Answer
$y = x^3 + 3x + 1$ satisfies $y''' + xy'' - 2y' = 0$.
Work Step by Step
If $y = x^3 + 3x + 1$, then
\begin{align}
y' &= 3x^{3-1} + 3x^{1-1} + 0 \\
&= 3x^2 + 3 \\
y'' &= (2)3x^{2-1} + 0 \\
&= 6x \\
y'' &= 6x^{1-1} \\
&= 6 \\
\end{align}
Substituting each of these into the differential equation, we obtain
$(6) + x(6x) - 2(3x^2 +3) = 6 + 6x^2 - 6x^2 - 6 = 0$
