Suppose a uniform beam of length $L$ is simply supported at $x = 0$ and $x = L$. If the load per unit length of the beam is given by w(x), $0, then the differential equation modeling the deflection of the beam y(x) is given by Because the beam is simply supported at both ends we know that
$y(0) =$ $y(L) =$
$y^{\prime\prime}(0) =$ $y^{\prime\prime}(L) =$

We are going to use a Fourier series to solve this equation. To satisfy the initial conditions we should use a expansion of $y(x)$:
Then
$\displaystyle\frac{dy}{dx} = \sum\limits_{n=0}^\infty$ $\displaystyle b_n \cos\!\left(\frac{n\pi x}{L}\right)$

$\displaystyle\frac{d^2y}{dx^2} = \sum\limits_{n=0}^\infty$ $\displaystyle b_n \sin\!\left(\frac{n\pi x}{L}\right)$

$\displaystyle\frac{d^3y}{dx^3} = \sum\limits_{n=0}^\infty$ $\displaystyle b_n \cos\!\left(\frac{n\pi x}{L}\right)$

$\displaystyle\frac{d^4y}{dx^4} = \sum\limits_{n=0}^\infty$ $\displaystyle b_n \sin\!\left(\frac{n\pi x}{L}\right)$

Plugging the last sum into the differential equation we get that
$\displaystyle\sum\limits_{n=0}^\infty$$\displaystyle b_n \sin\!\left(\frac{n\pi x}{L}\right) = w(x)$

Using a Fourier Sine series expansion on $[0,L]$we know that
 $b_n$ $=$ $\displaystyle\int$ $w(x)dx$

You can earn partial credit on this problem.