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<x<L , then the differential equation modeling the deflection of the beam y(x) is given by
EI \frac{d^4 y}{dx^4} = w(x)
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
Fourier Sine
Fourier Cosine
Fourier
expansion of y(x) :

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

We are going to use a Fourier series to solve this equation. To satisfy the initial conditions we should use a

Plugging the last sum into the differential equation we get that

Using a Fourier Sine series expansion on

You can earn partial credit on this problem.