Question

高数 微分方程y的二阶导+y的一介导=x²+1当y(0)=1 y一介导（0）=-2 求特解

Answer 1

y"+y' = x²+1,y(0) = 1,y'(0) = -2.

方程不含y,可先将z = y'作为未知函数求解.

由z'+z = x²+1,有(z·e^x)' = (x²+1)·e^x,积分得z·e^x = (x²-2x+3)·e^x+C.

由z(0) = y'(0) = -2得C = -5.

y' = z = x²-2x+3-5e^(-x),积分得y = x³/3-x²+3x+5e^(-x)+D.

由y(0) = 1得D = -4,y = x³/3-x²+3x-4+5e^(-x).