f(x)+2∫(0到x) f(t) dt=x²f'(x)+2f(x)=2x即y'+2y=2x...①y'+2y=0的通解是y=c₁e^(-2x)特解:y=ax+b,y'=a代入①得a+2(ax+b)=2x2ax+(a+2b)=2xax=2,a=1a+2b=0,b=-1/2∴f(x)=c₁e^(-2x)+x-1/2
