(i) Let $I[y]=\int_{0}^{1}\left(\left(y^{\prime}\right)^{2}-y^{2}\right) d x$, where $y$ is twice differentiable and $y(0)=y(1)=0$. Write down the associated Euler-Lagrange equation and show that the only solution is $y(x)=0$.

(ii) Let $J[y]=\int_{0}^{1}\left(y^{\prime}+y \tan x\right)^{2} d x$, where $y$ is twice differentiable and $y(0)=y(1)=$ 0 . Show that $J[y]=0$ only if $y(x)=0$.

(iii) Show that $I[y]=J[y]$ and deduce that the extremal value of $I[y]$ is a global minimum.

(iv) Use the second variation of $I[y]$ to verify that the extremal value of $I[y]$ is a local minimum.

(v) How would your answers to part (i) differ in the case $I[y]=\int_{0}^{2 \pi}\left(\left(y^{\prime}\right)^{2}-y^{2}\right) d x$, where $y(0)=y(2 \pi)=0$ ? Show that the solution $y(x)=0$ is not a global minimizer in this case. (You may use without proof the result $I[x(2 \pi-x)]=-\frac{8}{15}\left(2 \pi^{2}-5\right)$.) Explain why the arguments of parts (iii) and (iv) cannot be used.