(1) Noting that \(\displaystyle y y' y'' = 4x^3\) I looked for a polynomial solution. I got \(\displaystyle y = x^2\).

(2) Break this up into homogeneous and particular parts. So solve \(\displaystyle y_h^{(4)} - y_h = 0\). The characteristic equation is \(\displaystyle m^4 - 1 = 0\) which has solutions of m = \(\displaystyle \{ \pm 1, ~ \pm i \}\). The homogeneous solution is therefore \(\displaystyle y_h(x) = A e^{x} + B e^{-x} + C ~ \sin(x) + D ~ \cos(x)\).

Now find the particular solution...

-Dan