二阶常系数非齐次微分方程例题

$例：求y^{\prime \prime }+y=cos2x+2sinx的通解$
$解：\because {\beta }_1̸={\beta }_2$
$\therefore 将方程式y^{\prime \prime }+y=cos2x+2sinx$
   $拆成y^{\prime \prime }+y=cos2x与y^{\prime \prime }+y=2sinx两个二阶常系数非齐次微分方程。$

$\Rightarrow 其特征方程r^2=1=0的根为\pm i$
$易知：y^{\prime \prime }+y=0的通解为：Y=C_{1}cosx+C_{2}sinx$
$1)$ $y^{\prime \prime }+y=cos2x$
$\Rightarrow \alpha =0;\beta =2;s=max[m,n]=0$
$\because \alpha \pm \beta i=\pm 2i不是特征方程的根$
$\therefore 令:y\ast =a_{0}cos2\beta +b_{0}sin2\beta$
            $y{\ast }^{\prime }=-2a_{0}sin2\beta +2b_{0}cos2\beta$
            $y{\ast }^{\prime \prime }=-4a_{0}cos2\beta -4b_{0}sin2\beta$

$\Rightarrow 将y\ast ,y{\ast }^{\prime },y{\ast }^{\prime \prime }代入原方程求解得：a_0=\frac{1}{3};b_0=0$
$\therefore y\ast =\frac{1}{3}cos2x$
$2)y^{\prime \prime }+y=2sinx$

$\Rightarrow \alpha =0;\beta =1;s=max[m,n]=0$
$\because \alpha \pm \beta i=\pm i是特征方程的一对单共轭复根$
$\therefore 令:y\ast =x(a_{1}cos\beta +b_{1}sin\beta )$
$\Rightarrow 将y\ast ,y{\ast }^{\prime },y{\ast }^{\prime \prime }代入原方程求解得：a_0=-1;b_0=0$

$\therefore y\ast =-xcosx$
$综上：y^{\prime \prime }+y=cos2x+2sinx的通解为$
$$y=C_{1}cosx+C_{2}sinx+\frac{1}{3}cos2x+-xcosx$$