Example 2 Try to find the number of non-negative integer solutions $\varphi(n)$ of the indeterminate equation $5 x+2 y+z=10 n$ (where $n$ is a natural number).
Let $\left(x_{0}, y_{0}, z_{0}\right)$ be a non-negative integer solution of
$$
5 x+2 y+z=10 n
$$

then $\left(x_{0}+2, y_{0}, z_{0}\right)$ must be a solution of
$$
5 x+2 y+z=10(n+1)
$$

It can be seen that: (i) when $x \geqslant 2$, the number of non-negative integer solutions of (4) is $\varphi(n)$; (ii) when $x=0$, from (4) we get $0 \leqslant y=\frac{1}{2}[10(n+1)-z] \leqslant 5 n+5$, so the number of non-negative integer solutions of (4) is $5 n+6$; (iii) when $x=1$, $0 \leqslant y \leqslant 5 n+2$, so the number of non-negative integer solutions of (4) is $5 n+3$. Therefore, $\varphi(n+1)=\varphi(n)+10 n+9$.
It is also verified directly that $\varphi(1)=10$. Thus,
$$
\begin{array}{l}
\varphi(1)=10, \\
\varphi(2)-\varphi(1)=10 \cdot 1+9, \\
\varphi(3)-\varphi(2)=10 \cdot 2+9, \\
\cdots \cdots \cdots \\
\varphi(n)-\varphi(n-1)=10(n-1)+9 .
\end{array}
$$

Adding these equations successively, we get
$$
\varphi(n)=10[1+2+\cdots+(n-1)]+10+9(n-1)=5 n^{2}+4 n+1 .
$$