Deriving velocity as a function of displacement from ({velocity, displacement} as functions of time) I have been trying to solve this problem for a while to no avail.
Suppose we have
$$
m\frac{dv}{dt}=mg-bv\enspace,\qquad v(0)=v_0
$$
(Where, of course, $m$ is mass, $g$ and $b$ are constants.)
We derive the following expressions for velocity and displacement in terms of time (assuming $x(0)=0):$
$$
v(t)=\frac{mg}{b}+\left(v_0-\frac{mg}{b}\right)e^{-bt/m}\\
x(t)=\frac{mg}{b}t+\frac{m}{b}\left(v_0-\frac{mg}{b}\right)\left(1-e^{-bt/m}\right)
$$
Now, going from here, we are supposed to eliminate time, and relate velocity and displacement without $t$.
The hint given is that, if we let $v(x)=v(x(t))$ be the velocity with regard to displacement,
$$
\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}\cdot v
$$
Stepping a tiny bit further, I found that $V$ can be expressed as a derivative such that
$$
\frac{dv}{dt}=\frac{dv}{dx}\frac{d}{dv}\left[\frac{1}{2}v^2\right]=\frac{d}{dx}\left[\frac{1}{2}v^2\right]
$$
Plugging the force equation with which we started, that reads as
$$
\frac{dv}{dt}=g-\frac{b}{m}v=\frac{d}{dx}\left[\frac{1}{2}v^2\right]
$$
So
$$
g-\frac{b}{m}\frac{d}{dx}\left[\frac{1}{2}v^2\right]=\frac{d}{dx}\left[\frac{1}{2}v^2\right]\quad\Leftrightarrow
$$
$$
d\left[\frac{1}{2}v^2\right]=\frac{mg}{b+m}dx\quad\Leftrightarrow\quad\int d\left[\frac{1}{2}v^2\right]=\frac{mg}{b+m}\int dx \quad\Leftrightarrow
$$
$$
|v|=\sqrt{\frac{2mg}{b+m}x}+c_1\quad\Leftrightarrow\quad x=\frac{1}{2}\frac{b+m}{mg}v^2+c_2
$$
EDIT:
O.k., so I saw that substituting $\frac{1}{2}v^2$ wasn't necessary, and using $mg=bv_0$, I get
$$
\frac{b}{m}\left(v_0-v\right)=\frac{dv}{dx}v
$$
which is separable to
$$
\frac{b}{m}dx=\frac{vdv}{v_0-v}
$$
Solving for x, I get
$$
x = -\frac{m}{b}\left(v_0log\left(v-v_0\right)+v\right)-\frac{m}{b}v_0
$$
Sadly, I still can't connect the dots that would lead to the solution the book gives, which is
$$
e^{bv}|bv-mg|^{mg}=e^{v_0b}|bv_0-mg|^{mg}e^{-b^2x/m}
$$
 A: Let us rewrite your expressions for velocity and displacement in terms of time, setting:
$\ \frac{mg}{b}=v_{lim}$ and $\frac{m}{b}=\tau$, so we have:
$$\ v(t)=v_{lim}+(v_0-v_{lim})e^{-t/\tau}$$
$$\ x(t)=v_{lim}t+\tau(v_0-v_{lim})(1-e^{-t/\tau})$$
Now, deriving $\ t$ from the first expression:
$$\ \frac{v(t)-v_{lim}}{v_0-v_{lim}}=e^{-t/\tau}$$
$$\ t=-\tau\ln\bigg[\frac{v(t)-v_{lim}}{v_0-v_{lim}}\bigg]$$
Now plugging this into $\ x(t)$ gives you
$$\ x(t)=-\tau v_{lim}\ln\bigg[\frac{v(t)-v_{lim}}{v_0-v_{lim}}\bigg]+\tau (v_0-v_{lim})\bigg(1-\frac{v(t)-v_{lim}}{v_0-v_{lim}}\bigg)$$
Which simplifies as
$$\ x(t)=-\tau v_{lim}\ln\bigg[\frac{v(t)-v_{lim}}{v_0-v_{lim}}\bigg]+\tau(v_0-v(t))$$
Notice that for $\ t=0$ we still have $\ x(0)=0$, and for $\ t\to\infty$, $\ x\to +\infty$ as we expect.
