考研数学笔记 – 微分算子法(微分方程求特解)
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  • 微分算子法主要用于求微分方程的特解

【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】

  • 分母均不为0
  • 算子式和方程式没有乘法交换律,算子式和算子式有乘法交换律
  • 当(1)式的分母为0时,可用(3)式,此时f(x)=1
  • (4)式中 F(D) = F1(D) * F2(D)

【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】

  • λ可以是0
  • 考研中出现的微分方程,一般就是这三种题型
  • 如果有了新题型。。。立即推!放弃考研!

例题

型一、型二

考研数学笔记 - 微分算子法(微分方程求特解)

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