# 无论什么样的二阶微分方程问题，先求解特征根总没错

## 二、解析

$$y^{\prime \prime}+2 m y^{\prime}+n^{2} y=0 \Rightarrow$$

$$\lambda^{2}+2 m \lambda+n^{2}=0 \Rightarrow$$

$$\lambda=\frac{-2 m \pm \sqrt{4 m^{2}-4 n^{2}}}{2}=\frac{-2 m \pm 2 \sqrt{m^{2}-n^{2}}}{2} \Rightarrow$$

$$\lambda_{1}=-m+\sqrt{m^{2}-n^{2}<0}$$

$$\lambda_{2}=-m-\sqrt{m^{2}-n^{2}}<0$$

$$m>n>0 \Rightarrow$$

$$y=C_{1} e^{\lambda_{1} x}+C_{2} e^{\lambda_{2} x} \Rightarrow$$

$$y^{\prime}=C_{1} \lambda_{1} e^{\lambda_{1} x}+C_{2} \lambda_{2} e^{\lambda_{2} x} \Rightarrow$$

$$\int_{0}^{+\infty}\left[y^{\prime \prime}(x)+2 m y^{\prime}(x)+n^{2} y(x)\right] d x=0$$

$$\left.y^{\prime}(x)\right|_{0} ^{+\infty}+\left.2 m \cdot y(x)\right|_{0} ^{+\infty}+n^{2} \int y(x) d x=0 \Rightarrow$$

$$\lim \limits_{x \rightarrow+\infty} y^{\prime}(x)=0, \quad \lim \limits_{x \rightarrow+\infty} y(x)=0 \Rightarrow$$

$$0-y^{\prime}(0)+2 m[0-y(0)]+n^{2} \int_{0}^{+\infty} y(x) d x=0 \Rightarrow$$

$$-b-2 m a+n^{2}y(x) d x=0 \Rightarrow$$

$$n^{2} \int_{0}^{+\infty} y(x) d x=b+2 m a \Rightarrow$$

$$\int_{0}^{+\infty} y(x) d x=\frac{b+2 m a}{n^{2}}$$