Solving the ODE $\frac{dx}{dt} = \frac{1}{2}v_0 + \frac{x}{2t}$ Im given this differential equation $\frac{dx}{dt} = \frac{1}{2}v_0 + \frac{x}{2t}$, where $v_0$ is the maximum velocity of a car in a traffic flow, and $x$ is the directed distance of a car from a trafic light.If the car starts up from rest, then $x=-x_o$ for $t= \frac{x_0}{v_0}$
a) Find the solution to the initial value problem and prove that ${v_0}$ is in fact the maximum velocity for the car.
b) Find the ammount of time that the car needs to reach the traffic light and compare it with the ammount of time  that the car would need if it were advancing with maximum velocity the complete distance from where it started until it reached the traffic light. 
I have not too much idea of physics, so, firstly, it is not that natural to me how to identify the variables in the equation $\frac{dx}{dt} = \frac{1}{2}v_0 + \frac{x}{2t}$. I would suppose the independent variable is time, and the dependent variable is distance ($x$). And $\frac{dx}{dt}$ is the velocity? That means that i would have to solve for $x$, but that confuses me since I am being asked for velocity. How do I use the fact that says "If the car starts up from rest, then $x=-x_o$ for $t= \frac{x_0}{v_0}$"
 A: You say "I have not too much idea of Physics" but have you ever taken a Calculus class?  That is what is needed here.  The fact that you have "$\frac{dx}{dt}$ should tell you immediately that "t" is intended to be the independent variable and x the dependent variable.
Seeing that $\frac{x}{2t}$ the first thing that would occur to me would be to try the substitution $y= \frac{x}{t}$.  Then $ty= x$ and, differentiating with respect to t, $t\frac{dy}{dt}+ y= \frac{dx}{dt}$ so the equation becomes $t\frac{dy}{dt}= v_0+ y/2$.  That is a "separable equation".  We can separate the two variables as $\frac{2dy}{v_0- y}= \frac{dt}{t}$.  Integrate both sides of that.
Once you have found x(t), of course dx/dt is the derivative.
If the car starts up from rest, Then dv/dt(0)= 0.
A: to find velocity, we have to solve differential equation given to find $x$ as a function of time. then by differentiating we can find velocity. back to equation. it a first order linear differential equation. it can be solved easily. assume we multiply both sides by a function of time like $u$ and compare it to the RHS below:
$$\frac{d}{dt}(ux)=u\frac{dx}{dt}+x\frac{du}{dt}$$
compare it with LHS below:
$$\frac{dx}{dt}u -\frac{x}{2t}u= \frac{1}{2}v_0u$$
in order that LHS be a differential form like $\frac{d}{dt}(ux)$ then $$\frac{du}{dt}=\frac{-u}{2t} \Rightarrow \ln (u)=-\frac{1}{2} \ln(t) \Rightarrow u = \frac{1}{\sqrt{t}}$$ 
by substituting this in main equation we get:
$$\frac{d}{dt}(x\frac{1}{\sqrt{t}})=\frac{1}{2}v_0u=\frac{1}{2\sqrt{t}}v_0 \Rightarrow \frac{x}{\sqrt{t}}=v_0 \sqrt t +C \Rightarrow x=v_0t+C\sqrt t$$
now we use initial value condition to find the constant $C$:
$$x(t=\frac{x_0}{v_0})=-x_0 \Rightarrow C=-2\sqrt{\frac{v_0}{x_0}} \Rightarrow x=v_0t-2\sqrt{\frac{v_0}{x_0}t}$$
for finding maximum value of $x$, have to find critical points of $x(t)$. cause critical points are extremums of $x(t)$ usually(except if higher order differentials be zero at critical point too). to do this we have to find where 
$$\frac{dx}{dt}=0 \Rightarrow v_0-\sqrt{\frac{v_0}{x_0}}\frac{1}{\sqrt t}=0 \Rightarrow t=\frac{x_0}{v_0}$$
also for finding maximum value of velocity we have to find where $\frac{d^2x}{dt^2}=0$. so have to find:
$$\frac{d^2x}{dt^2}=\sqrt{\frac{v_0}{x_0}} \frac{1}{t\sqrt t} \neq 0 $$
so the extremums are in boundaries. indeed maximum velocity is at $t \to \infty$ which is $\frac{dx}{dt}_{max} = v_0$
to solve part b) we have to know where the light is in other words $x_{light}=?$ then by solving for $t$ in $x(t) = x_{light}$ we can solve second part.
