Which is the partial solution of the ode 4th order I want to solve the ode $$y''''(t)+3y''(t)+y(t)=e^{-t}+t^2, \ \forall t>0$$ Forst we have to find the solution for the homogeneous problem using the characteristic polynomial, right? 
We have the characteristic polynomial $$\lambda^4+3\lambda^2+1=0$$ which has the solutions $$\lambda=\pm i\sqrt{\frac{3-\sqrt{5}}{2}} \ \text{ and } \ \lambda=\pm i\sqrt{\frac{3+\sqrt{5}}{2}}$$ So we get the solutio of the homogenous part: $$y_h(t)=A\cos \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+A\cos \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )$$ Is everything correct so far? How can we find the partial solution? 

With $$\lambda=\pm i\sqrt{\frac{3-\sqrt{5}}{2}} \ \text{ and } \ \lambda=\pm i\sqrt{\frac{3+\sqrt{5}}{2}}$$
You have your $$y_c=A\cos \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+A\cos \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )$$
correct.
For the particular solution you consider $$y_p = Ae^{-t} + Bt^2+Ct+D$$
Plug in your equation and find the cofficients.
