Answer

问题及解答

线性微分算子 $L=a_n(x)D^n+a_{n-1}(x)D^{n-1}+\cdots+a_1(x)D+a_0(x)$

Posted by haifeng on 2023-08-12 11:15:07 last update 2023-08-12 11:15:31 | Edit | Answers (0)

\[L=a_n(x)D^n+a_{n-1}(x)D^{n-1}+\cdots+a_1(x)D+a_0(x),\]

其中 $D^n=\dfrac{\mathrm{d}^n}{\mathrm{d}x^n}$, $n=1,2,\ldots$. 证明 $L$ 满足线性性.

\[L(ay_1+by_2)=aL(y_1)+bL(y_2).\]