二阶非齐次常微分方程 $$u''(z) + p(z)u'(z) + q(z)u(z) = f(z)$$
假设以上方程对应的齐次方程的线性独立的解为 \(u_1(z)\)，\(u_2(z)\)
\(\begin{cases} u_1''(z) + p(z)u_1'(z) + q(z)u_1(z) = 0 \\ u_2''(z) + p(z)u_2'(z) + q(z)u_2(z) = 0 \end{cases}\)（齐次下方程解）
设该非齐次方程的特解为
\(u_0(z) = y_1(z)u_1(z) + y_2(z)u_2(z)\)
\(u_0'(z) = y_1'(z)u_1(z) + y_1(z)u_1'(z) + y_2'(z)u_2(z) + y_2(z)u_2'(z)\)
\(u_0'(z) = y_1(z)u_1'(z) + y_2(z)u_2'(z)\) \(\quad\) [因为假设 \(y_1'(z)u_1(z) + y_2'(z)u_2(z) = 0\)]
回代
\[y_1(z) [u_1''(z) + p(z)u_1'(z) + q(z)u_1(z)] + y_2(z) [u_2''(z) + p(z)u_2'(z) + q(z)u_2(z)]+ y_1'(z)u_1'(z) + y_2'(z)u_2'(z) = f(z) \]\(\begin{cases} y_1'(z)u_1(z) + y_2'(z)u_2(z) = 0 \\ y_1'(z)u_1'(z) + y_2'(z)u_2'(z) = f(z) \end{cases}\)
可得：
\(L(z) = \begin{vmatrix} u_1(z) & u_2(z) \\ u_1'(z) & u_2'(z) \end{vmatrix} \neq 0\) 即必有解
则
\[u(z) = a_1u_1(z) + a_2u_2(z) + u_0(z) = [a_1 + y_1(z)]u_1(z) + [a_2 + y_2(z)]u_2(z) \] 标签：begin,4.6,end,二阶,齐次,微分方程,cases From： https://www.cnblogs.com/RES-HON/p/18546464