When the Euler-Lagrange equation reduces to 0=0

Question

I've gotten the functional $$\int_a^b(y^2+2xyy')dx$$ with Dirichlet boundary conditions. Applying the Euler-Lagrange equation I get:

$$0=\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}= 2y+2xy' -\frac{d}{dx}[2xy] = 2y+2xy'-[2y+2xy']=0$$

So this just gives me $0=0$. Have I done something wrong or does this mean that every curve is a stationary curve? If it's the latter, what's the reason for this functional being path independent (I thought that only happened when the functional is indepedent of $y'$)?

Answer 1

$$ \frac{d}{dx} \left( xy^2 \right) = y^2 + 2xyy'. $$ Hence $$ \int_a^b \left( y^2 + 2xyy' \right) \, dx = b y(b)^2 - a y(a)^2. $$