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二阶常系数齐次线性微分方程

如果二阶齐次线性微分方程

$$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + P\left( x \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} + Q\left( x \right)y = 0$$

的系数$P(x)、Q(x)$均为常数，即

$$\frac{{{{\text{d}}^2}}}{{{\text{d}}{x^2}}}y + p\frac{{\text{d}}}{{{\text{d}}x}}y + qy = 0$$

则上式称为二阶常系数齐次线性微分方程.

我们知道，对于函数$y=f(x)$，

$$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {\frac{{\text{d}}}{{{\text{d}}x}}(y)} \right]$$

则二阶常系数齐次线性微分方程可以写为

$${\left( {\frac{{\text{d}}}{{{\text{d}}x}}} \right)^2}y + p\frac{{\text{d}}}{{{\text{d}}x}}y + qy = 0$$ $$\left[ {{{\left( {\frac{{\text{d}}}{{{\text{d}}x}}} \right)}^2} + p\frac{{\text{d}}}{{{\text{d}}x}} + q} \right]y = 0$$

令$\frac{{\text{d}}}{{{\text{d}}x}} = m$，则

$${\left( {\frac{{\text{d}}}{{{\text{d}}x}}} \right)^2} + p\frac{{\text{d}}}{{{\text{d}}x}} + q = 0$$

可表示为一个一元二次方程

$$m^2+pm+q=0$$

我们把上式称为二阶齐次线性微分方程

$$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + P\left( x \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} + Q\left( x \right)y = 0$$

的特征方程.

对于特征方程

$$m^2+pm+q=0$$

当$p^2-4q>0$时，方程有两个不相等的实数根

$${m_1} = \frac{{ - p + \sqrt {{p^2} - 4q} }}{2},$$ $${m_2} = \frac{{ - p - \sqrt {{p^2} - 4q} }}{2}$$

即

$$(m-m_1 )(m-m_2 )=0$$

则二阶常系数齐次线性微分方程写为

$$\eqalign{ & \left( {\frac{{\text{d}}}{{{\text{d}}x}} - {m_1}} \right)\left( {\frac{{\text{d}}}{{{\text{d}}x}} - {m_2}} \right)y = 0 \cr & \left( {\frac{{\text{d}}}{{{\text{d}}x}} - {m_2}} \right)y = 0 \cr}$$

或写为

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} - {m_2}y = 0$$

分离变量，得

$$\frac{{{\text{d}}y}}{y} = {m_2}{\text{d}}x$$

两端积分，得

$${\text{ln}}\left| y \right| = {m_2}x + K \Rightarrow y = \pm {{\text{e}}^{{m_2}x + K}}$$ $$= \pm {{\text{e}}^K} \cdot {{\text{e}}^{{m_2}x}}$$

$±e^K$是不等于0等任意常数，显然，$y=$0也是二阶齐次线性微分方程的一个解，因此

$$y = {C_1}{{\text{e}}^{{m_2}x}}$$

同理可得

$$y = {C_2}{{\text{e}}^{{m_1}x}}$$

也是原微分方程的另一个解，从而原微分方程的通解为

$$y = {C_1}{{\text{e}}^{{m_2}x}} + {C_2}{{\text{e}}^{{m_1}x}}$$

当$p^2-4q=0$时，方程有两个相等的实数根

$${m_1} = {m_2} = - \frac{p}{2}$$

此时原微分方程只有一个解

$${y_1} = C{{\text{e}}^{ - \frac{p}{2}x}}$$

为了得到原微分方程的通解，还需要寻找另一个解$y_2$，并且这两个解需要线性无关，即$\frac{{{y_2}}}{{{y_1}}}$不是常数，设

$$\eqalign{ & \frac{{{y_2}}}{{{y_1}}} = u(x) \cr & {y_2} = u(x){y_1} \cr}$$

求导，得

$$\eqalign{ & {y_2}' = u'\left( x \right){y_1} + u(x)y{'_1} \cr & {y_2}'' = u''\left( x \right){y_1} + 2u'\left( x \right)y{'_1} + u(x)y'{'_1} \cr}$$

将${y_2},{y_2}',{y_2}''$代入原微分方程

$$\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + p\frac{{{\text{d}}y}}{{{\text{d}}x}} + qy = 0$$

