Integral calculus question relating to particle motion

Question

"A particle of mass m is attracted toward a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If the particle is released from rest, at distance L, from 0, find the time for it to reach 0."

Here's what I know:

$F=-{k \over x}$, where $k$ is a proportionality constant.

From basic physics I also know $F=ma=m{dv \over dt}$.

I equate the two: $-{k \over x}=m{dv \over dt}$.

I can simplify this expression with the following: ${dv \over dt} = {dx \over dt}{dv \over dx} = v{dv \over dx}$, which I then substitute into the previous expresion.

Through separable variables, followed by solving for the constant c, I can solve for $v$. As $v = {dx \over dt}$, I obtain the following:

$${dx \over dt} = \sqrt{-{2k \over m} ln\left({x \over L}\right)}$$

Here's where I am stumped. I believe the correct way forward to use separable variables again to obtain the following expression:

$${L*du \over \sqrt{- ln(u)}} = \sqrt{{2k \over m}} dt$$ where $u=\frac xL$. I have seen the left hand side being integrated from u=0 to u=1, using the Gamma Function. Using the Gamma Function is also in line with what we are using in calculus class right now and I believe is the right way forward. However, I am confused to as how exactly I can solve for $x$ as a function of $t$. Can anyone help? I've been stuck at this problem for a few hours now with no luck, and would be very grateful to anyone with a solution to this problem.

Answer 1

The answer is $t = \sqrt{\pi}\sqrt{\frac{m}{2k}} L$. Given what you've written above, you should be able to make the last connection.

Answer 2

The differential equation being $${dx \over dt} = \sqrt{-{2k \over m} ln\left({x \over L}\right)}$$ you can look at $${dt \over dx} =\frac{1} {\sqrt{-{2k \over m} ln\left({x \over L}\right)}}$$ and make the change of variable $$\sqrt{-\frac{2 k \log \left(\frac{x}{L}\right)}{m}}=y$$ that is to say $$x=L e^{-\frac{m y^2}{2 k}}$$ Then $${dt \over dy}=-\frac{L m e^{-\frac{m y^2}{2 k}}}{k}$$ and so $$t=-\frac{\sqrt{\frac{\pi }{2}} L \sqrt{m} \text{erf}\left(\frac{\sqrt{m} y}{\sqrt{2} \sqrt{k}}\right)}{\sqrt{k}}+ C$$ If you replace $y$ by its definition , you should arrive to something like $$t(x)= c_1+\frac{\sqrt{\frac{\pi }{2}} L \sqrt{\log \left(\frac{x}{L}\right)} \text{erfi}\left(\sqrt{\log \left(\frac{x}{L}\right)}\right)}{\sqrt{-\frac{k \log \left(\frac{x}{L}\right)}{m}}}$$

I hope I did not make any mistake.