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.
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.