中值定理构造

• $f^{\prime }(\xi )+f(\xi )g(\xi )=0$，令 $F(x)=f(x)\cdot e^{\int g(x)\mathrm{d}x}$，只需证 $\mathrm{\exists }{F}^{\prime }(\xi )=0$
  • $\begin{array}{|cc|}\hline g(x)& \int g(x)\mathrm{d}x\\ x& {\displaystyle \frac{1}{2}}{x}^2\\ {\displaystyle \frac{1}{x}}& \ln x\\ a& ax\\ \hline\end{array}$
  • $F^{\prime }(x)=f^{\prime }(x)\cdot e^{\cdots }+f(x)\cdot e^{\cdots }\cdot g(x)=e^{\cdots }(f^{\prime }(x)+f(x)g(x))$
  • 举例
    • $\xi f^{\prime }(\xi )+2f(\xi )=0,F(x)=x^{2}f(x)$
    • $f^{\prime }(x)=f(x),F(x)=e^{-x}f(x)$
    • $f^{\prime }(x)+(1)f(x)=\lambda ,F(x)=e^x(f(x)-\lambda )$
    • $f^{\prime }(x)+f(x)=e^x⟹(f^{\prime }(x)-{\displaystyle \frac{e^x}{2}})+(f(x)-{\displaystyle \frac{e^x}{2}}),F(x)=e^x(f(x)-{\displaystyle \frac{e^x}{2}})$
    • $x{f}^{\prime }(x)+\alpha f(x)=0,F(x)=x^{\alpha }f(x)$
    • $f^{\prime }(x)+xf(x)=0,F(x)=f(x)e^{\frac{1}{2}{x}^2}$
• $f^{″}(x)+f^{\prime }(x)=0,F(x)=f(x)+f^{\prime }(x)\text{ or }F(x)=f^{\prime }(x)e^x$
• $f^{\prime }(x)+f^2(x)=0⟹{\displaystyle \frac{f^{\prime }(x)}{f^2(x)}}=-1,F(x)=-{\displaystyle \frac{1}{f(x)}}+x$
• 二阶导
  • $f^{″}(x)+f(x)=0,F(x)=f^2(x)+(f^{\prime }(x){)}^2,F^{\prime }(x)=2f^{\prime }(x)(f(x)+f^{″}(x))$
  • 补项消去
    • $f^{″}(x)-f(x)=0,F(x)=(f^{\prime }(x)+f(x))e^{-x}\text{ or }F(x)=(f^{\prime }(x)-f(x))e^x$
      • $F^{\prime }(x)=(f^{″}+f^{\prime }-f^{\prime }-f)e^{-x}\text{ or }(f^{\prime }-f+f^{″}-f^{\prime })e^x$
    • $f(x)g^{″}(x)-f^{″}(x)g(x)=0,F(x)=f(x)g^{\prime }(x)-f^{\prime }(x)g(x)$
• 分配项
  • $f^{″}(x)-3f^{\prime }(x)+2f(x)=0⟹f^{″}(x)-f^{\prime }(x)-2(f^{\prime }(x)-f(x))=0$
    • 令 $g(x)=f^{\prime }(x)-f(x)$，即求 $g^{\prime }(x)-2g(x)=0$

几个重要构造函数

• $x^2\sin {\displaystyle \frac{1}{x}}$
  • 常用于反例
    • [[MID2019A Collection.canvas|2019期中A T4]]

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y=x^{2}*\sin{\frac{1}{x}}