Construct an orthogonal system of eigen functions corresponding to the S-L BVP y''+dy=0, y(0)=0, y(pi)=0. Hence express the function f(x)=x² in terms of the orthogonal system over the interval (0, pi).

The given Sturm-Liouville Boundary Value Problem (S-L BVP) is $y'' + \lambda y = 0$ with boundary conditions $y(0) = 0$ and $y(\pi) = 0$.
We consider three cases for the eigenvalue $\lambda$:

• Case 1: $\lambda < 0$ Let $\lambda = -\mu^2$ for some $\mu > 0$. The characteristic equation is $r^2 - \mu^2 = 0$, so $r = \pm \mu$. The general solution is $y(x) = c_1 e^{\mu x} + c_2 e^{-\mu x}$. Applying $y(0) = 0 \implies c_1 + c_2 = 0 \implies c_2 = -c_1$. So $y(x) = c_1(e^{\mu x} - e^{-\mu x}) = 2c_1 \sinh(\mu x)$. Applying $y(\pi) = 0 \implies 2c_1 \sinh(\mu \pi) = 0$. Since $\mu > 0$ and $\pi > 0$, $\sinh(\mu \pi) \neq 0$. Thus, $c_1 = 0$, which implies $c_2 = 0$. This yields only the trivial solution $y(x) = 0$.

• Case 2: $\lambda = 0$ The differential equation is $y'' = 0$. Integrating twice gives $y(x) = c_1 x + c_2$. Applying $y(0) = 0 \implies c_2 = 0$. So $y(x) = c_1 x$. Applying $y(\pi) = 0 \implies c_1 \pi = 0 \implies c_1 = 0$. This yields only the trivial solution $y(x) = 0$.

• Case 3: $\lambda > 0$ Let $\lambda = \mu^2$ for some $\mu > 0$. The characteristic equation is $r^2 + \mu^2 = 0$, so $r = \pm i\mu$. The general solution is $y(x) = c_1 \cos(\mu x) + c_2 \sin(\mu x)$. Applying $y(0) = 0 \implies c_1 \cos(0) + c_2 \sin(0) = 0 \implies c_1 = 0$. So $y(x) = c_2 \sin(\mu x)$. Applying $y(\pi) = 0 \implies c_2 \sin(\mu \pi) = 0$. For a non-trivial solution, $c_2 \neq 0$, so we must have $\sin(\mu \pi) = 0$. This implies $\mu \pi = n \pi$ for $n = 1, 2, 3, \dots$. (We exclude $n=0$ as it leads to $\lambda=0$). Thus, $\mu_n = n$. The eigenvalues are $\lambda_n = n^2$ for $n = 1, 2, 3, \dots$. The corresponding eigenfunctions are $y_n(x) = \sin(nx)$.

The orthogonal system of eigenfunctions is $\boxed{\{\sin(nx)\}_{n=1}^{\infty}}$.

This means finding the Fourier sine series expansion of $f(x) = x^2$ on the interval $(0, \pi)$. The Fourier sine series is given by $f(x) = \sum_{n=1}^{\infty} b_n \sin(nx)$, where the coefficients $b_n$ are:

$$b_n = \frac{2}{\pi} \int_0^{\pi} f(x) \sin(nx) \,dx$$

Substitute $f(x) = x^2$:

$$b_n = \frac{2}{\pi} \int_0^{\pi} x^2 \sin(nx) \,dx$$

We evaluate the integral using integration by parts twice. Recall $\int u \,dv = uv - \int v \,du$.
First, let $u = x^2$ and $dv = \sin(nx) \,dx$. Then $du = 2x \,dx$ and $v = -\frac{1}{n}\cos(nx)$.

$$\int_0^{\pi} x^2 \sin(nx) \,dx = \left[-\frac{x^2}{n}\cos(nx)\right]_0^{\pi} - \int_0^{\pi} \left(-\frac{1}{n}\cos(nx)\right) (2x) \,dx$$ $$= \left(-\frac{\pi^2}{n}\cos(n\pi) - 0\right) + \frac{2}{n}\int_0^{\pi} x\cos(nx) \,dx$$ $$= -\frac{\pi^2}{n}(-1)^n + \frac{2}{n}\int_0^{\pi} x\cos(nx) \,dx$$

Next, evaluate $\int_0^{\pi} x\cos(nx) \,dx$. Let $u = x$ and $dv = \cos(nx) \,dx$. Then $du = dx$ and $v = \frac{1}{n}\sin(nx)$.

$$\int_0^{\pi} x\cos(nx) \,dx = \left[\frac{x}{n}\sin(nx)\right]_0^{\pi} - \int_0^{\pi} \frac{1}{n}\sin(nx) \,dx$$ $$= \left(\frac{\pi}{n}\sin(n\pi) - 0\right) - \frac{1}{n}\left[-\frac{1}{n}\cos(nx)\right]_0^{\pi}$$

Since $\sin(n\pi) = 0$ for integer $n$:

$$= 0 + \frac{1}{n^2}[\cos(nx)]_0^{\pi} = \frac{1}{n^2}(\cos(n\pi) - \cos(0)) = \frac{1}{n^2}((-1)^n - 1)$$

Substitute this result back into the expression for $b_n$:

$$b_n = \frac{2}{\pi} \left[ -\frac{\pi^2}{n}(-1)^n + \frac{2}{n}\left(\frac{1}{n^2}((-1)^n - 1)\right) \right]$$ $$b_n = \frac{2}{\pi} \left[ -\frac{\pi^2}{n}(-1)^n + \frac{2}{n^3}((-1)^n - 1) \right]$$ $$b_n = -\frac{2\pi}{n}(-1)^n + \frac{4}{\pi n^3}((-1)^n - 1)$$

We analyze $b_n$ based on the parity of $n$:

• If $n$ is even ($n=2k$ for $k \ge 1$), then $(-1)^n = 1$: $$b_n = -\frac{2\pi}{n}(1) + \frac{4}{\pi n^3}(1 - 1) = -\frac{2\pi}{n}$$
• If $n$ is odd ($n=2k-1$ for $k \ge 1$), then $(-1)^n = -1$: $$b_n = -\frac{2\pi}{n}(-1) + \frac{4}{\pi n^3}(-1 - 1) = \frac{2\pi}{n} + \frac{4}{\pi n^3}(-2) = \frac{2\pi}{n} - \frac{8}{\pi n^3}$$

The function $f(x) = x^2$ can be expressed in terms of the orthogonal system as:

$$\boxed{f(x) = \sum_{n=1}^{\infty} b_n \sin(nx), \quad \text{where } b_n = \begin{cases} -\frac{2\pi}{n} & \text{if } n \text{ is even} \\ \frac{2\pi}{n} - \frac{8}{\pi n^3} & \text{if } n \text{ is odd} \end{cases}}$$