Example 8 Try to find a 3rd degree polynomial $y(x)$, such that it satisfies $y(0)=y(1)=0, y(2)=1, y(3)=5$.
(Guangdong Province Olympiad 1987 Winter Training)
Solution: Let $y(x)=a x^{3}+b x^{2}+c x+d$. From the given information, we have
$$
\left\{\begin{array}{l}
d=0, \\
a+b+c+d=0, \\
8 a+4 b+2 c+d=1, \\
27 a+9 b+3 c+d=5 .
\end{array}\right.
$$

Solving, we get $a=\frac{1}{3}, b=-\frac{1}{2}, c=\frac{1}{6}, d=0$.
$$
\therefore y(x)=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}+\frac{1}{6} x \text {. }
$$