Find the value of a function with definite integrals I am trying to understand a paper of Maynard Smith (1974), that connects biology with game theory. I don't want to overwhelm you with useless stuff, but I have this definite integrals:
$$E(m)=\int_0^m (v-x)p(x) dx - \int_m^\infty mp(x) dx  \tag{1}$$
We want to choose $p(x)$ such that $E(m)$ is the same constant $C$ for all $m$. 
Now I'll copy exactly what he (Maynard Smith) says:

To find $p(x)$ we put $E(m) = E(m+\Delta m)$, so that
$$E(m)=\int_0^m (v-x)p(x) dx -\int_m^\infty mp(x) dx=\int_0^{m+\Delta m}(v-x)p(x)dx - \int_{m+\Delta m}^\infty( m+\Delta m) p(x) dx  \tag{2}$$
After a little manipulation, remembering that
  $E(m)=\int_0^\infty p(x) dx =1$ this gives
$$ vp(m)= 1-\int_0^m p(x) dx  \tag{3}$$
Equation $(3)$ is satisfied by the function 
  $$ p(x) = (1/v ) e^{-x/v}  \tag{4}$$
which is the equilibrium strategy we are seeking.

Since I am not a mathematician (but I have some knowledge of Calculus and Probability) I am having an hard time understanding this. I have no idea how we can go from $(2)$ to $(3)$ and from $(3)$ to $(4)$. Could anyone help? 
Thank you. 
 A: The easiest thing to do is to apply the condition
$$\int_0^{\infty} dx \: p(x) = 1$$
at the beginning.  This means rewriting $E(m)$ as
$$\begin{align}E(m) &= \int_0^m dx \: (v-x) p(x) - m + m \int_0^m dx \: p(x)\\ &= (v+m) \int_0^m dx \: p(x) - \int_0^m dx \: x p(x) - m\end{align}$$
Now write
$$\begin{align}\\ E(m+dm) &= (v+m+dm) \int_0^{m+dm} dx \:p(x) - \int_0^{m+dm} dx \: x p(x) - m-dm\\ &= (v+m+dm)\left[\int_0^m dx \: p(x) + \underbrace{\int_m^{m+dm} dx \:  p(x)}_{p(m) dm} \right]\\ &- \left[\int_0^m dx \: x p(x) + \underbrace{\int_m^{m+dm} dx \: x p(x)}_{m p(m) dm}  \right]-m-dm\\ &=  (v+m) \int_0^m dx \: p(x) - \int_0^m dx \: x p(x) - m \\ &+\left [\int_0^m dx \: p(x) + (v+m) p(m) -m p(m)-1 \right ] dm \\ &= E(m) + \left [\int_0^m dx \: p(x) + (v+m) p(m) -m p(m)-1 \right ] dm\end{align}$$
The quantity in brackets is zero because $dm$ is arbitrary:
$$\int_0^m dx \: p(x) + (v+m) p(m) -m p(m)-1=0$$
or
$$v p(m) = 1-\int_0^m dx \: p(x)$$
as stated in step (3).
To get step (4), differentiate both sides with respect to $m$:
$$v p'(m) = -p(m) \implies p'(m) = -\frac{1}{v} p(m)$$
The general solution to this equation is
$$p(m) = A e^{-m/v}$$
$A=p(0)$ is determined by setting $m=0$ in the equation in step (3):
$$0=1-v p(0) \implies p(0)=\frac{1}{v}$$
Therefore
$$p(m) = \frac{1}{v} e^{-m/v}$$
as stated in step (4).
