对应的齐次方程为y'+2y=0解得y*=C(1)[e^(-2x)]然后用常数变易法求原方程的解,设原方程的解为 y=C(x)[e^(-2x)]则 y'=C'(x)[e^(-2x)]+C(x)(-2)[e^(-2x)]代入原方程,则C'(x)[e^(-2x)]+C(x)(-2)[e^(-2x)]+2C(x)[e^(-2x)]...