考研数学笔记 —— 微分算子法(微分方程求特解)
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考研数学笔记 —— 微分算子法(微分方程求特解)

in 高数 with 0 comment, Views is 251

【公式】

$$ 设 \; D = \dfrac{d}{dx}, \; 则 \; y^\prime = D y, \; y^{\prime \prime} = D^2 y $$

$$ \left( 1 \right) \begin{cases} F_{\left( D \right)} e^{kx} = F_{\left( k \right)} e^{kx} \\ \\ \dfrac{1}{F_{\left( D \right)}} e^{kx} = \dfrac{1}{F_{\left( k \right)}} e^{kx} \end{cases} $$

$$ \left( 2 \right) \begin{cases} F_{\left( D^2 \right)} \sin ax = F_{\left( -a^2 \right)} \sin ax \\ \\ \dfrac{1}{F_{\left( D^2 \right)}} \sin ax = \dfrac{1}{F_{\left( -a^2 \right)}} \sin ax \end{cases} $$

$$ \left( 3 \right) \begin{cases} F_{\left( D \right)} e^{kx} f_{\left( x \right)} = e^{kx} F_{\left( D + k \right)} f_{\left( x \right)} \\ \\ \dfrac{1}{F_{\left( D \right)}} e^{kx} f_{\left( x \right)} = e^{kx} \dfrac{1}{F_{\left( D + k \right)}} f_{\left( x \right)} \end{cases} $$

$$ \left( 4 \right) \begin{cases} F_{\left( D \right)} f_{\left( x \right)} = F_{1 \left( D \right)} F_{2 \left( D \right)} f_{\left( x \right)} = F_{2 \left( D \right)} F_{1 \left( x \right)} f_{\left( x \right)} \\ \\ \dfrac{1}{F_{\left( D \right)}} f_{\left( x \right)} = \dfrac{1}{F_{1 \left( D \right)} F_{2 \left( D \right)}} f_{\left( x \right)} = \dfrac{1}{F_{2 \left( D \right)} F_{1 \left( x \right)}} f_{\left( x \right)} \end{cases} $$

【注】


【题型】

型一

$$ F_{\left( D \right)} y = f_{\left( x \right)} \; => \; y^* = \dfrac{1}{F_{\left( D \right)}} f_{\left( x \right)} $$

型二

$$ F_{\left( D \right)} y = e^{\lambda x} f_{\left( x \right)} \; => \; y^* = \dfrac{1}{F_{\left( D \right)}} e^{\lambda x} f_{\left( x \right)} = \begin{cases} \dfrac{1}{F_{\left( \lambda \right)}} e^{\lambda x} f_{\left( x \right)} & , \; F_{\left( \lambda \right)} \neq 0 \\ \\ e^{\lambda x} \dfrac{1}{F_{\left( D + \lambda \right)}} f_{\left( x \right)} & , \; F_{\left( \lambda \right)} = 0 \end{cases} $$

型三

$$ \begin{cases} F_{\left( D \right)} y = e^{\lambda x} f_{\left( x \right)} \sin ax \\ \\ F_{\left( D \right)} y = e^{\lambda x} f_{\left( x \right)} \cos ax \end{cases} $$

$$ 利用欧拉公式: \begin{cases} e^{iax} = \cos ax + i \sin ax \\ \\ i = \sqrt{-1} \; 或 \; i^2 = -1 \end{cases} $$

$$ 令 \; F_{\left( D \right)} y = e^{\left( \lambda + ia \right) x} f_{\left( x \right)} \; => \; y = e^{\left( \lambda + ia \right) x} \dfrac{1}{F_{\left( D + \lambda + ia \right)}} f_{\left( x \right)} $$

$$ 算出 \; y \; 后将欧拉公式回代,可得到带有(虚部)和(实部)的解 $$

$$ \begin{cases} F_{\left( D \right)} y = e^{\lambda x} f_{\left( x \right)} \sin ax => 取(虚部)为特解 \; y^* \\ \\ F_{\left( D \right)} y = e^{\lambda x} f_{\left( x \right)} \cos ax => 取(实部)为特解 \; y^* \end{cases} $$

【注】

例题

型一、型二

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