attention to that (following) bit where I have made changes :-). This version of the question should be error-free. – magna Nov 13 '11 at 17:49
@joriki: I have deleted the misleading "oops, error" :-) – magna Nov 13 '11 at 18:06
up vote 1 down vote accepted

Strictly speaking, the differential equation isn't defined at $x=0$. Thus the maximal domain on which you can consider this is $\mathbb R\setminus\{0\}$, and there's nothing to constrain the solutions on both sides of $0$ to have anything to do with each other; thus you can choose their parameters $a$ and $b$ independently of each other. You can also use $c_+\mathrm e^x+c_-\mathrm e^{-x}$ if you like, by the way.

Your guess $y(x)=b\sinh x$ is correct, except again you can choose different parameters $b$ to the left and right of $0$. For $x\to\infty$, the exponential representation above is more conducive for seeing the answer immediately.

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Thanks, @joriki ! – magna Nov 13 '11 at 20:42

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