Now let us derive the solution the book gives, assuming $\ v_0>v_{lim}$:
$$\ x=-\frac{m}{b}\frac{mg}{b}\ln\bigg[\frac{v-\frac{mg}{b}}{v_0-\frac{mg}{b}}\bigg]+\frac{m}{b}(v_0-v)$$
$$\implies x=-\frac{m}{b}\frac{mg}{b}\ln\bigg[\frac{bv-mg}{bv_0-mg}\bigg]+\frac{m}{b}(v_0-v)$$
Now multiplying by $\ -b^2/m$:
$$\ -\frac{b^2x}{m}=mg\ln\bigg[\frac{bv-mg}{bv_0-mg}\bigg]-b(v_0-v)$$
$$\ -\frac{b^2x}{m}=\ln\bigg[\frac{bv-mg}{bv_0-mg}\bigg]^{mg}-\ln\bigg[e^{b(v_0-v)}\bigg]$$
$$\ -\frac{b^2x}{m}=\ln\bigg[\bigg(\frac{bv-mg}{bv_0-mg}\bigg)^{mg}\cdot e^{b(v-v_0)}\bigg]$$
Now exponentiating both sides:
$$\ e^{-b^2x/m}=\bigg(\frac{bv-mg}{bv_0-mg}\bigg)^{mg}\cdot e^{b(v-v_0)}$$
Finally:
$$\ e^{-b^2x/m}(bv_0-mg)^{mg}e^{bv_0}=(bv-mg)^{mg}e^{bv}$$
A comment on your try: the expression 
$$\ x=\frac{1}{2}\frac{b+m}{mg}v^2+c_2$$
is physically wrong because $\ m$ and $\ b$ have different dimensions so you cannot sum them.
A: It isn't necessary to subsitute the solution for $t$ into $x$, it is possible to achieve the correct solution using the chain rule:
$$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$
$$mv\frac{dv}{dx}=mg-bv \quad v\left(0\right)=v_0$$
To simplify things i will make all variables dimensionless:
$$\tilde{v}=\frac{v}{\Delta v}=\frac{v}{v_{0}}\quad\tilde{x}=\frac{x}{\Delta x}=\frac{xg}{v_{0}^{2}}\quad\beta=\frac{bv_{0}}{mg}
 $$
To yield:
$$\tilde{v}\frac{d\tilde{v}}{d\tilde{x}}=1-\beta\tilde{v}\quad \tilde{v}\left(0\right)=1$$
You can interpret $\beta$ as the relative strength of drag vs gravity.
Integrating by parts:
$$\int\frac{\tilde{v}}{1-\beta\tilde{v}}d\tilde{v}={\it \int d\tilde{x}}$$
$$-\frac{1}{\beta^{2}}\ln\left|1-\beta\tilde{v}\right|-\frac{1}{\beta}\tilde{v}=\tilde{x}+K
$$
Using the initial condition: $\tilde{v}\left(0\right)=1\rightarrow K=-\frac{1}{\beta^{2}}\ln\left|1-\beta\right|-\frac{1}{\beta}
 $
Rearranging leads to:
$$-\frac{1}{\beta^{2}}\ln\left|\frac{1-\beta\tilde{v}}{1-\beta}\right|-\frac{1}{\beta}\left(\tilde{v}-1\right)=\tilde{x}$$
which in dimensional terms is:
$$-\frac{m^{2}g}{b^{2}}\ln\left|\frac{mg-bv}{mg-bv_{0}}\right|-\frac{m}{b}\left(v-v_{0}\right)=x$$
and can be rearranged to:
$$\ln\left|\frac{mg-bv}{mg-bv_{0}}\right|^{mg}+b\left(v-v_{0}\right)=-\frac{b^{2}}{m}x$$
Exponentiation of both sides of this equation and applying logarithm identies will lead from this equation to the solution as provided in your book.
Update: personally I think the final solution given in your book looks a bit ridiculous. I don't understand the necessity to raise to the power of $mg$. I much more prefer to rewrite it to:
$$\ln\left|\frac{1-\beta\tilde{v}}{1-\beta}\right|+\beta\left(\tilde{v}-1\right)=-\beta^{2}\tilde{x}$$
And use it in this form or perhaps exponentiate it to further 'simplify'.
