答：y''+3y'+2y=0特征方程a²+3a+2=0(a+1)(a+2)=0解得：a=-1或者a=-2通解y=C1e^(-x)+C2e^(-2x)y'(x)=-C1e^(-x)-2C2e^(-2x)所以：y(0)=C1+C2=-1y'(0)=-C1-2C2=0解得：C2=-1/3，C1=-2/3所以：特解为y=(-1/3)*[e^(-x)+2e^(-2x)]
