如果是y''-y=(sinx)^2原方程可化为y''-y=(1-cos2x)/2y''-y=1/2-(1/2)cos2x①y''-y=0特征方程：r^2-1=0r1=1,r2=-1y=C1*e^x+C2*e^(-x)②y''-y=1/2-(1/2)cos2x设特解为y=a+b*cos2x+c*sin2xy'=-2b*sin2x+2c*cos2xy''=-4b...