Find the Euler-Lagrange equation for $\int_0^4(tx'-(x')^2)dt$.
My attempt: $0 - \frac{d}{dt}(t-2x')=0$
From here, I am not sure what to do. Quite frankly, I don't even think I started the problem correctly. Any hints to guide me into the right direction would be much appreciated.
Your calculation is correct. From what you got $$-\frac{d}{dt}(t-2x^\prime)=0$$, we obtain $$\boxed{1-2x^{\prime\prime}=0}.$$ This is the answer for your question. On the other hand, if you want to solve the differential equation, it is better integrating $-\frac{d}{dt}(t-2x^\prime)=0$ to obtain $t-2x^\prime=C_1$, where $C_1$ is a constant. Integrating one more time, we get the solution $$x(t)=\frac{t}{2}\left(\frac{t}{2}-C_1\right)+C_2,$$ where $C_2$ is another constant.