Estimating age of the universe by Hubble's law? Hubble's law states that $v=Hx$
Age of the universe is calculated by $T= \frac{x}{v} = \frac{1}{H}$
but the velocity is not constant; it changes with distance, so I think that this equation cannot be applied because simply the velocity is NOT uniform
$v$ should have been replaced with $\frac{\mathrm{d}x}{\mathrm{d}t}$, so it gives this differential equation $\frac{\mathrm{d}x}{\mathrm{d}t}=Hx$.
Nevertheless, when this differential equation is solved, and we substitute for initial condition ($x=0$ when $t=0$), it appears that there are no solutions, because it will be an exponential function, which can't take the value $0$ ever. 
What is wrong with my understanding? 
 A: You're actually pretty close to the correct method for estimating the age of the Universe, but $H$ is not constant with time, it is $H=H(t)$.
One of the many ways of writing the equation to solve is:
$$t(z) = \frac{1}{H_0}\int_z^\infty \frac{dz}{(1+z)E(z)}$$
Here $z$ is redshift; $z\rightarrow\infty$ at the Big Bang, and $z=0$ now, so if you integrate from $0$ to $\infty$ you get $t(0)$ , the age of the Universe.
$E(z)$ is a function describing the relative content of the Universe at different redshifts, one example of $E(z)$ is the one for the $\rm\Lambda CDM$ model:
$$E(z) = \sqrt{\Omega_{\Lambda,0}+(1-\Omega_0)(1+z)^2+\Omega_{m,0}(1+z)^3+\Omega_{r,0}(1+z)^4}$$
The $\Omega$ are density parameters, which you can read more about here or here. The $0$ subscripts indicate that these should be the values measured at $z=0$, so these are constants and you can write the age of the Universe in terms of them (or you could, if that integral had an analytic solution - I don't suggest trying to solve it by hand, unless you make some simplifying assumptions or approximations first!).
The reason that $t(0)\approx1/H_0$ works reasonably well is that the Universe has been expanding at very nearly the rate it is expanding at now for the majority of the time since the Big Bang. So approximating $v$ as constant at the present value actually gets you pretty close to the right answer.