得

$$u''\left( x \right){y_1} + 2u'\left( x \right)y{'_1} + u\left( x \right)y{'_1}'$$ $$+ p\left[ {u'\left( x \right){y_1} + u\left( x \right)y{'_1}} \right] + qu\left( x \right){y_1} = 0$$

重新整理，得

$$u''\left( x \right){y_1} + 2u'\left( x \right)y{'_1} + u\left( x \right)y{'_1}'$$ $$+ pu'\left( x \right){y_1} + pu\left( x \right)y{'_1} + qu\left( x \right){y_1} = 0$$ $$\left[ {y{'_1}' + py{'_1} + q{y_1}} \right]u\left( x \right) + u''\left( x \right){y_1}$$ $$+ \left[ {2y{'_1} + p{y_1}} \right]u'\left( x \right) = 0$$

因为$y_1$是原微分方程的解，所以

$${y_1}'' + p{y_1}' + q{y_1} = 0,2{y_1}' + p{y_1} = 0$$

则

$$u''\left( x \right){y_1} = 0$$

因为$y_1≠0$，所以$u'' (x)=0$.

因为只需要$u(x)$不是常数，不妨取最简单的$u(x)=x$即可，则

$${y_2} = u\left( x \right){y_1} = x{{\text{e}}^{ - \frac{p}{2}x}}$$

从而微分方程的通解为

$$y = {C_1}{{\text{e}}^{ - \frac{p}{2}x}} + {C_2}x{{\text{e}}^{ - \frac{p}{2}x}}$$

即

$$y = ({C_1} + {C_2}x){{\text{e}}^{ - \frac{p}{2}x}}$$

当$\Delta = {p^2} - 4q < 0$时，特征方程有一对共轭复根：

$${m_1} = - \frac{p}{2} + \frac{{\sqrt {{p^2} - 4q} }}{2}{\text{i}},$$ $${m_2} = - \frac{p}{2} - \frac{{\sqrt {{p^2} - 4q} }}{2}{\text{i}}{\text{.}}$$

设

$$\alpha = - \frac{p}{2},\beta = \frac{{\sqrt {{p^2} - 4q} }}{2} \ne 0.$$

则$m_1=α+βi$，$m_2=α-βi$.则微分方程的两个解为

$${y_1} = {C_1}{{\text{e}}^{(\alpha + \beta {\text{i}})x}},{y_2} = {C_2}{{\text{e}}^{(\alpha - \beta {\text{i}})x}}$$

根据欧拉公式

$${{\text{e}}^{\theta {\text{i}}}} = \cos \theta + {\text{isin}}\theta$$

所以原微分方程的解为

$$y = {{\text{e}}^{\alpha x}}({C_1}{{\text{e}}^{{\text{i}}\beta x}} + {C_2}{{\text{e}}^{ - {\text{i}}\beta x}})$$

将解写为

$${y_3} = {{\text{e}}^{\alpha x}}\left[ {\left( {{C_1} + {C_2}} \right)\cos \beta x + {\text{i}}\left( {{C_1} + {C_2}} \right)\sin \beta x} \right]$$

显然

$${y_4} = {{\text{e}}^{\alpha x}}\left[ {\left( {{C_1} + {C_2}} \right)\cos \beta x - {\text{i}}\left( {{C_1} + {C_2}} \right)\sin \beta x} \right]$$

也是微分方程的解，由于微分方程的系数是实数，我们将复数解化为实数，上面两式相加除以2，得

$$\overline {{y_3}} = \frac{1}{2}\left( {{y_3} + {y_4}} \right) = {{\text{e}}^{\alpha x}}\cos \beta x$$ $$\overline {{y_4}} = \frac{1}{2}\left( {{y_3} - {y_4}} \right) = {{\text{e}}^{\alpha x}}\sin \beta x$$

且

$$\frac{{\overline {{y_3}} }}{{\overline {{y_4}} }} = \frac{{{{\text{e}}^{\alpha x}}\cos \beta x}}{{{{\text{e}}^{\alpha x}}\sin \beta x}} = \cot \beta x$$

不是常数，所以原微分方程的解为

$$y = {{\text{e}}^{\alpha x}}({C_1}\cos \beta x + {C_2}\sin \beta x)$$