I'm trying to find the extremals of the functional $$J[y] = \int_0^1 (y')^2 + y^2 + 4ye^x \, {\rm d}x,$$ imposed that $y(0) = 0$ and $y(1) = 1 $. I got that there can't be extremals, and that's weird since this is the first exercise the book gives, I would expect some nice answers. Can someone check my work, please? It's not long.
If $f(x,y,y') = (y')^2 + y^2 + 4ye^x$, then Euler-Lagrange becomes: $$\frac{\partial f}{\partial y} - \frac{\rm d}{{\rm d}x}\left(\frac{\partial f}{\partial y'}\right) = 0 \implies y'' - y = 2e^x.$$
I got $y = c_1e^x + c_2e^{-x} + e^x,$ $c_1,c_2 \in \Bbb R$ after solving the ODE. Then using the boundary conditions, I get: $$\begin{cases} c_1 + c_2 + 1 = 0 \\ ec_1 + ec_2 + e = 1 \end{cases},$$ but that system has no solutions.
My initial conclusion is right? Or have I made some mistake somewhere? I'm self-learning here. Thanks.
After reviewing the calculations and talking to a colleague, it was a distraction of my part. The solution to the ODE is actually $$y = c_1e^x + c_2e^{-x} + xe^x,$$ I had solved it correctly before, but I don't know why I forgot that $x$ after. Then the system becomes $$\begin{cases} c_1 + c_2 = 0 \\ ec_1 + \frac{c_2}{e} + e = 1 \end{cases}.$$
(note my second mistake here, $ec_2$ instead of $c_2/e$ before)
Then the solutions are $c_1 = -c_2 = -e/(1+e)$, if I'm not wrong (yet again). And that's pretty much it.