Physical meaning of linear ODE $xy''+2y' + \lambda^2 x y = 0$

It has a physical meaning. It is the radial wave equations for the case $\ell = 0$.

The radial wave equation is written as

$$ \left[ \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} + k^2 - \frac{\ell(\ell+1)}{r^2} \right] \psi = 0, $$

which can also be written as

$$ \frac{d^2\psi}{dr^2} + \frac{2}{r} \frac{d\psi}{dr} + k^2 - \frac{\ell(\ell+1)}{r^2} \psi = 0. $$

The case $\ell = 0$ can be written as

$$ r \frac{d^2\psi}{dr^2} + 2 \frac{d\psi}{dr} + k^2 r \psi = 0, $$

which is the form

$$ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} + \lambda^2 x y = 0. $$

The equation can be though of as the 3D spherical symmetric version of the equation

$$\nabla^2y + \lambda^2 y = 0$$

since the Laplacian $\nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr}$ in this case. Likewise it can also be though of as the static spherical symmetric version of the (relativistic) equation $\square y + \lambda^2 y = 0$ where $\square = -\frac{d^2}{dt^2} + \nabla^2$ is the wave operator.

If $\lambda^2 > 0$ then the physical solution, which is regular at $r=0$, is $y \propto \frac{\sin(\lambda r)}{r}$. As mentioned in the other answer this can be though of as the monopole component of the radial wave-equation.

If $\lambda^2 < 0$ then the physical solution $y \propto \frac{1}{r}e^{-\lambda r}$ can for example describe the potential for two particles interacting throught the exchange of a particle with mass $\lambda$. This is known as the Yukawa interaction. If the graviton (or the photon) has a mass then the force potential $\frac{1}{r}$ leading to the inverse square law force of gravity and electromagnetism would be replaced by $\frac{1}{r}e^{-\lambda r}$ for which the force is exponentially suppressed at large distances $r > 1/\lambda$.