Question

设函数f(x)具有连续一阶导数,且满足f(x)=∫(上限是x下限是0)(x^2-t^2)f^,(t)dt+x^2求f(x

Answer 1

f(x)=∫(上限是x下限是0)(x^2-t^2)f'(t)dt+x^2 所以f(0)=0,

又函数f(x)具有连续一阶导数,对上式两边求导得;

f'(x)=)=∫(上限是x下限是0)2xf'(t)dt+2x=2xf(x)+2x=2x(f(x)+1)

dy/(y+1)=2xdx 解得f(x)=e^x^2-1.