求解常微分方程!y'=xy-x^2+1,且y(0)=1.0

求解常微分方程!y'=xy-x^2+1,且y(0)=1.0
求解常微分方程!y'=xy-x^2+1,且y(0)=1.0

求解常微分方程!y'=xy-x^2+1,且y(0)=1.0
y'=xy-x^2+1,y'-xy=-x^2+1,解齐次的先y'-xy=0
y=C(x)e^(x^2/2)
带入C'(x)e^(x^2/2)+C(x)xe^(x^2/2)-C(x)xe^(x^2/2)=-x^2+1
C'(x)e^(x^2/2)=-x^2+1求出C(x)=∫e^(-x^2/2)(-x^2+1)dx
C(x)=xe^(-x^2/2)+C1
y=(xe^(-x^2/2)+C1)e^(x^2/2)且根据条件y(0)=1.0