How to solve this Nonhomogeneous ODE problem of beam deflection and find particular solution Problem states that the load on the beam having length L and fixed on both end is;
$\omega(x)=w_0\frac{x}{L}$
Function of the deflection of the beam is given as;
$EI\frac{d^4y}{dx^4}=w_0\frac{x}{L}$ where $E, \quad I \quad and \quad w_0$ are constants.
a) find the half-sine fourier series representation of $w(x)$
b) using the series representation found in a, find the particular solution of diff equation.
 A: Firstly, load are usually taken to be positive down, so I'm going to change your differential equation to
$$EI\frac{d^4y}{dx^4}=-w(x)=-w_0\frac xL$$
Then we want eigenfunctions that are zero at both ends so those would be
$$y_n(x)=\sin\frac{n\pi x}L$$
Then your load becomes
$$w(x)=\sum_{n=1}^{\infty}w_n\sin\frac{n\pi x}L$$
We can find
$$\begin{align}\int_0^Lw_0\frac xL\sin\frac{n\pi x}Ldx&=\frac{w_0}L\left[-\frac{Lx}{n\pi}\cos\frac{n\pi x}L+\frac{L^2}{n^2\pi^2}\sin\frac{n\pi x}L\right]_0^L\\
&=(-1)^{n+1}\frac{L^2}{n\pi}=\sum_{k=1}^{\infty}w_k\int_0^L\sin\frac{k\pi x}L\sin\frac{n\pi x}Ldx\\
&=\sum_{k=1}^{\infty}w_k\frac L2\delta_{kn}=\frac L2w_n\end{align}$$
Then we let
$$y(x)=\sum_{n=1}^{\infty}b_n\sin\frac{n\pi x}L$$
Inserting into the differential equation,
$$\sum_{n=1}^{\infty}b_nEI\frac{n^4\pi^2}{L^4}\sin\frac{n\pi x}L=-\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2w_0}{n\pi}\sin\frac{n\pi x}L$$
So our solution is
$$y(x)=\sum_{n=1}^{\infty}(-1)^n\frac{2L^4w_0}{n^5\pi^5EI}\sin\frac{n\pi x}L$$
Now, if you solve that differential equation the old fashioned way, applying the boundary conditions $V(0^+)=\frac16w_0L$, $M(0^+)=0$, and $y(0)=y(L)=0$, you get
$$y(x)=\frac{w_0}{360EIL}(-3x^5+10L^2x^3-7L^4x)$$
And this does in fact have the same Fourier series, but I don't know how you are supposed get from the Fourier series to the polynomial.
