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? 
 A: For the particular solutions of equation $$y''''(t)+3y''(t)+y(t)=e^{-t}+t^2$$, you can separate this equation into two equations: $$y''''(t)+3y''(t)+y(t)=e^{-t}$$ and $$y''''(t)+3y''(t)+y(t)=t^2,$$ and find their particular solutions separately.
For the first one, you can suppose its particular solution is $y_1=a e^{-t}$,
and you can get $a=\frac{1}{5}$.
 For the second one, you can suppose its special solution is $y_2=at^2+b t+c$,
and  you can get your result. Please do it by yourself.
Then $$y=y_1+y_2$$ is the particular solution of your equation!
A: 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.
A: I think the general solution of the homogenous part should be
$$y_h(t)=A\cos \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+C\cos \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )+D\sin \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )$$
Every linear combination of the fundamental solution is itself a solution.
You may find a particular solution of the inhomogen equation by using variation of parameters. I denote your four fundamental solutions by $u_1(t), u_2(t), u_3(t)$ and $u_4(t)$. Use the ansatz $$y_p(t) =A(t)\cos \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+B(t)\sin \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+C(t)\cos \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )+D(t)\sin \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )$$
Since you have $4$ unknowns, you need additional conditions. The following will do it:
$$A'(t) u_1(t) + B'(t) u_2(t) + C'(t) u_3(t) + D'(t) u_4(t) = 0$$
$$A'(t) u'_1(t) + B'(t) u'_2(t) + C'(t) u'_3(t) + D'(t) u'_4(t) = 0$$
$$A'(t) u''_1(t) + B'(t) u''_2(t) + C'(t) u''_3(t) + D'(t) u''_4(t) = 0$$
Together with the inhomogen equation, you get that $y_p(t)$ is a solution if 
\begin{align*}A'u_1 + B'u_2 + C' u_3 + D'u_4&=0 \\\\
A'u'_1 + B'u'_2 + C' u'_3 + D'u'_4&=0 \\\\
A'u''_1 + B'u''_2 + C' u''_3 + D'u''_4&=0 \\\\
A'u'''_1 + B'u'''_2 + C' u'''_3 + D'u'''_4&=\mathrm{e}^{-t}+t^2 \end{align*}
I didn't try to solve this, but I hope you can :)
