Calculus of variations and integrating $\left(\frac{dy}{dx}\right)^2=\alpha^2(1+x)$ I'm trying to wrap my head around a step in the solution of an exercise that has me stumped. The exercise is the following :
Given the functional
$$S[y]=\int_0^{1}dx\sqrt{1+x+y'^2}, \quad y(0)=x_0, \quad y(1)=x_1$$
it is asked to show that $y(x)$ defined by
$$y'(x)=k\sqrt{1+x+y'(x)^2},$$ where $k$ is a constant, makes the functional stationary. Then, by expressing $y'(x)$ in terms of $x$, one should show that the solution is
$$y(x)=x_0+\frac{x_1-x_0}{2^{3/2}-1}\left((1+x)^{3/2}-1\right).$$
It has been shown previously in the course that given the functional $S[y]=\int_a^{b}dxF(x,y'),\quad y(a)=A, \quad y(b)=B,$ we obtain the following differential equation for the stationary path $y$ of $S$:
$$\frac{\partial}{\partial y'}F(x,y')=k, \quad y(a)=A, \quad y(b)=B,$$
where $k$ is a constant.
Now, the solution of the initial problem goes as follows:
We have $F(x,v)=\sqrt{1+x+v^2}$ and the general equation (above) becomes
$$v=k\sqrt{1+x+v^2}, \quad \text{where}\quad v=y'(x).$$
Rearranging and squaring, we obtain
$$\left(\frac{dy}{dx}\right)^2=\alpha^2(1+x), \quad \alpha^2=\frac{k^2}{1-k^2}.$$
Integrating gives the solution
$$y(x)-x_0=\alpha\int_0^xdx\sqrt{1+x}=\frac{2\alpha}{3}\left((1+x)^{3/2}-1\right).$$
It then goes on to find the value of $\alpha$ with the boundary conditions.
My questioning mostly pertains to the integration part. I feel like it is very vague, and I'm not sure how to go about such an integration and come out with the desired result. What about the integration limits $0$ and $x$? Secondarily, I also wonder about the need to introduce the variable $v$. Any help would be tremendously appreciated.
 A: If I understand you correctly, the part you're confused about is how we can go from
$$(y'(x))^2=\alpha^2(1+x)$$
to
$$y(x)-x_0=\alpha\int_0^x\sqrt{1+x}~\mathrm{d}x=\frac{2\alpha}{3}\left((1+x)^{3/2}-1\right).$$
So what we start by doing is to just rewrite the first equation as
$$y'(x)=\alpha\sqrt{1+x}$$
(note the absence of a $\pm$ in front of it; that is because we can just bake the sign into $\alpha$ and determine it later). Now recall that by the fundamental theorem of calculus,
$$\int_a^b y'(t)~\mathrm{d}t=y(b)-y(a),$$
and so in particular, since we have the value $y(0)=x_0$, we can choose $a=0$ (but we could just as well have chosen some other value if that was more convenient), and as we wish to find an expression for $y(x)$, we set $b=x$, so that
$$\int_0^x y'(t)~\mathrm{d}t=y(x)-x_0.$$
This means that, using the equality we had above,
$$y(x)-x_0=\int_0^x y'(t)~\mathrm{d}t=\int_0^x \alpha\sqrt{1+t}~\mathrm{d}t.$$
Now the right hand side we can calculate as
$$\int_0^x \alpha\sqrt{1+t}~\mathrm{d}t=\alpha\int_1^{x+1} u^{1/2}~\mathrm{d}u=\alpha\biggl[\frac{2u^{3/2}}{3}\biggr]_1^{x+1}=\frac{2\alpha}{3}\left((1+x)^{3/2}-1\right),$$
where we made the substitution $u=1+t$ as an intermediate step. This now gives the desired equality.
As for your second question, no, there is really no need to introduce $v=y'(x)$ other than, perhaps, to make it easier to solve $y'(x)=\alpha\sqrt{1+x+y'(x)^2}$ for $y'(x)$.
