# 这道题没说函数可导，所以就不能求导了嘛？

## 二、解析

$$f(x)=\cos 2 x-4 \int_{0}^{x}(x-t) f(t) \mathrm{~ d} t \Rightarrow$$

$$x=0 \Rightarrow f(0)=1$$

$$f(x)=\cos 2 x-4 x \int_{0}^{x} f(t) \mathrm{~ d} t+4 \int_{0}^{x} t f(t) \mathrm{~ d} t \Rightarrow$$

$$f^{\prime}(x)=-2 \sin 2 x-4 \int_{0}^{x} f(t) \mathrm{~ d} t-4 x f(x) + 4 x f(x) \Rightarrow$$

$$f^{\prime}(x)=-2 \sin 2 x-4 \int_{0}^{x} f(t) \mathrm{~ d} t \Rightarrow$$

$$x=0 \Rightarrow f^{\prime}(0)=0 \Rightarrow$$

$$f^{\prime \prime}(x)=-4 \cos 2 x-4 f(x) \Rightarrow$$

$$y^{\prime \prime}+4 y=-4 \cos 2 x \tag{1}$$

$$\lambda^{2}+4=0 \Rightarrow \lambda=0 \pm 2 i$$

$$y^{*}=x^{k} e^{2 x}\left(Q_{n}(x) \cos \beta x+W_{n}(x) \sin \beta x\right) \Rightarrow$$

$$k=1, \ \alpha=0, \ \beta=2, \ Q_{n}(x)=A, \ W_{n}(x)=B \Rightarrow$$

$$y ^{*} = x(A \cos 2 x+B \sin 2 x)$$

$$y^{*}=x \bar{y} \Rightarrow$$

$$y^{* \prime}=\bar{y} + x \bar{y}^{\prime} \Rightarrow$$

$$y^{* \prime \prime}=\bar{y}^{\prime}+\bar{y}^{\prime}+x \bar{y}^{\prime \prime}=2 \bar{y}^{\prime \prime}+x \bar{y}^{\prime \prime}$$

$$(y^{*})^{\prime \prime}+4\left(y^{*}\right)=-4 \cos 2 x \Rightarrow$$

$$2 \bar{y}^{\prime}+x \bar{y}^{\prime \prime}+4 x \bar{y}^{\prime} = -4 \cos 2 x \Rightarrow$$

$$x\left(\bar{y}^{\prime \prime}+4 \bar{y}\right)+2 \bar{y}^{\prime} = -4 \cos 2 x.$$

$$\bar{y}^{\prime}=-2 A \sin 2 x+2 B \cos 2 x.$$

$$\bar{y}^{\prime \prime}=-4 A \cos 2 x – 4 B \sin 2 x .$$

$$4 \bar{y}^{\prime}=4 A \cos 2 x+4 B \sin 2 x .$$

$$x\left(\bar{y}^{\prime \prime}+4 \bar{y}\right)+2 \bar{y}^{\prime}=2 \bar{y}^{\prime} \Rightarrow$$

$$2 \bar{y}^{\prime}=-4 \cos 2 x \Rightarrow$$

$$2(-2 A \sin 2 x+2 B \cos 2 x)=-4 \cos 2 x \Rightarrow$$

$$A=0, \quad B=-1$$

$$Y = -x \sin 2 x+C_{1} \cos 2 x + C_{2} \sin 2 x$$

$$f(0) = y(0)=1 \Rightarrow C_{1}=1$$

$$f^{\prime}(0) = y^{\prime}(0)=0 \Rightarrow 2 C_{2} \cos 2 x=0 \Rightarrow C_{2}=0$$

$$f(x) = Y = \cos 2 x – x \sin 2 x$$