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Let $f:\mathbb R \to \mathbb C$ satisfies the following differential equation (DE): $$f''(x)+x^{-2}f(x)=0 $$

Questions:

If $f$ satisfies the above (DE), then can we say $f_{x_0}(x)=f(x-x_0)$ ($0\neq x_0 \in \mathbb R$) also satisfies the above (DE)?

If $f$ satisfies the above (DE), then what can we say about $f$? (Is there any, method to obtain $f$ from the (DE)?)

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    This is known as a [Second Order Linear Homogenous ODE with non-constant coefficients](http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt1.pdf). – 2017-01-03
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    [This page](http://www.sosmath.com/diffeq/second/euler/euler.html) with $b = 0, c = 1$. The page derives a closed form for $f$, which I encourage you to follow along with. – 2017-01-03

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No, the DE is not translation invariant in the sense you specified.

Fix $a \in \mathbb R$, $a \ne 0$, and let $g(x) = f(x-a)$.

Then $g(x) = f(x-a) \implies g'(x) = f'(x-a) \implies g''(x) = f''(x-a)$.

Then

$$f''(x)+x^{-2}f(x)=0, \forall x \in \mathbb R$$

$$\implies f''(x-a)+(x-a)^{-2}f(x-a)=0, \forall x \in \mathbb R$$

$$\implies g''(x)+(x-a)^{-2}g(x)=0, \forall x \in \mathbb R$$

which is not the same as

$$g''(x)+x^{-2}g(x)=0, \forall x \in \mathbb R$$

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    To be strickt: it's the same if and only if $g(x) \equiv 0$. – 2017-01-03
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    Agreed. Although, in that case, translation by a doesn't yield a new function, thus in all cases, no new solutions can be had by translation. – 2017-01-03
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$$\frac{d^2f}{dx^2}+\frac{f}{x^2}=0$$ Of course, the ODE isn't translation invariant, but the ODE is expansion or contraction invariant. The change of variable $\quad x=at\quad$ leads to : $$\frac{d^2f}{dt^2}+\frac{f}{t^2}=0$$ This mean that the ODE is invariant (Of course, the solution isn't invariant : the coefficients in the solution change with $a$, but the analytic form of the solution doesn't change).

Analytic solving of the ODE :

Change of variable $\quad x=e^X \quad\to\quad \frac{d^2f}{dX^2}-\frac{df}{dX}+f=0$ $$f=c_1e^{X/2}\cos\left(\frac{\sqrt{3}}{2}X\right)+c_2e^{X/2}\sin\left(\frac{\sqrt{3}}{2}X\right)$$ $$f(x)=c_1x^{1/2}\cos\left(\frac{\sqrt{3}}{2}\ln|x|\right)+c_2x^{1/2}\sin\left(\frac{\sqrt{3}}{2}\ln|x|\right)$$ The change of $x$ to $ax$ changes $\quad \ln|x|\quad$ to $\quad\ln|x|+\ln|a|.\quad$ A few trigonometric transformations leads to the same form of solution, but with different constants $C_1$ , $C_2$ instead of $c_1$ , $c_2$.