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What is the differential equation of $$y = A\cos{(\log{x})} + B\sin{(\log{x})}$$

There are two arbitrary constants, so I think I have to differentiate $y$ twice. But I'm getting a very ugly equation upon differentiating, and I don't think I can't eliminate the arbitrary constants $A$ and $B$ using those.

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    Hint: find a relationship between the terms $x^2y''$, $xy'$ and $y$. This kind of differential equation is famous and named after Euler. – 2017-01-05
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    It might help, after you've found the first derivative $y'=\frac{1}{x}\left[-A\sin(\log x)+B\cos(\log x)\right]$ to multiply both sides of the equation by $x$ to ease the differentiation. – 2017-01-05

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\begin{array}{lrl}&(y)'&=&\frac1x\left[-A\sin{(\ln{x})}+B\cos{(\ln{x})}\right]\\[0.2cm]\implies& xy'&=&-A\sin{(\ln{x})}+B\cos{(\ln{x})}\\[0.2cm]\implies& (xy')'&=&\frac1x\left[-A\cos{(\ln{x})}-B\sin{(\ln{x})}\right]=\frac1x(-y)\\[0.2cm]\implies& y+xy'+x^2y''&=&0\end{array} which is called the Cauchy-Euler or most commonly just Euler's differential equation.

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    Thanks! Didn't know what Cauchy-Euler was! – 2017-01-05