Answer
$y_1(t)= t/3$ and $y_2= e^{-t} +t/3$ are solutions to the differential equation $y''''+4y'''+3y = t$.
Work Step by Step
To confirm that $y_{1} = t/3$ and $y_2 = e^{-t}+t/3$ are solutions to the differential equation $y''''+4y'''+3y = t$ we need only substitute the functions $y_1$ and $y_2$, respectively, for the variable $y$ in the given equation and perform the indicated operations in the l.h.s. (left hand side) of the equation and see if we obtain $t$ on the l.h.s. to be the same as the r.h.s. (right hand side), i.e., $t$.
Thus, for $y_1 = t/3$,
$y_1''''+4y_1'''+3y_1$
$=0+0+3(t/3)$
$=t$
So $y_1(t) = t/3$ is a solution.
For $y_2 = e^{-t}+t/3$,
$y_2''''+4y_2'''+3y_2$
$=(e^{-t})+4(-e^{-t})+3(e^{-t}+t/3)$
$=t$.
So $y_2(t) = e^{-t}+t/3$ is a solution for the given differential equation.
