Spring-mass function Here is the question that I'm stuck on:

What is the smallest value of $b$ for which we
  get solutions that, when viewed in the position-velocity plane, lie along a straight line? Algebraically
  support your conclusion.

I'm given Newton's Law of motion equation which is
$$ m \frac{d^2x}{dt^2}+b \frac{dx}{dt} + kx = 0$$
where $x$ is the position of an object (a block, for example) that's attached to the end of the spring, $m$ is the mass of the object, $b$ is the firction parameter (damping coefficient) and $k$ is the spring constant. I'm not sure how to approach the question that I imposed above, but I do know that 
$$\frac{dx}{dt} = y$$ where $y$ is the velocity. Also, I know that 
$$\frac{dy}{dt} = \frac{d^2x}{dt^2}$$
I could then re-write my  second order linear differential equation as a system of first order linear differential equations, which is 
$$\frac{dx}{dt} = y$$
$$\frac{dy}{dt} = -\frac{k}{m}x - \frac{b}{m}y$$
Even if the work above doesn't pertain to how to solve the problem, this is what I know so far. 
 A: Define the velocity $v = dx/dt$. If the solution in the position-velocity plane is a straight line, then we have $v \;=\; c\,x$ for some unknown, real constant $c$. We know it must have this form, as opposed to a more general form like $v \;=\; c\, x \;+\; a$, because $x \;=\; 0$, $v \;=\; 0$ is an equilibrium point of this damped system, so it must lie along the line.
Newton's law for this system is:
$$
m\frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x \;=\;
m\frac{d v}{dt} + b v + k x \;=\;
0\, .
$$
The fact that $v = c\,x$ means that $dv/dt = c\, dx/dt = c\,v$, so:
$$
m\,c\,v \;+\; b\,v + k\,x \;=\; 0\; \rightarrow\;
v \;=\; \frac{-k}{mc \,+\, b}\, x
$$
However, we already know that $v = c\, x$, which implies that
$$
c \;=\; \frac{-k}{mc \,+\, b} \; \rightarrow\;
m\, c^2 \;+\; b\, c \;+\; k \;=\; 0
$$
Solving this quadratic equation for $c$ yields
$$
c \;=\;
\frac{-b \;\pm\; \sqrt{b^2 \;-\; 4\,m\,k}}{2 m}\, .
$$
Since $m$, $k$, and $b$ are all positive quantities, the only way that $c$ can be real is if the quantity inside the square root is greater than or equal to $0$, i.e. if $b \ge \sqrt{4 m k}$. This answers the original question.
Such a condition on $b$, leading to a straight line in the $v-x$ plane, is known as overdamping.
