Let $u:[a,b]\to \mathbb{R}$ be differentiable, $$\cal{F}(u):=\int_a^b\sqrt{1+u'(t)^2}dt.$$ Find $u$ with $u(a)=u_0, u(b)=u_1$.
Idea: Minimalizing. It is $0=\frac{d}{ds}\cal{F}(u+s\phi)|_{s=0}=\int_a^b\frac{u'(t)\cdot \phi(t)}{\sqrt{1+u'(t)^2}}$, so integration by parts and the fundamental lemma gives $$\left(\frac{u'(t)}{\sqrt{1+u'(t)^2}}\right)'=0$$ Now I need to find $u$ which satisfies the initial conditions. It is $\frac{u'(t)}{\sqrt{1+u'(t)^2}}$ constant so I thought $$\frac{u'(t)}{\sqrt{1+u'(t)^2}} \equiv c \Rightarrow u'(t)=\frac{c}{\sqrt{1+c^2}}$$ Integrating gives $$u(t)=\frac{c}{\sqrt{1+c^2}}\cdot t+c_0$$ with $u(a)=\frac{ac}{\sqrt{1+c^2}}+c_0=u_0$ and $u(b)=\frac{bc}{\sqrt{1+c^2}}+c_0=u_1$ to find $c$ and $c_0$. Correct?
Your last line does not make sense. Why do you still have $t$ in thjose equations? And why does $c_0$ appear twice?
You can simplify your equations byreplacing $\frac c{\sqrt{1+c^2}}$ by some new constant $d$. Then $$u(t)=d\cdot t+c_0$$so $$u(a)=d\cdot a+c_0=u_0\\u(b)=d\cdot b+c_0=u_1$$ Subtracting these two equations you get $$d(a-b)=u_0-u_1$$ or $$d=\frac{u_1-u_0}{b-a}$$ and then use either of those equations to find $c_0$.