Calculation of the second variation of the functional $I(y)=\int_{-1}^1 [x^2(y')^2+x(y')^3]\,dx$

Question

My question:

I don't understand the last equation about second variation. According to definition, shouldn't it be $\int_{-1}^1 [2x^2+6xy'] (\eta)^2$?

Can anyone help me with this? Where am I wrong?

Answer 1

Replacing $y$ with $y+\eta$ yields $$ \int_{-1}^1 [x^2(y'+\eta')^2+x(y'+\eta')^3]\,dx $$ where the quadratic term is $$ \frac12 \delta^2I_y (\eta)=\int_{-1}^1 [x^2 (\eta')^2+3xy'(\eta')^2]\,dx $$ The author considers second variation at $y=0$, where it takes the form $$ \frac12 \delta^2I_0 (\eta)=\int_{-1}^1 x^2 (\eta')^2\,dx $$