第十三讲 二阶非齐次常系数线性ODE的特解

一,二阶非齐次常系数线性ODE的标准形式:

  • {y}''+A{y}'+By=f(x)

二,通解:

  •  y=y_{p}+{\color{Magenta} c_{1}y_{1}+c_{2}y_{2}}

三,将方程化成特殊形式:

  • 设方程右边的输入项为“纯振荡”:f(x)=e^{{\color{Red} \alpha }x}=e^{(a+i\omega ) x}=e^{ax}(cos(\omega x)+isin(\omega x))a< 0{\color{Red} \alpha}表示复数
  • 方程左边换成线性算子式:{y}''+A{y}'+By=(D^{2}+AD+B)y=p(D)y
  • {\color{Red} (D^{2}+AD+B)y=e^{\alpha x}}

四,代换法则:

  • {\color{Red} p(D)e^{\alpha x}=p(\alpha )e^{\alpha x}}
  • 证明:p(D)e^{\alpha x}=(D^{2}+AD+B)e^{\alpha x}=D^{2}e^{\alpha x}+ADe^{\alpha x}+Be^{\alpha x}
  • \becauseD表示对函数求导:D^{2}e^{\alpha x}=\alpha ^{2}e^{\alpha x}De^{\alpha x}=\alpha e^{\alpha x}
  • \thereforeD^{2}e^{\alpha x}+ADe^{\alpha x}+Be^{\alpha x}=\alpha ^{2}e^{\alpha x}+A\alpha e^{\alpha x}+Be^{\alpha x}=p(\alpha )e^{\alpha x}

五,指数输入定理(当p(\alpha )\neq 0时):

  • (D^{2}+AD+B)y=e^{\alpha x}的特解为{\color{Red} y_{p}=\frac{e^{\alpha x}}{p(\alpha )}}p(\alpha )是常数
  • 证明:将特解代入原方程,利用代换法则p(D)\frac{e^{\alpha x}}{p(\alpha )}=\frac{p(D)e^{\alpha x}}{p(\alpha )}=\frac{p(\alpha )e^{\alpha x}}{p(\alpha )}=e^{\alpha x}

六,求特解例题1:

  • {y}''-{y}'+2y=10e^{-x}sinx
  • sinx复数化:sinxe^{ix}的虚部,将e^{ix}代替sinxe^{-x}sinx=e^{-x+ix}=e^{(-1+i)x}
  • 方程左边换成线性算子式:(D^{2}-D+2)yp(D)=D^{2}-D+2
  • 原方程化简为:(D^{2}-D+2)\widetilde{y}=10e^{(-1+i)x}\widetilde{y}表示复数解,\alpha =-1+i
  • 利用指数输入定理\widetilde{y}_{p}=\frac{10e^{(-1+i)x}}{p(-1+i)}=\frac{10e^{(-1+i)x}}{(-1+i)^{2}-(-1+i)+2}=\frac{10e^{(-1+i)x}}{3(1-i)}
  • 分子分母同乘以1+i得:\frac{5}{3}(1+i)e^{(-1+i)x}=\frac{5}{3}(1+i)e^{-x}(cosx+isinx)
  • 取出虚部:y_{p}=Im(\widetilde{y}_{p})=\frac{5}{3}e^{-x}(cosx+sinx)=\frac{5}{3}e^{-x}\sqrt{2}cos(x-\frac{\pi }{4})\phi =\frac{\pi }{4}

七,指数位移法则:

  • {\color{Red} p(D)[e^{\alpha x}u(x)]=e^{\alpha x}p(D+\alpha )u(x)}
  • 证明:假设p(D)=D
  • D[e^{\alpha x}u(x)]={(e^{\alpha x}u(x))}'=\alpha e^{\alpha x}u(x)+e^{\alpha x}{u}'(x)=e^{\alpha x}(\alpha u(x)+{u}'(x))
  • e^{\alpha x}(\alpha u(x)+{u}'(x))=e^{\alpha x}(\alpha u(x)+Du(x))=e^{\alpha x}(D+\alpha )u(x)
  • 假设p(D)=D^{2}
  • D^{2}[e^{\alpha x}u(x)]=D[D[e^{\alpha x}u(x)]]=D[e^{\alpha x}(D+\alpha )u(x)]=e^{\alpha x}(D+\alpha )^{2}u(x)
  • 假设p(D)=D^{n}
  • D^{n}[e^{\alpha x}u(x)]=e^{\alpha x}(D+\alpha )^{n}u(x)

八,指数输入定理(当p(\alpha )=0时):

  • 如果\alpha是特征方程的单根\alpha _{1}\neq \alpha _{2}{p}'(\alpha )\neq 0p(D)y=e^{\alpha x}的特解为{\color{Red} y_{p}=\frac{xe^{\alpha x}}{{p}'(\alpha )}}
  • 证明{p}'(\alpha )\neq 0
  1. p(D)=(D-\alpha )(D-\beta )\alpha \neq \beta
  2. {p}'(D)=(D-\beta )+(D-\alpha )
  3. {p}'(\alpha )=\alpha -\beta\neq 0
  • 证明p(D)y=e^{\alpha x}的特解为y_{p}=\frac{xe^{\alpha x}}{{p}'(\alpha )}
  1. 将特解代入原方程p(D)\frac{xe^{\alpha x}}{{p}'(\alpha )}=e^{\alpha x}
  2. 利用指数位移法则p(D)\frac{xe^{\alpha x}}{{p}'(\alpha )}=\frac{p(D)[e^{\alpha x}x]}{{p}'(\alpha )}=\frac{e^{\alpha x}p(D+\alpha )x}{{p}'(\alpha )}
  3. p(D)=(D-\alpha )(D-\beta )\alpha \neq \beta
  4. p(D+\alpha )=D(D+\alpha -\beta )
  5. \frac{e^{\alpha x}p(D+\alpha )x}{{p}'(\alpha )}=\frac{e^{\alpha x}(D+\alpha-\beta )Dx}{{p}'(\alpha )}
  6. \because Dx=1(D+\alpha -\beta )Dx=\alpha -\beta{p}'(\alpha )=\alpha -\beta
  7. \therefore \frac{e^{\alpha x}(D+\alpha-\beta )Dx}{{p}'(\alpha )}=e^{\alpha x}
  • 如果\alpha是特征方程的二重根\alpha _{1}= \alpha _{2}{p}''(\alpha )\neq 0p(D)y=e^{\alpha x}的特解为{\color{Red} y_{p}=\frac{x^{2}e^{\alpha x}}{{p}''(\alpha )}}
  • 如果\alpha是特征方程的n重根\alpha _{1}= \alpha _{2}= \alpha _{n}p^{(n)}(\alpha )\neq 0p(D)y=e^{\alpha x}的特解为{\color{Red} y_{p}=\frac{x^{n}e^{\alpha x}}{p^{(n)}(\alpha )}}

九,求特解例题2:

  • {y}''-3{y}'+2y=e^{x}
  • 化简:(D^{2}-3D+2)y=e^{x}
  • p(D)=D^{2}-3D+2=(D-2)(D-1)
  • p(D)=0时:\alpha =1\alpha是特征方程的单根
  • {p}'(D)=D^{2}-3D+2=2D-3
  • {p}'(\alpha )={p}'(1)=2-3=-1
  • 直接指数输入定理y_{p}=\frac{xe^{x}}{{p}'(1)}=-xe^{x}

 

你可能感兴趣的:(微分方程)