无论什么样的二阶微分方程问题，先求解特征根总没错

一、题目

已知，$y=y(x)$ 是二阶常系数线性微分方程 $y^{\prime \prime }+2m{y}^{\prime }+n^{2}y=0$ 满足 $y(0)=a$ 与 $y^{\prime }(0)$ $=b$ 的特解，其中 $m>n>0$, 则 ${\int }_0^{+\mathrm{\infty }}y(x)\mathrm{d}x=?$

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二、解析

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

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

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

$${\lambda }_1=-m+\sqrt{m^2-n^2<0}$$

$${\lambda }_2=-m-\sqrt{m^2-n^2}<0$$

$$m>n>0\Rightarrow$$

$$y=C_1{e}^{{\lambda }_{1}x}+C_2{e}^{{\lambda }_{2}x}\Rightarrow$$

$$y^{\prime }=C_1{\lambda }_1{e}^{{\lambda }_{1}x}+C_2{\lambda }_2{e}^{{\lambda }_{2}x}\Rightarrow$$

$${\int }_0^{+\mathrm{\infty }}[y^{\prime \prime }(x)+2m{y}^{\prime }(x)+n^{2}y(x)]dx=0$$

$${y^{\prime }(x)|}_0^{+\mathrm{\infty }}+{2m\cdot y(x)|}_0^{+\mathrm{\infty }}+n^2\int y(x)dx=0\Rightarrow$$

$$\underset{x\to +\mathrm{\infty }}{lim}{y}^{\prime }(x)=0,\underset{x\to +\mathrm{\infty }}{lim}y(x)=0\Rightarrow$$

$$0-y^{\prime }(0)+2m[0-y(0)]+n^2{\int }_0^{+\mathrm{\infty }}y(x)dx=0\Rightarrow$$

$$-b-2ma+n^{2}y(x)dx=0\Rightarrow$$

$$n^2{\int }_0^{+\mathrm{\infty }}y(x)dx=b+2ma\Rightarrow$$

$${\int }_0^{+\mathrm{\infty }}y(x)dx=\frac{b+2ma}{n^2}$$

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