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Let \(f = y^{'2}\) and \(h = y^2\).

Then

\begin{align*}

G[y(x)] &= \int_0^1[f - \lambda h]dx\\

&= \int_0^1\left[y^{'2} - \lambda y^2\right]dx

\end{align*}

The Euler-Lagrange equation is then

\begin{align*}

-2y\lambda - \frac{d}{dx}\left(2y'\right) &= 0\\

y\lambda + y'' &= 0

\end{align*}

We have the equation for a simple harmonic oscillators,

\(y(x) = A\cos\left(x\sqrt{\lambda}\right) +

B\sin\left(x\sqrt{\lambda}\right)\).

\begin{alignat*}{3}

y(0) &= A &{}= 0\\

y(1) &= B\sin\big(\sqrt{\lambda}\big) &{}= 0\\

\sin\big(\sqrt{\lambda_n}\big) &= 0\\

\lambda_n &= \pi^2n^2 &&\quad (\text{where \(n\in\mathbb{Z}\).})

\end{alignat*}

So our equation is \(y_n(x) = B\sin\big(\pi nx\big)\).

We can now use our constraint integral.

\begin{align*}

\int_0^1B\sin^2\big(\pi nx\big)dx &= 1\\

\frac{B}{2}\int_0^1(1 - \cos\big(2\pi nx\big))dx &=

\frac{B}{2}\left[x - \frac{1}{\pi n}\sin\big(2\pi n x\big)\right|_0^1\\

B &= 2

\end{align*}

Therefore, our equation is \(y_n(x) = 2\sin\big(\pi n x\big)\).

So then the minimum is

\[

\int_0^1y^{'2}dx = 2\pi^2n^2.

\]

Is this correct?