y'''' + 2y'' + y = 3t + 4,
y(0)= y'(0) = 0,
y''(0) = y'''(0)=1
The charachteristic equation is,
$\displaystyle k^4 + 2k^2 + 1 = 0 \implies (k^2 + 1)^2 = 0 \implies (k-i)^2(k+i)^2=0$
Because the roots have multiplicity it means the general solution is,
$\displaystyle y = c_1\sin (t) + c_2 \cos (t) + c_3 t\sin (t) + c_4 t\cos (t)$
However, this only solves the equation $\displaystyle y'''' + 2y'' + y = 0$.
To finish the solution look for a particular solution which happens to be (for example) $\displaystyle 3t+4$.
Thus, the full solution is, $\displaystyle y = c_1\sin (t) + c_2 \cos (t) + c_3 t\sin (t) + c_4 t\cos (t) + 3t+4$