# 对于二阶常系数非齐次微分方程，当需要直接求函数解时可以用公式法，当需要用到中间的某些量时可以用常数变易法

## 一、题目

$$\lim \limits_{x \rightarrow+\infty} y^{\prime}(x)=?$$

$$\lim \limits_{x \rightarrow+\infty} y^{\prime \prime}(x)=?$$

## 二、解析

$$y(x) = [\int f(x) \cdot e^{\int a \mathrm{~ d} x} \mathrm{~ d} x + C] \cdot e^{- \int a \mathrm{~ d} x}$$

$$y^{\prime \prime}+a y^{\prime}=f(x) \Rightarrow$$

$$e^{\int a \mathrm{~ d} x}\left[y^{\prime \prime}+a y^{\prime}\right]=e^{\int a \mathrm{~ d} x} \cdot f(x) \Rightarrow$$

$$y^{\prime \prime} \cdot e^{a x}+a y^{\prime} \cdot e^{a x}=e^{a x} \cdot f(x) \Rightarrow$$

$${\left[y^{\prime} \cdot e^{a x}\right]_{x}^{\prime}=e^{a x} \cdot f(x) \Rightarrow}$$

$$\int\left[y^{\prime} \cdot e^{a x}\right]_{x}^{\prime} \mathrm{~ d} x=\int e^{a x} \cdot f(x) \mathrm{~ d} x \Rightarrow$$

$$y^{\prime} \cdot e^{a x}+C_{0}=\int e^{a x} \cdot f(x) \mathrm{~ d} x \Rightarrow$$

$$y^{\prime}=C_{1} e^{-a x}+\frac{\int e^{a x} \cdot f(x) \mathrm{~ d} x}{e^{a x}} \Rightarrow$$

$$\lim \limits_{x \rightarrow+\infty} y^{\prime}=\lim \limits_{x \rightarrow+\infty} C_{1} e^{-a x}+\lim \limits_{x \rightarrow+\infty} \frac{\int e^{a x} \cdot f(x) \mathrm{~ d} x }{e^{a x}}$$

$$\lim \limits_{x \rightarrow+\infty} y^{\prime}=0+\lim \limits_{x \rightarrow+\infty} \frac{e^{a x} \cdot f(x)}{a e^{a x}} \Rightarrow$$

$$\lim \limits_{x \rightarrow+\infty} y^{\prime}=\frac{b}{a}$$

$$\lim \limits_{x \rightarrow+\infty} y^{\prime}=\frac{b}{a} \Rightarrow \lim \limits_{x \rightarrow+\infty}\left[y^{\prime \prime} \cdot e^{a x}+a y^{\prime} \cdot e^{a x}=e^{a x} f(x)\right.$$

$$\lim \limits_{x \rightarrow+\infty}\left[y^{\prime \prime} \cdot e^{a x}+a \cdot \frac{b}{a} e^{a x}=e^{a x} \cdot b\right] \Rightarrow$$

$$\lim \limits_{x \rightarrow+\infty}\left[y^{\prime \prime}+b=b\right] \Rightarrow \lim \limits_{x \rightarrow+\infty} y^{\prime \prime}=0$$