I'm trying to find the minimum to the functional $$J[y]=\int_0^1 \left(\frac{1}{2}y'^2+yy'+y'+y\right)\, \mathrm dx$$ where the values for $y$ at the boundary values are not specified.
My first idea was to find the partial derivatives of $L(x,y,y')=\frac{1}{2}y'^2+yy'+y'+y$ which are \begin{align} \frac{\partial L}{\partial y}&=y'+1, \\ \frac{\partial L}{\partial y'}&=y'+y+1. \end{align} Now I applied the Euler-Lagrange Equation to get $$0=\frac{\partial L}{\partial y}-\frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)=y'+1-\frac{d}{dx}(y'+y+1)=y'+1-y''-y'=1-y''$$ and hence $0=1-y''$ and $y''=1$. Integrating twice we get $y=x^2+Ax+B$ with $A,B \in \mathbb{R}$.
The thing that is confusing me is that the book that I have the problem from gives the solution $y=\frac{1}{2}(x^2-3x+1)$. Where did I do a mistake? What's the full way to the problem?
Satisfying the Euler-Lagrange equations is necessary for an extremum, so any minimiser is of the form $y = \frac 12 x^2 + ax + b$. With this, \begin{align*} f(a,b) = J[y] &= \int_0^1 \left(\frac 12(x+a)^2 + \left(\frac{x^2}{2} + ax + b \right)(x+a) + (x+a) + \left(\frac{x^2}{2} + ax + b \right)\right) \, \mathrm dx \\ &= \int_0^1 \left( \frac 12 x^3 + \left(\frac 12 + \frac{3a}{2} + \frac 12\right)x^2 + \left(a + a^2 + b + 1 + a\right)x + \left(\frac 12 a^2 + ab + a + b\right)\right) \\ &= \left[\frac 18 x^4 + \frac 13 \left(1 + \frac{3a}{2}\right)x^3 + \frac 12\left(a^2 + 2a + b + 1\right)x^2 + \left(\frac 12 a^2 + ab + a + b\right)x\right]_0^1 \\ &= \frac 18 + \frac 13 + \frac{a}{2} + \frac{a^2}{2} + a + \frac{b}{2} + \frac 12 + \frac{a^2}2 + ab + a + b \\ &= a^2 + \frac{5}{2} a + ab + \frac{3}{2} b + \frac{23}{24}. \end{align*} This is minimal when $f_a(a,b) = 0 \implies 2a + \frac 52 + b = 0$ and $f_b(a,b) = 0 \implies \ a + \frac 32 = 0$. Thus $a = -\frac 32$ and $b = \frac 12$, and $$y = \frac 12 x^2 - \frac 32 x + \frac 12,$$ as required.